Django模型“未声明显式的app_label”

问题:Django模型“未声明显式的app_label”

我机智的结束了。经过十多个小时的故障排除(可能还有更多),我认为自己终于可以从事业务了,但是后来我得到了:

Model class django.contrib.contenttypes.models.ContentType doesn't declare an explicit app_label 

网络上对此信息很少,目前还没有解决方案解决了我的问题。任何建议将不胜感激。

我正在使用Python 3.4和Django 1.10。

在我的settings.py中:

INSTALLED_APPS = [
    'DeleteNote.apps.DeletenoteConfig',
    'LibrarySync.apps.LibrarysyncConfig',
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.sessions',
    'django.contrib.messages',
    'django.contrib.staticfiles',
]

我的apps.py文件如下所示:

from django.apps import AppConfig


class DeletenoteConfig(AppConfig):
    name = 'DeleteNote'

from django.apps import AppConfig


class LibrarysyncConfig(AppConfig):
    name = 'LibrarySync'

I’m at wit’s end. After a dozen hours of troubleshooting, probably more, I thought I was finally in business, but then I got:

Model class django.contrib.contenttypes.models.ContentType doesn't declare an explicit app_label 

There is SO LITTLE info on this on the web, and no solution out there has resolved my issue. Any advice would be tremendously appreciated.

I’m using Python 3.4 and Django 1.10.

From my settings.py:

INSTALLED_APPS = [
    'DeleteNote.apps.DeletenoteConfig',
    'LibrarySync.apps.LibrarysyncConfig',
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.sessions',
    'django.contrib.messages',
    'django.contrib.staticfiles',
]

And my apps.py files look like this:

from django.apps import AppConfig


class DeletenoteConfig(AppConfig):
    name = 'DeleteNote'

and

from django.apps import AppConfig


class LibrarysyncConfig(AppConfig):
    name = 'LibrarySync'

回答 0

您是否缺少将应用程序名称放入设置文件中的信息?这myAppNameConfig.manage.py createapp myAppName命令在apps.py生成的默认类。其中myAppName是您的应用程序的名称。

settings.py

INSTALLED_APPS = [
'myAppName.apps.myAppNameConfig',
'django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles',
]

这样,设置文件可以找出您要调用的应用程序。您可以通过在以下代码中添加以下代码来更改apps.py文件中的外观:

myAppName / apps.py

class myAppNameConfig(AppConfig):
    name = 'myAppName'
    verbose_name = 'A Much Better Name'

Are you missing putting in your application name into the settings file? The myAppNameConfig is the default class generated at apps.py by the .manage.py createapp myAppName command. Where myAppName is the name of your app.

settings.py

INSTALLED_APPS = [
'myAppName.apps.myAppNameConfig',
'django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles',
]

This way, the settings file finds out what you want to call your application. You can change how it looks later in the apps.py file by adding the following code in

myAppName/apps.py

class myAppNameConfig(AppConfig):
    name = 'myAppName'
    verbose_name = 'A Much Better Name'

回答 1

我收到相同的错误,但我不知道如何解决此问题。我花了很多时间才注意到,我有一个与django的manage.py相同的initcopy文件。

之前:

|-- myproject
  |-- __init__.py
  |-- manage.py
  |-- myproject
    |-- ...
  |-- app1
    |-- models.py
  |-- app2
    |-- models.py

后:

|-- myproject
  |-- manage.py
  |-- myproject
    |-- ...
  |-- app1
    |-- models.py
  |-- app2
    |-- models.py

您会收到此“未声明显式app_label”错误,这非常令人困惑。但是删除此初始化文件解决了我的问题。

I get the same error and I don´t know how to figure out this problem. It took me many hours to notice that I have a init.py at the same direcory as the manage.py from django.

Before:

|-- myproject
  |-- __init__.py
  |-- manage.py
  |-- myproject
    |-- ...
  |-- app1
    |-- models.py
  |-- app2
    |-- models.py

After:

|-- myproject
  |-- manage.py
  |-- myproject
    |-- ...
  |-- app1
    |-- models.py
  |-- app2
    |-- models.py

It is quite confused that you get this “doesn’t declare an explicit app_label” error. But deleting this init file solved my problem.


回答 2

使用PyCharm运行测试时,我遇到了完全相同的错误。我已经通过显式设置DJANGO_SETTINGS_MODULE环境变量来修复它。如果您使用的是PyCharm,只需点击编辑配置按钮,然后选择环境变量

将变量设置为your_project_name.settings,这应该可以解决问题。

似乎发生此错误,因为PyCharm使用自己的运行测试manage.py

I had exactly the same error when running tests with PyCharm. I’ve fixed it by explicitly setting DJANGO_SETTINGS_MODULE environment variable. If you’re using PyCharm, just hit Edit Configurations button and choose Environment Variables.

Set the variable to your_project_name.settings and that should fix the thing.

It seems like this error occurs, because PyCharm runs tests with its own manage.py.


回答 3

我在使用时得到了这个,./manage.py shell 然后我不小心从根项目级别目录中导入了

# don't do this
from project.someapp.someModule import something_using_a_model
# do this
from someapp.someModule import something_using_a_model

something_using_a_model()

I got this one when I used ./manage.py shell then I accidentally imported from the root project level directory

# don't do this
from project.someapp.someModule import something_using_a_model
# do this
from someapp.someModule import something_using_a_model

something_using_a_model()

回答 4

作为使用Python3的菜鸟,我发现它可能是导入错误而不是Django错误

错误:

from someModule import someClass

对:

from .someModule import someClass

这种情况发生在几天前,但我真的无法复制它…我认为只有Django新手才可能遇到这种情况。

尝试在admin.py中注册模型:

from django.contrib import admin
from user import User
admin.site.register(User)

尝试运行服务器,错误看起来像这样

some lines...
File "/path/to/admin.py" ,line 6
tell you there is an import error
some lines...
Model class django.contrib.contenttypes.models.ContentType doesn't declare an explicit app_label

更改user.user,问题已解决

as a noob using Python3 ,I find it might be an import error instead of a Django error

wrong:

from someModule import someClass

right:

from .someModule import someClass

this happens a few days ago but I really can’t reproduce it…I think only people new to Django may encounter this.here’s what I remember:

try to register a model in admin.py:

from django.contrib import admin
from user import User
admin.site.register(User)

try to run server, error looks like this

some lines...
File "/path/to/admin.py" ,line 6
tell you there is an import error
some lines...
Model class django.contrib.contenttypes.models.ContentType doesn't declare an explicit app_label

change user to .user ,problem solved


回答 5

我刚才有同样的问题。我通过在应用程序名称上添加命名空间来修复我的问题。希望有人觉得这有帮助。

apps.py

from django.apps import AppConfig    

class SalesClientConfig(AppConfig):
        name = 'portal.sales_client'
        verbose_name = 'Sales Client'

I had the same problem just now. I’ve fixed mine by adding a namespace on the app name. Hope someone find this helpful.

apps.py

from django.apps import AppConfig    

class SalesClientConfig(AppConfig):
        name = 'portal.sales_client'
        verbose_name = 'Sales Client'

回答 6

我在测试中导入模型时遇到此错误,即鉴于此Django项目结构:

|-- myproject
    |-- manage.py
    |-- myproject
    |-- myapp
        |-- models.py  # defines model: MyModel
        |-- tests
            |-- test_models.py

test_models.pyMyModel以这种方式导入的文件中:

from models import MyModel

如果以这种方式导入,则问题已解决:

from myapp.models import MyModel

希望这可以帮助!

PS:也许这有点晚了,但是在我的代码中没有其他人找到如何解决此问题的答案,我想分享我的解决方案。

I got this error on importing models in tests, i.e. given this Django project structure:

|-- myproject
    |-- manage.py
    |-- myproject
    |-- myapp
        |-- models.py  # defines model: MyModel
        |-- tests
            |-- test_models.py

in file test_models.py I imported MyModel in this way:

from models import MyModel

The problem was fixed if it is imported in this way:

from myapp.models import MyModel

Hope this helps!

PS: Maybe this is a bit late, but I not found in others answers how to solve this problem in my code and I want to share my solution.


回答 7

在继续遇到这个问题并继续回到这个问题之后,我想我会分享我的问题所在。

@Xeberdee正确的所有内容,请按照以下说明操作,看看是否可以解决问题,如果不是,这是我的问题:

在我的apps.py中,这就是我拥有的:

class AlgoExplainedConfig(AppConfig):
    name = 'algo_explained'
    verbose_name = "Explain_Algo"
    ....

我所做的就是在我的应用名称之前添加了项目名称,如下所示:

class AlgoExplainedConfig(AppConfig):
name = '**algorithms_explained**.algo_explained'
verbose_name = "Explain_Algo"

这样就解决了我的问题,之后我就可以运行makemigrations和migration命令!祝好运

After keep on running into this issue and keep on coming back to this question I thought I’d share what my problem was.

Everything that @Xeberdee is correct so follow that and see if that solves the issue, if not this was my issue:

In my apps.py this is what I had:

class AlgoExplainedConfig(AppConfig):
    name = 'algo_explained'
    verbose_name = "Explain_Algo"
    ....

And all I did was I added the project name in front of my app name like this:

class AlgoExplainedConfig(AppConfig):
name = '**algorithms_explained**.algo_explained'
verbose_name = "Explain_Algo"

and that solved my problem and I was able to run the makemigrations and migrate command after that! good luck


回答 8

我今天在尝试运行Django测试时遇到此错误,因为我from .models import *在其中一个文件中使用了速记语法。问题是我的文件结构如下:

    apps/
      myapp/
        models/
          __init__.py
          foo.py
          bar.py

models/__init__.py我使用速记语法导入模型时:

    from .foo import *
    from .bar import *

在我的应用程序中,我正在导入如下模型:

    from myapp.models import Foo, Bar

这导致了Django model doesn't declare an explicit app_label运行时./manage.py test

要解决此问题,我必须从完整路径中显式导入models/__init__.py

    from myapp.models.foo import *
    from myapp.models.bar import *

那解决了错误。

H / t https://medium.com/@michal.bock/fix-weird-exceptions-when-running-django-tests-f58def71b59a

I had this error today trying to run Django tests because I was using the shorthand from .models import * syntax in one of my files. The issue was that I had a file structure like so:

    apps/
      myapp/
        models/
          __init__.py
          foo.py
          bar.py

and in models/__init__.py I was importing my models using the shorthand syntax:

    from .foo import *
    from .bar import *

In my application I was importing models like so:

    from myapp.models import Foo, Bar

This caused the Django model doesn't declare an explicit app_label when running ./manage.py test.

To fix the problem, I had to explicitly import from the full path in models/__init__.py:

    from myapp.models.foo import *
    from myapp.models.bar import *

That took care of the error.

H/t https://medium.com/@michal.bock/fix-weird-exceptions-when-running-django-tests-f58def71b59a


回答 9

就我而言,这是因为我在项目级urls.py中使用了相对模块路径,INSTALLED_APPSapps.py不是植根于项目根目录中。即,绝对模块路径始终存在,而不是相对模块路径+ hack。

无论我在应用程序中INSTALLED_APPSapps.py应用程序中弄乱了多少,我都无法同时使用它们runserverpytest直到这三个都植根于项目根目录为止。

资料夹结构:

|-- manage.py
|-- config
    |-- settings.py
    |-- urls.py
|-- biz_portal
    |-- apps
        |-- portal
            |-- models.py
            |-- urls.py
            |-- views.py
            |-- apps.py

用下面的,我可以运行manage.py runserver,并与WSGI和使用gunicorn portal应用看法,并没有麻烦,但pytest将与错误ModuleNotFoundError: No module named 'apps',尽管DJANGO_SETTINGS_MODULE被正确配置。

config / settings.py:

INSTALLED_APPS = [
    ...
    "apps.portal.apps.PortalConfig",
]

biz_portal / apps / portal / apps.py:

class PortalConfig(AppConfig):
    name = 'apps.portal'

config / urls.py:

urlpatterns = [
    path('', include('apps.portal.urls')),
    ...
]

改变应用程序的参考配置/ settings.py来biz_portal.apps.portal.apps.PortalConfigPortalConfig.namebiz_portal.apps.portal允许pytest来运行(我没有做检查portal的意见还),但runserver将与错误

RuntimeError:模型类apps.portal.models.Business没有声明显式的app_label,并且不在INSTALLED_APPS中的应用程序中

最后,我摸索apps.portal着查看仍然在使用相对路径的内容,并发现config / urls.py也应该使用biz_portal.apps.portal.urls

In my case, this was happening because I used a relative module path in project-level urls.py, INSTALLED_APPS and apps.py instead of being rooted in the project root. i.e. absolute module paths throughout, rather than relative modules paths + hacks.

No matter how much I messed with the paths in INSTALLED_APPS and apps.py in my app, I couldn’t get both runserver and pytest to work til all three of those were rooted in the project root.

Folder structure:

|-- manage.py
|-- config
    |-- settings.py
    |-- urls.py
|-- biz_portal
    |-- apps
        |-- portal
            |-- models.py
            |-- urls.py
            |-- views.py
            |-- apps.py

With the following, I could run manage.py runserver and gunicorn with wsgi and use portal app views without trouble, but pytest would error with ModuleNotFoundError: No module named 'apps' despite DJANGO_SETTINGS_MODULE being configured correctly.

config/settings.py:

INSTALLED_APPS = [
    ...
    "apps.portal.apps.PortalConfig",
]

biz_portal/apps/portal/apps.py:

class PortalConfig(AppConfig):
    name = 'apps.portal'

config/urls.py:

urlpatterns = [
    path('', include('apps.portal.urls')),
    ...
]

Changing the app reference in config/settings.py to biz_portal.apps.portal.apps.PortalConfig and PortalConfig.name to biz_portal.apps.portal allowed pytest to run (I don’t have tests for portal views yet) but runserver would error with

RuntimeError: Model class apps.portal.models.Business doesn’t declare an explicit app_label and isn’t in an application in INSTALLED_APPS

Finally I grepped for apps.portal to see what’s still using a relative path, and found that config/urls.py should also use biz_portal.apps.portal.urls.


回答 10

当我尝试为单个应用生成迁移时遇到了此错误,该应用由于git合并而存在现有的格式错误的迁移。例如

manage.py makemigrations myapp

当我删除它的迁移然后运行:

manage.py makemigrations

不会发生该错误,并且迁移成功生成。

I ran into this error when I tried generating migrations for a single app which had existing malformed migrations due to a git merge. e.g.

manage.py makemigrations myapp

When I deleted it’s migrations and then ran:

manage.py makemigrations

the error did not occur and the migrations generated successfully.


回答 11

我有一个类似的问题,但是我可以通过在模型类中使用Meta类明确指定app_label来解决我的问题

class Meta:
    app_label  = 'name_of_my_app'

I had a similar issue, but I was able to solve mine by specifying explicitly the app_label using Meta Class in my models class

class Meta:
    app_label  = 'name_of_my_app'

回答 12

尝试将Django Rest Framework应用程序升级到DRF 3.6.3和Django 1.11.1时出现此错误。

对于这种情况下的其他人,我在GitHub问题中找到了解决方案,该问题UNAUTHENTICATED_USERDRF设置中取消设置

# webapp/settings.py
...
REST_FRAMEWORK = {
    ...
    'UNAUTHENTICATED_USER': None
    ...
}

I got this error while trying to upgrade my Django Rest Framework app to DRF 3.6.3 and Django 1.11.1.

For anyone else in this situation, I found my solution in a GitHub issue, which was to unset the UNAUTHENTICATED_USER setting in the DRF settings:

# webapp/settings.py
...
REST_FRAMEWORK = {
    ...
    'UNAUTHENTICATED_USER': None
    ...
}

回答 13

我只是遇到了这个问题,并弄清楚出了什么问题。由于以前没有答案描述发生在我身上的问题,因此我将其发布给其他人:

  • 问题出python migrate.py startapp myApp在我的项目根文件夹中,然后将myApp移到一个子文件夹中mv myApp myFolderWithApps/
  • 我写了myApp.models并运行 python migrate.py makemigrations。一切顺利。
  • 然后我对另一个从myApp导入模型的应用做了同样的操作。b!执行迁移时,我遇到了这个错误。那是因为我不得不使用myFolderWithApps.myApp引用我的应用程序,但是我却忘记了更新MyApp / apps.py。因此,我在第二个应用程序中更正了myApp / apps.py,设置/ INSTALLED_APPS和导入路径。
  • 但是随后错误不断发生:原因是我进行了迁移,试图使用错误的路径从myApp导入模型。我试图更正迁移文件,但我发现重置数据库和删除迁移从头开始更容易。

简而言之:-问题最初是由于myApp的apps.py,第二个应用程序的设置和导入路径中错误的应用程序名称引起的。-但这不足以更正这三个位置的路径,因为迁移是使用导入引用了错误的应用程序名称创建的。因此,相同的错误在迁移过程中始终发生(这次迁移除外)。

所以…请检查您的迁移情况,祝您好运!

I just ran into this issue and figured out what was going wrong. Since no previous answer described the issue as it happened to me, I though I would post it for others:

  • the issue came from using python migrate.py startapp myApp from my project root folder, then move myApp to a child folder with mv myApp myFolderWithApps/.
  • I wrote myApp.models and ran python migrate.py makemigrations. All went well.
  • then I did the same with another app that was importing models from myApp. Kaboom! I ran into this error, while performing makemigrations. That was because I had to use myFolderWithApps.myApp to reference my app, but I had forgotten to update MyApp/apps.py. So I corrected myApp/apps.py, settings/INSTALLED_APPS and my import path in my second app.
  • but then the error kept happening: the reason was that I had migrations trying to import the models from myApp with the wrong path. I tried to correct the migration file, but I went at the point where it was easier to reset the DB and delete the migrations to start from scratch.

So to make a long story short: – the issue was initially coming from the wrong app name in apps.py of myApp, in settings and in the import path of my second app. – but it was not enough to correct the paths in these three places, as migrations had been created with imports referencing the wrong app name. Therefore, the same error kept happening while migrating (except this time from migrations).

So… check your migrations, and good luck!


回答 14

在Django rest_framework中构建API时遇到类似的错误。

RuntimeError:模型类apps.core.models.University未声明显式> app_label,并且不在INSTALLED_APPS中的应用程序中。

luke_aus的答案通过纠正我的问题帮助了我 urls.py

from project.apps.views import SurgeryView

from apps.views import SurgeryView

I’ve got a similar error while building an API in Django rest_framework.

RuntimeError: Model class apps.core.models.University doesn’t declare an explicit > app_label and isn’t in an application in INSTALLED_APPS.

luke_aus’s answer helped me by correcting my urls.py

from

from project.apps.views import SurgeryView

to

from apps.views import SurgeryView

回答 15

就我而言,将代码从Django 1.11.11移植到Django 2.2时出现此错误。我正在定义一个自定义FileSystemStorage派生类。在Django 1.11.11中,我在models.py中包含以下行:

from django.core.files.storage import Storage, DefaultStorage

后来在文件中,我有了类定义:

class MyFileStorage(FileSystemStorage):

但是,在Django 2.2中,我需要FileSystemStorage在导入时显式引用类:

from django.core.files.storage import Storage, DefaultStorage, FileSystemStorage

和贴吧,错误消失了。

请注意,每个人都在报告Django服务器随地吐痰的错误消息的最后一部分。但是,如果向上滚动,则会在该错误的中间部分找到原因。

In my case I got this error when porting code from Django 1.11.11 to Django 2.2. I was defining a custom FileSystemStorage derived class. In Django 1.11.11 I was having the following line in models.py:

from django.core.files.storage import Storage, DefaultStorage

and later in the file I had the class definition:

class MyFileStorage(FileSystemStorage):

However, in Django 2.2 I need to explicitly reference FileSystemStorage class when importing:

from django.core.files.storage import Storage, DefaultStorage, FileSystemStorage

and voilà!, the error dissapears.

Note, that everyone is reporting the last part of the error message spitted by Django server. However, if you scroll up you will find the reason in the middle of that error mambo-jambo.


回答 16

就我而言,我能够找到一个修复程序,并且通过查看其他人的代码也可能是相同的问题。文件。

希望这对某人有帮助。这是我对编码社区的第一个贡献

in my case I was able to find a fix and by looking at the everyone else’s code it may be the same issue.. I simply just had to add ‘django.contrib.sites’ to the list of installed apps in the settings.py file.

hope this helps someone. this is my first contribution to the coding community


回答 17

TL; DR:添加空白__init__.py为我解决了此问题。

我在PyCharm中遇到此错误,并意识到我的设置文件根本没有被导入。没有明显的错误告诉我,但是当我在settings.py中放入一些废话时,它没有引起错误。

我在local_settings文件夹中有settings.py。但是,我希望在同一文件夹中包含__init__.py以便导入。一旦我添加了这个,错误就消失了。

TL;DR: Adding a blank __init__.py fixed the issue for me.

I got this error in PyCharm and realised that my settings file was not being imported at all. There was no obvious error telling me this, but when I put some nonsense code into the settings.py, it didn’t cause an error.

I had settings.py inside a local_settings folder. However, I’d fogotten to include a __init__.py in the same folder to allow it to be imported. Once I’d added this, the error went away.


回答 18

如果所有配置正确,则可能只是导入混乱。密切关注您如何导入违规模型。

以下无效from .models import Business。请使用完整的导入路径:from myapp.models import Business

If you have got all the config right, it might just be an import mess. keep an eye on how you are importing the offending model.

The following won’t work from .models import Business. Use full import path instead: from myapp.models import Business


回答 19

如果所有其他方法均失败,并且在尝试导入PyCharm“ Python控制台”(或“ Django控制台”)时遇到此错误:

尝试重新启动控制台。

这真是令人尴尬,但是花了我一段时间我才意识到自己忘记这样做了。

这是发生了什么:

添加了一个新的应用程序,然后添加了一个最小模型,然后尝试在Python / Django控制台中导入模型(PyCharm pro 2019.2)。这引发了doesn't declare an explicit app_label错误,因为我没有将新应用添加到中INSTALLED_APPS。因此,我将应用添加到了INSTALLED_APPS,再次尝试了导入,但是仍然遇到相同的错误。

来到这里,阅读所有其他答案,但似乎没有合适的答案。

最后,令我惊讶的是,在将新应用添加到后,我还没有重新启动Python控制台INSTALLED_APPS

注意:在将新对象添加到模块后,无法重新启动PyCharm Python控制台也是一种使人非常困惑的好方法 ImportError: Cannot import name ...

If all else fails, and if you are seeing this error while trying to import in a PyCharm “Python console” (or “Django console”):

Try restarting the console.

This is pretty embarassing, but it took me a while before I realized I had forgotten to do that.

Here’s what happened:

Added a fresh app, then added a minimal model, then tried to import the model in the Python/Django console (PyCharm pro 2019.2). This raised the doesn't declare an explicit app_label error, because I had not added the new app to INSTALLED_APPS. So, I added the app to INSTALLED_APPS, tried the import again, but still got the same error.

Came here, read all the other answers, but nothing seemed to fit.

Finally it hit me that I had not yet restarted the Python console after adding the new app to INSTALLED_APPS.

Note: failing to restart the PyCharm Python console, after adding a new object to a module, is also a great way to get a very confusing ImportError: Cannot import name ...


回答 20

O … M … G我也遇到了这个错误,我花了将近2天的时间,现在终于设法解决了。老实说…错误与问题无关。就我而言,这只是语法问题。我试图独立运行一个在django上下文中使用某些django模型的python模块,但该模块本身不是django模型。但是我宣布全班不对

而不是

class Scrapper:
    name = ""
    main_link= ""
    ...

我在做

class Scrapper(Website):
    name = ""
    main_link= ""
    ...

这显然是错误的。该消息是如此令人误解,以至于我忍不住想,但我认为这是配置方面的问题,或者只是以错误的方式使用django,因为我对此很陌生。

我将在这里与新手分享,因为我经历同样的愚蠢期盼能解决他们的问题。

O…M…G I was getting this error too and I spent almost 2 days on it and now I finally managed to solve it. Honestly…the error had nothing to do with what the problem was. In my case it was a simple matter of syntax. I was trying to run a python module standalone that used some django models in a django context, but the module itself wasn’t a django model. But I was declaring the class wrong

instead of having

class Scrapper:
    name = ""
    main_link= ""
    ...

I was doing

class Scrapper(Website):
    name = ""
    main_link= ""
    ...

which is obviously wrong. The message is so misleading that I couldn’t help myself but think it was some issue with configuration or just using django in a wrong way since I’m very new to it.

I’ll share this here for someone newbie as me going through the same silliness can hopefully solve their issue.


回答 21

SECRET_KEY从环境变量中移出并在运行应用程序时忘记对其进行设置后,收到了此错误。如果你有这样的事情settings.py

SECRET_KEY = os.getenv('SECRET_KEY')

然后确保您实际上在设置环境变量。

I received this error after I moved the SECRET_KEY to pull from an environment variable and forgot to set it when running the application. If you have something like this in your settings.py

SECRET_KEY = os.getenv('SECRET_KEY')

then make sure you are actually setting the environment variable.


回答 22

很可能您有依赖进口

在我的情况下,我在模型中使用了序列化程序类作为参数,而序列化程序类则在使用以下模型:serializer_class = AccountSerializer

from ..api.serializers import AccountSerializer

class Account(AbstractBaseUser):
    serializer_class = AccountSerializer
    ...

并在“序列化器”文件中:

from ..models import Account

class AccountSerializer(serializers.ModelSerializer):
    class Meta:
        model = Account
        fields = (
            'id', 'email', 'date_created', 'date_modified',
            'firstname', 'lastname', 'password', 'confirm_password')
    ...

Most probably you have dependent imports.

In my case I used a serializer class as a parameter in my model, and the serializer class was using this model: serializer_class = AccountSerializer

from ..api.serializers import AccountSerializer

class Account(AbstractBaseUser):
    serializer_class = AccountSerializer
    ...

And in the “serializers” file:

from ..models import Account

class AccountSerializer(serializers.ModelSerializer):
    class Meta:
        model = Account
        fields = (
            'id', 'email', 'date_created', 'date_modified',
            'firstname', 'lastname', 'password', 'confirm_password')
    ...

回答 23

我今天遇到了这个错误,并在谷歌搜索后最终到达了这里。现有的答案似乎都与我的情况无关。我唯一需要做的就是从__init__.py应用程序顶层的文件中导入模型。我必须使用模型将导入移动到函数中。

Django似乎有一些奇怪的代码,在许多不同的情况下都可能会失败!

I got this error today and ended up here after googling. None of the existing answers seem relevant to my situation. The only thing I needed to do was to import a model from my __init__.py file in the top level of an app. I had to move my imports into the functions using the model.

Django seems to have some weird code that can fail like this in so many different scenarios!


回答 24

我今天也遇到了这个错误。该消息引用了INSTALLED_APPS我的应用程序中的某些特定应用程序。但实际上,它与此特定的应用程序无关。我使用了一个新的虚拟环境,却忘记安装一些在该项目中使用的库。在我安装了附加库之后,它开始工作了。

I got this error also today. The Message referenced to some specific app of my apps in INSTALLED_APPS. But in fact it had nothing to do with this specific App. I used a new virtual Environment and forgot to install some Libraries, that i used in this project. After i installed the additional Libraries, it worked.


回答 25

对于PyCharm用户:使用“不干净”的项目结构时出现错误。

是:

project_root_directory
└── src
    ├── chat
       ├── migrations
       └── templates
    ├── django_channels
    └── templates

现在:

project_root_directory
├── chat
   ├── migrations
   └── templates
       └── chat
├── django_channels
└── templates

这里有很多好的解决方案,但是我认为,首先,您应该在设置之前清理项目结构或调整PyCharm Django设置 DJANGO_SETTINGS_MODULE变量等。

希望对别人有帮助。干杯。

For PyCharm users: I had an error using not “clean” project structure.

Was:

project_root_directory
└── src
    ├── chat
    │   ├── migrations
    │   └── templates
    ├── django_channels
    └── templates

Now:

project_root_directory
├── chat
│   ├── migrations
│   └── templates
│       └── chat
├── django_channels
└── templates

Here is a lot of good solutions, but I think, first of all, you should clean your project structure or tune PyCharm Django settings before setting DJANGO_SETTINGS_MODULE variables and so on.

Hope it’ll help someone. Cheers.


回答 26

问题是:

  1. 您已经对模型文件进行了修改,但尚未将其添加到数据库中,但是您正在尝试运行Python manage.py runserver。

  2. 运行Python manage.py makemigrations

  3. Python manage.py迁移

  4. 现在,Python manage.py runserver和一切都应该没问题。

The issue is that:

  1. You have made modifications to your models file, but not addedd them yet to the DB, but you are trying to run Python manage.py runserver.

  2. Run Python manage.py makemigrations

  3. Python manage.py migrate

  4. Now Python manage.py runserver and all should be fine.


Pipenv:找不到命令

问题:Pipenv:找不到命令

我是Python开发的新手,正在尝试使用pipenv。我运行了命令pip install pipenv,该命令成功运行:

...
Successfully built pipenv pathlib shutilwhich pythonz-bd virtualenv-clone
Installing collected packages: virtualenv, pathlib, shutilwhich, backports.shutil-get-terminal-size, pythonz-bd, virtualenv-clone, pew, first, six, click, pip-tools, certifi, chardet, idna, urllib3, requests, pipenv
...

但是,当我pipenv install在新的根项目目录中运行命令时,收到以下消息:-bash: pipenv: command not found。我怀疑可能需要修改.bashrc,但是不清楚要添加到文件中的内容还是不清楚是否需要修改。

I’m new to Python development and attempting to use pipenv. I ran the command pip install pipenv, which ran successfully:

...
Successfully built pipenv pathlib shutilwhich pythonz-bd virtualenv-clone
Installing collected packages: virtualenv, pathlib, shutilwhich, backports.shutil-get-terminal-size, pythonz-bd, virtualenv-clone, pew, first, six, click, pip-tools, certifi, chardet, idna, urllib3, requests, pipenv
...

However, when I run the command pipenv install in a fresh root project directory I receive the following message: -bash: pipenv: command not found. I suspect that I might need to modify my .bashrc, but I’m unclear about what to add to the file or if modification is even necessary.


回答 0

发生这种情况是因为您没有在全局范围内安装它(系统范围内)。为了使它可用,您path需要使用进行安装sudo,如下所示:

$ sudo pip install pipenv

That happens because you are not installing it globally (system wide). For it to be available in your path you need to install it using sudo, like this:

$ sudo pip install pipenv

回答 1

这为我解决了:

sudo -H pip install -U pipenv

This fixed it for me:

sudo -H pip install -U pipenv

回答 2

如果完成了用户安装,则需要将正确的文件夹添加到PATH变量中。

PYTHON_BIN_PATH="$(python3 -m site --user-base)/bin"
PATH="$PATH:$PYTHON_BIN_PATH"

看到 pipenv的安装说明

If you’ve done a user installation, you’ll need to add the right folder to your PATH variable.

PYTHON_BIN_PATH="$(python3 -m site --user-base)/bin"
PATH="$PATH:$PYTHON_BIN_PATH"

See pipenv’s installation instructions


回答 3

我尝试了这个:

python -m pipenv # for python2

python3 -m pipenv # for python3

希望这可以帮到你。

I tried this:

python -m pipenv # for python2

python3 -m pipenv # for python3

Hope this can help you.


回答 4

pipenv在Mac OS X 10.13 High Seirra上也有同样的问题,另一台Mac可以正常工作。我使用Heroku部署Django服务器,其中一些部署在2.7中,有些部署在3.6中。因此,我需要2.7和3.6。当HomeBrew安装Python时,它将python指向原始2.7,并且python3指向3.6。

问题可能是由于$ pip install pipenv。我检查了/ usr / local / bin,pipenv不存在。因此,我尝试了完全卸载:

$ pip uninstall pipenv

Cannot uninstall requirement pipenv, not installed
You are using pip version 9.0.1, however version 10.0.1 is available.
You should consider upgrading via the 'pip install --upgrade pip' command.

$ pip3 uninstall pipenv
Skipping pipenv as it is not installed.

然后重新安装即可立即使用:

$ pip3 install pipenv
Collecting pipenv

I have same problem with pipenv on Mac OS X 10.13 High Seirra, another Mac works just fine. I use Heroku to deploy my Django servers, some in 2.7 and some in 3.6. So, I need both 2.7 and 3.6. When HomeBrew install Python, it keeps python points to original 2.7, and python3 points to 3.6.

The problem might due to $ pip install pipenv. I checked /usr/local/bin and pipenv isn’t there. So, I tried a full uninstall:

$ pip uninstall pipenv

Cannot uninstall requirement pipenv, not installed
You are using pip version 9.0.1, however version 10.0.1 is available.
You should consider upgrading via the 'pip install --upgrade pip' command.

$ pip3 uninstall pipenv
Skipping pipenv as it is not installed.

Then reinstall and works now:

$ pip3 install pipenv
Collecting pipenv

回答 5

OSX的家伙,在这里!!!

正如@charlax回答的那样(对我而言,这是最好的),您可以使用更动态的命令来设置PATH,但对于mac用户而言,buuut 无法正常工作,有时从站点获取的USER_BASE路径错误,因此您需要找出您的位置python安装是。

$ which python3
/usr/local/bin/python3.6

您将获得一个符号链接,然后需要找到源的符号链接。

$ ls -la /usr/local/bin/python3.6
  lrwxr-xr-x  1 root  wheel  71 Mar 14 17:56 /usr/local/bin/python3.6 -> ../../../Library/Frameworks/Python.framework/Versions/3.6/bin/python3.6

(这../../../意味着根)

因此,您找到了python路径(/Library/Frameworks/Python.framework/Versions/3.6/bin/python3.6),然后只需将〜/ .bashrc放入其中,如下所示:

export PATH="$PATH:/Library/Frameworks/Python.framework/Versions/3.6/bin"

OSX GUYS, OVER HERE!!!

As @charlax answered (for me the best one), you can use a more dynamic command to set PATH, buuut for mac users this could not work, sometimes your USER_BASE path got from site is wrong, so you need to find out where your python installation is.

$ which python3
/usr/local/bin/python3.6

you’ll get a symlink, then you need to find the source’s symlink.

$ ls -la /usr/local/bin/python3.6
  lrwxr-xr-x  1 root  wheel  71 Mar 14 17:56 /usr/local/bin/python3.6 -> ../../../Library/Frameworks/Python.framework/Versions/3.6/bin/python3.6

(this ../../../ means root)

So you found the python path (/Library/Frameworks/Python.framework/Versions/3.6/bin/python3.6), then you just need to put in you ~/.bashrc as follows:

export PATH="$PATH:/Library/Frameworks/Python.framework/Versions/3.6/bin"


回答 6

通过覆盖全局/系统管理的pip安装,全局安装pipenv可能会产生不利影响,从而在尝试运行pip时导致导入错误。

您可以在用户级别安装pipenv:

pip install --user pipenv

这应该在/home/username/.local中的用户级别安装pipenv,以使其与pip的全局版本不冲突。就我而言,在运行’–user’开关后仍然无法正常工作,因此我运行了更长的“修复我搞砸的东西”命令一次,以恢复系统管理的环境:

sudo python3 -m pip uninstall pip && sudo apt install python3-pip --reinstall

^在这里找到:升级pip后出错:无法导入名称’main’

然后执行以下操作:

mkdir /home/username/.local …如果还不存在

export PYTHONUSERBASE=/home/username/.local

确保导出生效(在此过程中,只对我进行一次咬合):

echo $PYTHONUSERBASE

然后,我跑了pip install --user pipenv,一切都很好。然后,我可以从CLI运行pipenv,它不会覆盖全局/系统管理的pip模块。当然,这是特定于用户的,因此您要确保在以希望使用pipenv的用户身份工作时以这种方式安装pipenv。

参考文献:

https://pipenv.readthedocs.io/en/latest/diagnose/#no-module-named-module-name https://pipenv.readthedocs.io/en/latest/install/#pragmatic-installation-of-pipenv https://pip.pypa.io/zh_CN/stable/user_guide/#user-installs

Installing pipenv globally can have an adverse effect by overwriting the global/system-managed pip installation, thus resulting in import errors when trying to run pip.

You can install pipenv at the user level:

pip install --user pipenv

This should install pipenv at a user-level in /home/username/.local so that it does not conflict with the global version of pip. In my case, that still did not work after running the ‘–user’ switch, so I ran the longer ‘fix what I screwed up’ command once to restore the system managed environment:

sudo python3 -m pip uninstall pip && sudo apt install python3-pip --reinstall

^ found here: Error after upgrading pip: cannot import name ‘main’

and then did the following:

mkdir /home/username/.local … if it doesn’t already exist

export PYTHONUSERBASE=/home/username/.local

Make sure the export took effect (bit me once during this process):

echo $PYTHONUSERBASE

Then, I ran the pip install --user pipenv and all was well. I could then run pipenv from the CLI and it did not overwrite the global/system-managed pip module. Of course, this is specific to the user so you want to make sure you install pipenv this way while working as the user you wish to use pipenv.

References:

https://pipenv.readthedocs.io/en/latest/diagnose/#no-module-named-module-name https://pipenv.readthedocs.io/en/latest/install/#pragmatic-installation-of-pipenv https://pip.pypa.io/en/stable/user_guide/#user-installs


回答 7

Python存储包的位置

在进入将要安装的命令之前pipenv,值得了解pip Python软件包的安装位置

全局站点软件包是Python安装软件包的地方,这些软件包将可供系统上的所有用户和所有Python应用程序使用。您可以使用以下命令检查全局站点包

python -m site

例如,在具有Python 3.7的Linux上,路径通常是

/usr/lib/python3.7/dist-packages/setuptools

用户站点软件包是Python安装仅对您可用的软件包的地方。但是,这些包对您创建的所有Python项目仍然可见。您可以使用以下方法

python -m site --user-base

在具有Python 3.7的Linux上,路径通常是

~/.local/lib/python3.7/site-packages

使用Python 3.x

在大多数Linux和其他Unices上,通常并行安装Python 2和Python 3。默认的Python 3可执行文件几乎总是python3pip根据您的Linux发行版本,可能提供以下任一版本

pip3
python3-pip
python36-pip
python3.6-pip

的Linux

避免pipsudo!是的,这是安装Python软件包的最方便的方法,可执行文件位于/usr/local/bin/pipenv,但也意味着特定的软件包始终对所有用户和您创建的所有Python项目可见。而是使用按用户站点包,而不是--user

pip3 install --user pipenv

pipenv 在以下位置可用

~/.local/bin/pipenv

苹果系统

在macOS上,建议使用Homebrew来安装Python。您可以轻松升级Python,安装多个版本的Python,以及使用Homebrew在版本之间切换。

如果您使用的是自制软件的Python,则已pip install --user禁用。全局站点包位于

/usr/local/lib/python3.y/site-packages

并且您可以在此处安全地安装Python软件包。Python 3.y还在以下位置搜索模块:

 /Library/Python/3.y/site-packages
 ~/Library/Python/3.y/lib/python/site-packages

视窗

出于传统原因,Python安装在中C:\Python37。Python可执行文件通常被命名为py.exe,并且可以pip使用来运行py -m pip

全局站点包安装在

C:\Python37\lib\site-packages

由于通常不共享Windows设备,因此也可以在全局范围内安装软件包

py -m pip install pipenv

pipenv 现在可以在

C:\Python37\Scripts\pipenv.exe

我不建议使用Windows在Windows中安装Python软件包--user,因为默认的用户site-package目录位于Windows漫游配置文件中

C:\Users\user\AppData\Roaming\Python\Python37\site-packages 

在终端服务(远程桌面,Citrix等)中以及在公司环境中登录/注销时,将使用漫游配置文件。大型漫游配置文件可能导致Windows中缓慢的登录,注销和重新启动。

Where Python store packages

Before jumping into the command that will install pipenv, it is worth understanding where pip installs Python packages.

Global site-packages is where Python installs packages that will be available to all users and all Python applications on the system. You can check the global site package with the command

python -m site

For example, on Linux with Python 3.7 the path is usually

/usr/lib/python3.7/dist-packages/setuptools

User site-packages is where Python installs packages available only for you. But the packages will still be visible to all Python projects that you create. You can get the path with

python -m site --user-base

On Linux with Python 3.7 the path is usually

~/.local/lib/python3.7/site-packages

Using Python 3.x

On most Linux and other Unices, usually Python 2 and Python 3 is installed side-by-side. The default Python 3 executable is almost always python3. pip may be available as either of the following, depending on your Linux distribution

pip3
python3-pip
python36-pip
python3.6-pip

Linux

Avoid using pip with sudo! Yes, it’s the most convenient way to install Python packages and the executable is available at /usr/local/bin/pipenv, but it also mean that specific package is always visible for all users, and all Python projects that you create. Instead, use per-user site packages instead with --user

pip3 install --user pipenv

pipenv is available at

~/.local/bin/pipenv

macOS

On macOS, Homebrew is the recommended way to install Python. You can easily upgrade Python, install multiple versions of Python and switch between versions using Homebrew.

If you are using Homebrew’ed Python, pip install --user is disabled. The global site-package is located at

/usr/local/lib/python3.y/site-packages

and you can safely install Python packages here. Python 3.y also searches for modules in:

 /Library/Python/3.y/site-packages
 ~/Library/Python/3.y/lib/python/site-packages

Windows

For legacy reasons, Python is installed in C:\Python37. The Python executable is usually named py.exe, and you can run pip with py -m pip.

Global site packages is installed in

C:\Python37\lib\site-packages

Since you don’t usually share your Windows devices, it is also OK to install a package globally

py -m pip install pipenv

pipenv is now available at

C:\Python37\Scripts\pipenv.exe

I don’t recommend install Python packages in Windows with --user, because the default user site-package directory is in your Windows roaming profile

C:\Users\user\AppData\Roaming\Python\Python37\site-packages 

The roaming profile is used in Terminal Services (Remote Desktop, Citrix, etc) and when you log on / off in a corporate environment. Slow login, logoff and reboot in Windows can be caused by a large roaming profile.


回答 8

我已将其固定为:

sudo -H pip install -U pipenv

This is fixed for me to:

sudo -H pip install -U pipenv

回答 9

安装pipenv(sudo pip install pipenv)后,尝试运行Windows XP时,始终出现“找不到命令”错误pipenv shell命令。

我终于用以下代码修复了它:

pip3 install pipenv
pipenv shell

After installing pipenv (sudo pip install pipenv), I kept getting the “Command Not Found” error when attempting to run the pipenv shell command.

I finally fixed it with the following code:

pip3 install pipenv
pipenv shell

回答 10

您可以考虑pipenv通过安装pipsi

curl https://raw.githubusercontent.com/mitsuhiko/pipsi/master/get -pipsi.py | python3
pipsi install pew
pipsi install pipenv

不幸的是有一些问题与写作时的MacOS + python3,看到12。就我而言,我必须将bashprompt更改为#!/Users/einselbst/.local/venvs/pipsi/bin/python

You might consider installing pipenv via pipsi.

curl https://raw.githubusercontent.com/mitsuhiko/pipsi/master/get -pipsi.py | python3
pipsi install pew
pipsi install pipenv

Unfortunately there are some issues with macOS + python3 at the time of writing, see 1, 2. In my case I had to change the bashprompt to #!/Users/einselbst/.local/venvs/pipsi/bin/python


回答 11

在旧pip版本的某些情况下:

sudo easy_install pip
sudo pip install pipenv

In some cases of old pip version:

sudo easy_install pip
sudo pip install pipenv

回答 12

对于Windows用户,这可能是由于与virtualenv的安装冲突。对我来说,它在我先卸载virtualenv和pipenv,然后仅安装pipenv时起作用。

pip uninstall virtualenv
pip uninstall pipenv
pip install pipenv

现在pipenv install xxx为我工作

For window users this may be due to conflicting installation with virtualenv. For me it worked when I uninstalled virtualenv and pipenv first, and then install only pipenv.

pip uninstall virtualenv
pip uninstall pipenv
pip install pipenv

Now pipenv install xxx worked for me


回答 13

在Mac OS X Catalina上,它似乎遵循Linux路径。使用以下任何一种:

pip install pipenv
pip3 install pipenv
sudo pip install pipenv
sudo pip3 install pipenv

本质上在这里安装pipenv:

/Users/mike/Library/Python/3.7/lib/python/site-packages/pipenv

但是它不是可执行文件,因此永远找不到。唯一对我有用的是

pip install --user pipenv

这似乎导致__init__.py上述目录中的文件包含正确显示pipenv命令的内容。

当有关此问题的所有其他发布和评论建议都失败时,一切都开始起作用。

pipenv软件包当然看起来很挑剔。

On Mac OS X Catalina it appears to follow the Linux path. Using any of:

pip install pipenv
pip3 install pipenv
sudo pip install pipenv
sudo pip3 install pipenv

Essentially installs pipenv here:

/Users/mike/Library/Python/3.7/lib/python/site-packages/pipenv

But its not the executable and so is never found. The only thing that worked for me was

pip install --user pipenv

This seems to result in an __init__.py file in the above directory that has contents to correctly expose the pipenv command.

and everything started working, when all other posted and commented suggestions on this question failed.

The pipenv package certainly seems quite picky.


pip或pip3为Python 3安装软件包?

问题:pip或pip3为Python 3安装软件包?

我有一台配备OS X El Captain的Macbook。我认为那Python 2.7是预装的。但是,我也安装Python 3.5了。开始使用时Python 3,我读到要安装软件包,应输入:

pip3 install some_package

无论如何,现在当我使用

pip install some_package

some_package安装了Python 3。我的意思是我可以导入它,并且可以毫无问题地使用它。而且,当我只pip3输入时Terminal,我得到以下关于用法的消息:

Usage:   
  pip <command> [options]

这是我输入just时得到的相同信息pip

这是否意味着在previos版本,事情是不同的,现在pippip3可以互换使用?如果是这样,并且为了参数起见,我该如何安装软件包Python 2而不是Python 3

I have a Macbook with OS X El Captain. I think that Python 2.7 comes preinstalled on it. However, I installed Python 3.5 too. When I started using Python 3, I read that if I want to install a package, I should type:

pip3 install some_package

Anyway, now when I use

pip install some_package

I get some_package installed for Python 3. I mean I can import it and use it without problems. Moreover, when I type just pip3 in Terminal, I got this message about the usage:

Usage:   
  pip <command> [options]

which is the same message I get when I type just pip.

Does it mean that in previos versions, things were different, and now pip and pip3 can be used interchangeably? If so, and for the sake of argument, how can I install packages for Python 2 instead of Python 3?


回答 0

pip是使用链接到同一可执行文件路径的软链接pip3。您可以使用以下命令来检查您的真实路径pippip3实际路径:

$ ls -l `which pip`
$ ls -l `which pip3`

您也可以使用以下命令了解更多详细信息:

$ pip show pip
$ pip3 show pip

当我们安装不同版本的python时,我们可能会创建以下软链接到

  • 将默认点设置为某些版本。
  • 为不同的版本创建不同的链接。

这是同样的情况pythonpython2python3

如果您对在不同情况下如何发生感兴趣,可以在下面获取更多信息:

Your pip is a soft link to the same executable file path with pip3. you can use the commands below to check where your pip and pip3 real paths are:

$ ls -l `which pip`
$ ls -l `which pip3`

You may also use the commands below to know more details:

$ pip show pip
$ pip3 show pip

When we install different versions of python, we may create such soft links to

  • set default pip to some version.
  • make different links for different versions.

It is the same situation with python, python2, python3

More information below if you’re interested in how it happens in different cases:


回答 1

如果您安装了python 2.x,然后安装了python3,则您的pip将指向pip3。您可以通过键入来验证pip --version与相同pip3 --version

在您的系统上,现在有了pip,pip2和pip3

如果需要,可以将pip更改为指向pip2而不是pip3。

If you had python 2.x and then installed python3, your pip will be pointing to pip3. you can verify that by typing pip --version which would be the same as pip3 --version.

On your system you have now pip, pip2 and pip3.

If you want you can change pip to point to pip2 instead of pip3.


回答 2

安装后python3,即pip3会安装。如果你没有其他的Python安装(如python2.7),然后链接创建这点pippip3

所以pip是一个链接pip3,如果有Python安装的任何其他版本(比python3等)。 pip通常指向首次安装。

When you install python3, pip3 gets installed. And if you don’t have another python installation(like python2.7) then a link is created which points pip to pip3.

So pip is a link to to pip3 if there is no other version of python installed(other than python3). pip generally points to the first installation.


回答 3

这是一个棘手的问题。最后,如果您调用pip它,则将调用pip2pip3,具体取决于您如何设置系统。

This is a tricky subject. In the end, if you invoke pip it will invoke either pip2 or pip3, depending on how you set your system up.


回答 4

我认为pippip2并且pip3不是指向同一可执行文件路径的软链接。请注意以下命令并在我的linux终端中显示结果:

mrz@mrz-pc ~ $ ls -l `which pip`
-rwxr-xr-x 1 root root 292 Nov 10  2016 /usr/bin/pip
mrz@mrz-pc ~ $ ls -l `which pip2`
-rwxr-xr-x 1 root root 283 Nov 10  2016 /usr/bin/pip2
mrz@mrz-pc ~ $ ls -l `which pip3`
-rwxr-xr-x 1 root root 293 Nov 10  2016 /usr/bin/pip3
mrz@mrz-pc ~ $ pip -V
pip 9.0.1 from /home/mrz/.local/lib/python2.7/site-packages (python 2.7)
mrz@mrz-pc ~ $ pip2 -V
pip 8.1.1 from /usr/lib/python2.7/dist-packages (python 2.7)
mrz@mrz-pc ~ $ pip3 -V
pip 9.0.1 from /home/mrz/.local/lib/python3.5/site-packages (python 3.5)

如您所见,它们存在于不同的路径中。

pip3始终仅在Python3环境上运行,就像pip2在Python2上一样。pip在适合上下文的任何环境下运行。例如,如果您在Python3平台上,pip将在Python3环境下运行。

I think pip, pip2 and pip3 are not soft links to the same executable file path. Note these commands and results in my Linux terminal:

mrz@mrz-pc ~ $ ls -l `which pip`
-rwxr-xr-x 1 root root 292 Nov 10  2016 /usr/bin/pip
mrz@mrz-pc ~ $ ls -l `which pip2`
-rwxr-xr-x 1 root root 283 Nov 10  2016 /usr/bin/pip2
mrz@mrz-pc ~ $ ls -l `which pip3`
-rwxr-xr-x 1 root root 293 Nov 10  2016 /usr/bin/pip3
mrz@mrz-pc ~ $ pip -V
pip 9.0.1 from /home/mrz/.local/lib/python2.7/site-packages (python 2.7)
mrz@mrz-pc ~ $ pip2 -V
pip 8.1.1 from /usr/lib/python2.7/dist-packages (python 2.7)
mrz@mrz-pc ~ $ pip3 -V
pip 9.0.1 from /home/mrz/.local/lib/python3.5/site-packages (python 3.5)

As you see they exist in different paths.

pip3 always operates on the Python3 environment only, as pip2 does with Python2. pip operates in whichever environment is appropriate to the context. For example, if you are in a Python3 venv, pip will operate on the Python3 environment.


回答 5

通过插图:

pip --version
  pip 19.0.3 from /usr/lib/python3.7/site-packages/pip (python 3.7)

pip3 --version
  pip 19.0.3 from /usr/lib/python3.7/site-packages/pip (python 3.7)

python --version
  Python 3.7.3

which python
  /usr/bin/python

ls -l '/usr/bin/python'
  lrwxrwxrwx 1 root root 7 Mar 26 14:43 /usr/bin/python -> python3

which python3
  /usr/bin/python3

ls -l /usr/bin/python3
  lrwxrwxrwx 1 root root 9 Mar 26 14:43 /usr/bin/python3 -> python3.7

ls -l /usr/bin/python3.7
  -rwxr-xr-x 2 root root 14120 Mar 26 14:43 /usr/bin/python3.7

因此,在我的默认系统python(Python 3.7.3)中,my pip pip3

By illustration:

pip --version
  pip 19.0.3 from /usr/lib/python3.7/site-packages/pip (python 3.7)

pip3 --version
  pip 19.0.3 from /usr/lib/python3.7/site-packages/pip (python 3.7)

python --version
  Python 3.7.3

which python
  /usr/bin/python

ls -l '/usr/bin/python'
  lrwxrwxrwx 1 root root 7 Mar 26 14:43 /usr/bin/python -> python3

which python3
  /usr/bin/python3

ls -l /usr/bin/python3
  lrwxrwxrwx 1 root root 9 Mar 26 14:43 /usr/bin/python3 -> python3.7

ls -l /usr/bin/python3.7
  -rwxr-xr-x 2 root root 14120 Mar 26 14:43 /usr/bin/python3.7

Thus, my in my default system python (Python 3.7.3), pip is pip3.


回答 6

如果您安装了Python 2.7,我想您可以使用pip2pip2.7安装专门针对Python 2的软件包,例如

pip2 install some_pacakge

要么

pip2.7 install some_package

您可以使用pip3pip3.5安装专门用于Python 3的pacakges。

If you installed Python 2.7, I think you could use pip2 and pip2.7 to install packages specifically for Python 2, like

pip2 install some_pacakge

or

pip2.7 install some_package

And you may use pip3 or pip3.5 to install pacakges specifically for Python 3.


回答 7

在我的Windows实例上-并且我不完全了解我的环境-使用pip3安装kaggle-cli程序包有效-而pip无效。我在conda环境中工作,环境似乎有所不同。

(fastai)C:\ Users \ redact \ Downloads \ fast.ai \ deeplearning1 \ nbs> pip –version

来自C:\ ProgramData \ Anaconda3 \ envs \ fastai \ lib \ site-packages的pip 9.0.1(python 3.6)

(fastai)C:\ Users \ redact \ Downloads \ fast.ai \ deeplearning1 \ nbs> pip3 –version

来自c:\ users \ redact \ appdata \ local \ programs \ python \ python36 \ lib \ site-packages(python 3.6)的pip 9.0.1

On my Windows instance – and I do not fully understand my environment – using pip3 to install the kaggle-cli package worked – whereas pip did not. I was working in a conda environment and the environments appear to be different.

(fastai) C:\Users\redact\Downloads\fast.ai\deeplearning1\nbs>pip –version

pip 9.0.1 from C:\ProgramData\Anaconda3\envs\fastai\lib\site-packages (python 3.6)

(fastai) C:\Users\redact\Downloads\fast.ai\deeplearning1\nbs>pip3 –version

pip 9.0.1 from c:\users\redact\appdata\local\programs\python\python36\lib\site-packages (python 3.6)


回答 8

somepath / venv中激活了Python 3.6 virtualenv后,以下别名解决了macOS Sierra上的各种问题,其中pip坚持指向Apple的2.7 Python。

alias pip='python somepath/venv/lib/python3.6/site-packages/pip/__main__.py'

当我不得不做的时候,这不是很好,sudo pip因为root用户对我的别名或virtualenv一无所知,所以我也不得不添加一个额外的别名来处理这个问题。这是一个hack,但是可以用,我知道它的作用:

alias sudopip='sudo somepath/venv/bin/python somepath/venv/lib/python3.6/site-packages/pip/__main__.py'

背景:

pip3不存在,无法启动(找不到命令),which pip它将返回Apple Python的/opt/local/Library/Frameworks/Python.framework/Versions/2.7/bin/pip

Python 3.6是通过macports安装的。

激活我想使用的3.6 virtualenv后,which python将返回somepath / venv / bin / python

以某种方式pip install可以做正确的事情并击中我的virtualenv,但pip list会使Python 2.7包感到不安。

对于Python而言,这对初学者友好程度比我期望的要低。

Given an activated Python 3.6 virtualenv in somepath/venv, the following aliases resolved the various issues on a macOS Sierra where pip insisted on pointing to Apple’s 2.7 Python.

alias pip='python somepath/venv/lib/python3.6/site-packages/pip/__main__.py'

This didn’t work so well when I had to do sudo pip as the root user doesn’t know anything about my alias or the virtualenv, so I had to add an extra alias to handle this as well. It’s a hack, but it works, and I know what it does:

alias sudopip='sudo somepath/venv/bin/python somepath/venv/lib/python3.6/site-packages/pip/__main__.py'

background:

pip3 did not exist to start (command not found) with and which pip would return /opt/local/Library/Frameworks/Python.framework/Versions/2.7/bin/pip, the Apple Python.

Python 3.6 was installed via macports.

After activation of the 3.6 virtualenv I wanted to work with, which python would return somepath/venv/bin/python

Somehow pip install would do the right thing and hit my virtualenv, but pip list would rattle off Python 2.7 packages.

For Python, this is batting way beneath my expectations in terms of beginner-friendliness.


为什么Python 3.6.1抛出AttributeError:模块’enum’没有属性’IntFlag’?

问题:为什么Python 3.6.1抛出AttributeError:模块’enum’没有属性’IntFlag’?

我刚刚为MacOS X安装了Python 3.6.1

当我尝试运行控制台(或使用Python3运行任何命令)时,抛出此错误:

  AttributeError: module 'enum' has no attribute 'IntFlag'

$ /Library/Frameworks/Python.framework/Versions/3.6/bin/python3  
Failed to import the site module  
Traceback (most recent call last):  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 544, in <module>  
    main()  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 530, in main  
    known_paths = addusersitepackages(known_paths)  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 282, in addusersitepackages  
    user_site = getusersitepackages()  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 258, in getusersitepackages  
    user_base = getuserbase() # this will also set USER_BASE  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 248, in getuserbase  
    USER_BASE = get_config_var('userbase')  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/sysconfig.py", line 601, in get_config_var  
    return get_config_vars().get(name)  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/sysconfig.py", line 580, in get_config_vars  
    import _osx_support  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/_osx_support.py", line 4, in <module>  
    import re  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/re.py", line 142, in <module>  
    class RegexFlag(enum.IntFlag):  
AttributeError: module 'enum' has no attribute 'IntFlag'  

IntFlag类存在于enum.py中。那么,为什么会抛出AttributeError?

I just installed Python 3.6.1 for MacOS X

When I attempt to run the Console(or run anything with Python3), this error is thrown:

  AttributeError: module 'enum' has no attribute 'IntFlag'

$ /Library/Frameworks/Python.framework/Versions/3.6/bin/python3  
Failed to import the site module  
Traceback (most recent call last):  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 544, in <module>  
    main()  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 530, in main  
    known_paths = addusersitepackages(known_paths)  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 282, in addusersitepackages  
    user_site = getusersitepackages()  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 258, in getusersitepackages  
    user_base = getuserbase() # this will also set USER_BASE  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/site.py", line 248, in getuserbase  
    USER_BASE = get_config_var('userbase')  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/sysconfig.py", line 601, in get_config_var  
    return get_config_vars().get(name)  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/sysconfig.py", line 580, in get_config_vars  
    import _osx_support  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/_osx_support.py", line 4, in <module>  
    import re  
  File "/usr/local/Cellar/python3/3.6.1/Frameworks/Python.framework/Versions/3.6/lib/python3.6/re.py", line 142, in <module>  
    class RegexFlag(enum.IntFlag):  
AttributeError: module 'enum' has no attribute 'IntFlag'  

The class IntFlag exists within enum.py. So, why is the AttributeError being thrown?


回答 0

这是因为您enum不是标准库enum模块。您可能已经enum34安装了软件包。

检查是否属于这种情况的一种方法是检查物业 enum.__file__

import enum
print(enum.__file__)  
# standard library location should be something like 
# /usr/local/lib/python3.6/enum.py

从python 3.6开始,enum34库不再与标准库兼容。该库也是不必要的,因此您可以简单地将其卸载。

pip uninstall -y enum34

如果您需要代码在<= 3.4和> 3.4的python版本上运行,则可以enum-compat根据需要尝试。它仅enum34针对没有标准库枚举的python较旧版本安装。

It’s because your enum is not the standard library enum module. You probably have the package enum34 installed.

One way check if this is the case is to inspect the property enum.__file__

import enum
print(enum.__file__)  
# standard library location should be something like 
# /usr/local/lib/python3.6/enum.py

Since python 3.6 the enum34 library is no longer compatible with the standard library. The library is also unnecessary, so you can simply uninstall it.

pip uninstall -y enum34

If you need the code to run on python versions both <=3.4 and >3.4, you can try having enum-compat as a requirement. It only installs enum34 for older versions of python without the standard library enum.


回答 1

不知道您是否仍然有此问题。我遇到了类似的问题,只需取消设置即可解决PYTHONPATH

$ unset PYTHONPATH

Not sure whether you still have this issue. I had a similar issue and I was able to resolve it simply by unsetting PYTHONPATH

$ unset PYTHONPATH


回答 2

对我来说,此错误是在安装gcloud组件app-engine-python以便集成到pycharm后发生的。即使pycharm现在不上传到app-engine,卸载模块也有帮助。

For me this error occured after installing of gcloud component app-engine-python in order to integrate into pycharm. Uninstalling the module helped, even if pycharm is now not uploading to app-engine.


回答 3

如果有人在PyCharm中运行Google App Engine Python 3.7标准环境项目时由于遇到此错误而来这里,那么您所需要做的就是

  • 确保要运行的配置适用于Flask,而不适用于Google App Engine配置。
  • 然后在偏好设置>>语言和框架>> Google App Engine下禁用Google App Engine支持

原因是根据此链接

总体目标是您的应用程序应具有完全可移植性,并可以在任何标准Python环境中运行。您编写的是标准Python应用程序,而不是App Engine Python应用程序。作为此转变的一部分,您不再需要为应用程序的核心功能使用专有的App Engine API和服务。目前,Python 3.7运行时中不提供App Engine API。

我猜想当我们在PyCharm中将python 3.7项目创建为Google应用引擎项目时,它仍会尝试以与python2.7应用相同的方式进行操作

If anyone coming here because of getting this error while running a google app engine Python 3.7 standard environment project in PyCharm then all you need to do is

  • Make sure the configuration to run is for Flask, not Google App Engine configuration.
  • Then disable Google App Engine support under Preferences >> Languages & Framework >> Google App Engine

The reason being as per this link

The overall goal is that your app should be fully portable and run in any standard Python environment. You write a standard Python app, not an App Engine Python app. As part of this shift, you are no longer required to use proprietary App Engine APIs and services for your app’s core functionality. At this time, App Engine APIs are not available in the Python 3.7 runtime.

I guess when we create a python 3.7 project in PyCharm as a Google app engine project it still tries to do the same way it does for a python2.7 app


回答 4

免责声明:请@ juanpa.arrivillaga,如果您看到此答案,请随时写自己的答案,我将删除此帖子。

@ juanpa.arrivillaga 上面已经提到过

您的工作目录中是否有文件名enum.py?

这就是我遇到的问题。我当时还不知道python上的enum模块,并已命名我的测试文件enum.py

由于文件名模块名,因此存在冲突。有关模块的更多信息,请参见:https : //docs.python.org/2/tutorial/modules.html

DISCLAIMER: Please, @juanpa.arrivillaga, if you see this answer, feel free to write your own and I will remove this post.

@juanpa.arrivillaga had mentioned above:

Is there a file name enum.py in your working directory, by any chance?

This was the issue I encountered. I was not aware of the enum module on python at the time and had named my test file enum.py.

Since the file name is the module name, there was a conflict. More info on modules here: https://docs.python.org/2/tutorial/modules.html


回答 5

在pycharm首选项中禁用“ Google App Engine支持”对我来说解决了这个问题。

Disabling “Google App Engine Support” in pycharm preferences fixed this issue for me.


回答 6

HåkenLid的回答帮助解决了我的问题(谢谢!),在本例中,在运行于Docker容器中的Flask的Python3.7中FROM tiangolo/uwsgi-nginx-flask:python3.7-alpine3.7

就我而言,enum34是由另一个库(pip install smartsheet-python-sdk)安装的。对于那些遇到类似Docker容器问题的人,这是我的最终Dockerfile(跳至相关行):

FROM tiangolo/uwsgi-nginx-flask:python3.7-alpine3.7
...
RUN pip install -r requirements.txt
RUN pip uninstall -y enum34
...

Håken Lid’s answer helped solved my problem (thanks!) , in my case present in Python3.7 running Flask in a Docker container (FROM tiangolo/uwsgi-nginx-flask:python3.7-alpine3.7).

In my case, enum34 was being installed by another library (pip install smartsheet-python-sdk). For those coming with a similar Docker container problem, here is my final Dockerfile (stripped to the relevant lines):

FROM tiangolo/uwsgi-nginx-flask:python3.7-alpine3.7
...
RUN pip install -r requirements.txt
RUN pip uninstall -y enum34
...

回答 7

如果必须同时为python2和python3保留PYTHONPATH,则可以编写别名语句以在bash_profile中设置适当的PYTHONPATH:

在〜/ .bash_profile中,对PYTHONPATH2和PYTHONPATH3变量进行硬编码,并在其末尾添加以下别名:

alias python='export PYTHONPATH=${PYTHONPATH2};python'
alias python3='export PYTHONPATH=${PYTHONPATH3};python3'

python(指python2),因为我更经常使用python2。

In case you have to keep PYTHONPATH for both python2 and python3, you can write alias statements to set the proper PYTHONPATH in your bash_profile:

Hardcode your PYTHONPATH2, and PYTHONPATH3 variables in your ~/.bash_profile, and add the following aliases at the end of it:

alias python='export PYTHONPATH=${PYTHONPATH2};python'
alias python3='export PYTHONPATH=${PYTHONPATH3};python3'

My python (refers to python2) as I use python2 more often.


回答 8

每当我遇到这个问题时:

AttributeError:模块“枚举”没有属性“ IntFlag”

只需首先运行命令:

unset PYTHONPATH 

然后运行我想要的命令,然后在其中成功。

When ever I got this problem:

AttributeError: module ‘enum’ has no attribute ‘IntFlag’

simply first i run the command:

unset PYTHONPATH 

and then run my desired command then got success in that.


回答 9

使用pip install <required-library> --ignore-installed enum34
安装所需的库后,在构建过程中查找警告。我收到这样的错误:
Using legacy setup.py install for future, since package 'wheel' is not installed
ERROR: pyejabberd 0.2.11 has requirement enum34==1.1.2, but you'll have enum34 1.1.10 which is incompatible.

要解决此问题,请运行以下命令:pip freeze | grep enum34。这将为您提供已安装的enum34的版本。现在将其卸载,pip uninstall enum34然后重新安装所需的版本
pip install "enum34==1.1.20"

I did by using pip install <required-library> --ignore-installed enum34
Once your required library is installed, look for warnings during the build. I got an Error like this:
Using legacy setup.py install for future, since package 'wheel' is not installed
ERROR: pyejabberd 0.2.11 has requirement enum34==1.1.2, but you'll have enum34 1.1.10 which is incompatible.

To fix this issue now run the command: pip freeze | grep enum34. This will give you the version of the installed enum34. Now uninstall it by pip uninstall enum34 and reinstall the required version as
pip install "enum34==1.1.20"


回答 10

我的计算机上安装了Python 2和Python 3。由于某些奇怪的原因,当调用re模块时,我在Python 3的sys.path中也有一个指向Python2的sitepackage库目录的路径。如果我运行Python 3并导入枚举,print(enum.__file__)并且系统未显示此Python 2站点包路径。因此,一个非常粗糙和肮脏的技巧是直接修改导入枚举的模块(遵循回溯路径)并在导入枚举之前插入以下代码:

import sys
for i, p in enumerate(sys.path):
    if "python27" in p.lower() or "python2.7" in p.lower(): sys.path.pop(i)
import enum

那解决了我的问题。

I have Python 2 and Python 3 installed on my computer. For some strange reason I have in the sys.path of Python 3 also a path to the sitepackage library directory of Python2 when the re module is called. If I run Python 3 and import enum and print(enum.__file__) the system does not show this Python 2 path to site-packages. So a very rough and dirty hack is, to directly modify the module in which enum is imported (follow the traceback paths) and insert the following code just before importing enum:

import sys
for i, p in enumerate(sys.path):
    if "python27" in p.lower() or "python2.7" in p.lower(): sys.path.pop(i)
import enum

That solved my problem.


回答 11

安装enum34的1.1.8版本对我有用。

我能够通过向pyproject.toml添加enum34 =“ == 1.1.8”来解决此问题。显然enum34在v1.1.8中具有避免此错误的功能,但在v1.1.9 +中已退步。不过,这只是一种解决方法。更好的解决方案是使程序包使用环境标记,因此除非需要,您根本不需要安装enum34。

来源:https//github.com/python-poetry/poetry/issues/1122

Installing version 1.1.8 of enum34 worked for me.

I was able to fix this by adding enum34 = “==1.1.8” to pyproject.toml. Apparently enum34 had a feature in v1.1.8 that avoided this error, but this regressed in v1.1.9+. This is just a workaround though. The better solution would be for packages to use environment markers so you don’t have to install enum34 at all unless needed.

Source: https://github.com/python-poetry/poetry/issues/1122


回答 12

即使我在运行python -m grpc_tools.protoc –version时遇到了这个问题,也必须设置PYTHONPATH直到站点包并关闭所有命令提示符窗口,然后它才能工作。希望它对gRPC用户有所帮助。

Even I had this issue while running python -m grpc_tools.protoc –version Had to set the PYTHONPATH till site-packages and shutdown all the command prompt windows and it worked. Hope it helps for gRPC users.


回答 13

我在使用python3.8和tensorflow 2.2.0的虚拟环境内核中的jupyterlab的ubuntu20.04中遇到了这个问题。错误消息原为

 Traceback (most recent call last):
  File "/usr/lib/python2.7/runpy.py", line 174, in _run_module_as_main
    "__main__", fname, loader, pkg_name)
  File "/usr/lib/python2.7/runpy.py", line 72, in _run_code
    exec code in run_globals
  File "/home/hu-mka/.local/lib/python2.7/site-packages/ipykernel_launcher.py", line 15, in <module>
    from ipykernel import kernelapp as app
  File "/home/hu-mka/.local/lib/python2.7/site-packages/ipykernel/__init__.py", line 2, in <module>
    from .connect import *
  File "/home/hu-mka/.local/lib/python2.7/site-packages/ipykernel/connect.py", line 13, in <module>
    from IPython.core.profiledir import ProfileDir
  File "/home/hu-mka/.local/lib/python2.7/site-packages/IPython/__init__.py", line 48, in <module>
    from .core.application import Application
  File "/home/hu-mka/.local/lib/python2.7/site-packages/IPython/core/application.py", line 23, in <module>
    from traitlets.config.application import Application, catch_config_error
  File "/home/hu-mka/.local/lib/python2.7/site-packages/traitlets/__init__.py", line 1, in <module>
    from .traitlets import *
  File "/home/hu-mka/.local/lib/python2.7/site-packages/traitlets/traitlets.py", line 49, in <module>
    import enum
ImportError: No module named enum

问题是在/ usr / bin / python中的符号链接中指向python2。解:

cd /usr/bin/
sudo ln -sf python3 python

我希望python2快死了!马库斯Terveisin

I had this problem in ubuntu20.04 in jupyterlab in my virtual env kernel with python3.8 and tensorflow 2.2.0. Error message was

 Traceback (most recent call last):
  File "/usr/lib/python2.7/runpy.py", line 174, in _run_module_as_main
    "__main__", fname, loader, pkg_name)
  File "/usr/lib/python2.7/runpy.py", line 72, in _run_code
    exec code in run_globals
  File "/home/hu-mka/.local/lib/python2.7/site-packages/ipykernel_launcher.py", line 15, in <module>
    from ipykernel import kernelapp as app
  File "/home/hu-mka/.local/lib/python2.7/site-packages/ipykernel/__init__.py", line 2, in <module>
    from .connect import *
  File "/home/hu-mka/.local/lib/python2.7/site-packages/ipykernel/connect.py", line 13, in <module>
    from IPython.core.profiledir import ProfileDir
  File "/home/hu-mka/.local/lib/python2.7/site-packages/IPython/__init__.py", line 48, in <module>
    from .core.application import Application
  File "/home/hu-mka/.local/lib/python2.7/site-packages/IPython/core/application.py", line 23, in <module>
    from traitlets.config.application import Application, catch_config_error
  File "/home/hu-mka/.local/lib/python2.7/site-packages/traitlets/__init__.py", line 1, in <module>
    from .traitlets import *
  File "/home/hu-mka/.local/lib/python2.7/site-packages/traitlets/traitlets.py", line 49, in <module>
    import enum
ImportError: No module named enum

problem was that in symbolic link in /usr/bin/python was pointing to python2. Solution:

cd /usr/bin/
sudo ln -sf python3 python

Hopefully Python 2 usage will drop off completely soon.


回答 14

如果在运行时有这个问题测试PyCharm,确保第二个框选中的配置。

If you having this issue when running tests in PyCharm, make sure the second box is unchecked in the configurations.


回答 15

如果有人有试图从virtualenv中运行Jupyter内核的时候这个问题,只需添加正确的PYTHONPATHkernel.json您的virtualenv内核(Python 3的例子)的:

{
 "argv": [
  "/usr/local/Cellar/python/3.6.5/bin/python3.6",
  "-m",
  "ipykernel_launcher",
  "-f",
  "{connection_file}"
 ],
 "display_name": "Python 3 (TensorFlow)",
 "language": "python",
 "env": {
     "PYTHONPATH":     "/Users/dimitrijer/git/mlai/.venv/lib/python3.6:/Users/dimitrijer/git/mlai/.venv/lib/python3.6/lib-dynload:/usr/local/Cellar/python/3.6.5/Frameworks/Python.framework/Versions/3.6/lib/python3.6:/Users/dimitrijer/git/mlai/.venv/lib/python3.6/site-packages"
}
}

If anyone is having this problem when trying to run Jupyter kernel from a virtualenv, just add correct PYTHONPATH to kernel.json of your virtualenv kernel (Python 3 in example):

{
 "argv": [
  "/usr/local/Cellar/python/3.6.5/bin/python3.6",
  "-m",
  "ipykernel_launcher",
  "-f",
  "{connection_file}"
 ],
 "display_name": "Python 3 (TensorFlow)",
 "language": "python",
 "env": {
     "PYTHONPATH":     "/Users/dimitrijer/git/mlai/.venv/lib/python3.6:/Users/dimitrijer/git/mlai/.venv/lib/python3.6/lib-dynload:/usr/local/Cellar/python/3.6.5/Frameworks/Python.framework/Versions/3.6/lib/python3.6:/Users/dimitrijer/git/mlai/.venv/lib/python3.6/site-packages"
}
}

numpy dot()和Python 3.5+矩阵乘法@之间的区别

问题:numpy dot()和Python 3.5+矩阵乘法@之间的区别

我最近使用Python 3.5,注意到新的矩阵乘法运算符(@)有时与numpy点运算符的行为有所不同。例如,对于3d阵列:

import numpy as np

a = np.random.rand(8,13,13)
b = np.random.rand(8,13,13)
c = a @ b  # Python 3.5+
d = np.dot(a, b)

@运算符返回形状的阵列:

c.shape
(8, 13, 13)

np.dot()函数返回时:

d.shape
(8, 13, 8, 13)

如何用numpy点重现相同的结果?还有其他重大区别吗?

I recently moved to Python 3.5 and noticed the new matrix multiplication operator (@) sometimes behaves differently from the numpy dot operator. In example, for 3d arrays:

import numpy as np

a = np.random.rand(8,13,13)
b = np.random.rand(8,13,13)
c = a @ b  # Python 3.5+
d = np.dot(a, b)

The @ operator returns an array of shape:

c.shape
(8, 13, 13)

while the np.dot() function returns:

d.shape
(8, 13, 8, 13)

How can I reproduce the same result with numpy dot? Are there any other significant differences?


回答 0

@运营商称阵列的__matmul__方法,而不是dot。此方法在API中也作为函数存在np.matmul

>>> a = np.random.rand(8,13,13)
>>> b = np.random.rand(8,13,13)
>>> np.matmul(a, b).shape
(8, 13, 13)

从文档中:

matmul区别在于dot两个重要方面。

  • 标量不能相乘。
  • 将矩阵堆栈一起广播,就好像矩阵是元素一样。

最后一点很清楚,当传递3D(或更高维)数组时,dotmatmul方法的行为会有所不同。从文档中引用更多内容:

对于matmul

如果任何一个参数为ND,N> 2,则将其视为驻留在最后两个索引中的一组矩阵,并进行相应广播。

对于np.dot

对于2-D数组,它等效于矩阵乘法,对于1-D数组,其等效于向量的内积(无复共轭)。对于N维,它是a的最后一个轴和b的倒数第二个轴的和积

The @ operator calls the array’s __matmul__ method, not dot. This method is also present in the API as the function np.matmul.

>>> a = np.random.rand(8,13,13)
>>> b = np.random.rand(8,13,13)
>>> np.matmul(a, b).shape
(8, 13, 13)

From the documentation:

matmul differs from dot in two important ways.

  • Multiplication by scalars is not allowed.
  • Stacks of matrices are broadcast together as if the matrices were elements.

The last point makes it clear that dot and matmul methods behave differently when passed 3D (or higher dimensional) arrays. Quoting from the documentation some more:

For matmul:

If either argument is N-D, N > 2, it is treated as a stack of matrices residing in the last two indexes and broadcast accordingly.

For np.dot:

For 2-D arrays it is equivalent to matrix multiplication, and for 1-D arrays to inner product of vectors (without complex conjugation). For N dimensions it is a sum product over the last axis of a and the second-to-last of b


回答 1

@ajcr的答案说明了dotand matmul(由@符号调用)之间的区别。通过看一个简单的例子,可以清楚地看到两者在“矩阵堆栈”或张量上进行操作时的行为有何不同。

为了弄清差异,采用4×4数组,然后将dot乘积和matmul乘积返回3x4x2的“矩阵堆栈”或张量。

import numpy as np
fourbyfour = np.array([
                       [1,2,3,4],
                       [3,2,1,4],
                       [5,4,6,7],
                       [11,12,13,14]
                      ])


threebyfourbytwo = np.array([
                             [[2,3],[11,9],[32,21],[28,17]],
                             [[2,3],[1,9],[3,21],[28,7]],
                             [[2,3],[1,9],[3,21],[28,7]],
                            ])

print('4x4*3x4x2 dot:\n {}\n'.format(np.dot(fourbyfour,twobyfourbythree)))
print('4x4*3x4x2 matmul:\n {}\n'.format(np.matmul(fourbyfour,twobyfourbythree)))

每个操作的结果如下所示。注意点积如何

… a的最后一个轴与b的倒数第二个和的乘积

以及如何通过一起广播矩阵来形成矩阵乘积。

4x4*3x4x2 dot:
 [[[232 152]
  [125 112]
  [125 112]]

 [[172 116]
  [123  76]
  [123  76]]

 [[442 296]
  [228 226]
  [228 226]]

 [[962 652]
  [465 512]
  [465 512]]]

4x4*3x4x2 matmul:
 [[[232 152]
  [172 116]
  [442 296]
  [962 652]]

 [[125 112]
  [123  76]
  [228 226]
  [465 512]]

 [[125 112]
  [123  76]
  [228 226]
  [465 512]]]

The answer by @ajcr explains how the dot and matmul (invoked by the @ symbol) differ. By looking at a simple example, one clearly sees how the two behave differently when operating on ‘stacks of matricies’ or tensors.

To clarify the differences take a 4×4 array and return the dot product and matmul product with a 3x4x2 ‘stack of matricies’ or tensor.

import numpy as np
fourbyfour = np.array([
                       [1,2,3,4],
                       [3,2,1,4],
                       [5,4,6,7],
                       [11,12,13,14]
                      ])


threebyfourbytwo = np.array([
                             [[2,3],[11,9],[32,21],[28,17]],
                             [[2,3],[1,9],[3,21],[28,7]],
                             [[2,3],[1,9],[3,21],[28,7]],
                            ])

print('4x4*3x4x2 dot:\n {}\n'.format(np.dot(fourbyfour,threebyfourbytwo)))
print('4x4*3x4x2 matmul:\n {}\n'.format(np.matmul(fourbyfour,threebyfourbytwo)))

The products of each operation appear below. Notice how the dot product is,

…a sum product over the last axis of a and the second-to-last of b

and how the matrix product is formed by broadcasting the matrix together.

4x4*3x4x2 dot:
 [[[232 152]
  [125 112]
  [125 112]]

 [[172 116]
  [123  76]
  [123  76]]

 [[442 296]
  [228 226]
  [228 226]]

 [[962 652]
  [465 512]
  [465 512]]]

4x4*3x4x2 matmul:
 [[[232 152]
  [172 116]
  [442 296]
  [962 652]]

 [[125 112]
  [123  76]
  [228 226]
  [465 512]]

 [[125 112]
  [123  76]
  [228 226]
  [465 512]]]

回答 2

仅供参考,@其numpy的等价物dot,并matmul都大致一样快。(用我的一个项目perfplot创建的图。)

复制剧情的代码:

import perfplot
import numpy


def setup(n):
    A = numpy.random.rand(n, n)
    x = numpy.random.rand(n)
    return A, x


def at(data):
    A, x = data
    return A @ x


def numpy_dot(data):
    A, x = data
    return numpy.dot(A, x)


def numpy_matmul(data):
    A, x = data
    return numpy.matmul(A, x)


perfplot.show(
    setup=setup,
    kernels=[at, numpy_dot, numpy_matmul],
    n_range=[2 ** k for k in range(12)],
    logx=True,
    logy=True,
)

Just FYI, @ and its numpy equivalents dot and matmul are all equally fast. (Plot created with perfplot, a project of mine.)

Code to reproduce the plot:

import perfplot
import numpy


def setup(n):
    A = numpy.random.rand(n, n)
    x = numpy.random.rand(n)
    return A, x


def at(data):
    A, x = data
    return A @ x


def numpy_dot(data):
    A, x = data
    return numpy.dot(A, x)


def numpy_matmul(data):
    A, x = data
    return numpy.matmul(A, x)


perfplot.show(
    setup=setup,
    kernels=[at, numpy_dot, numpy_matmul],
    n_range=[2 ** k for k in range(15)],
)

回答 3

在数学上,我认为numpy中的更有意义

(a,b)_ {i,j,k,a,b,c} =

因为当a和b是向量时它给出点积,或者当a和b是矩阵时给出矩阵乘积


对于numpy中的matmul操作,它由结果的一部分组成,可以定义为

> matmul(a,b)_ {i,j,k,c} =

因此,您可以看到matmul(a,b)返回的数组形状较小,从而减少了内存消耗,并在应用程序中更有意义。特别是结合广播,您可以获得

matmul(a,b)_ {i,j,k,l} =

例如。


从以上两个定义中,您可以看到使用这两个操作的要求。假设a.shape =(s1,s2,s3,s4)b.shape =(t1,t2,t3,t4)

  • 要使用点(a,b),您需要

    1. t3 = s4 ;
  • 要使用matmul(a,b),您需要

    1. t3 = s4
    2. t2 = s2或t2和s2之一为1
    3. t1 = s1或t1和s1之一为1

使用以下代码说服自己。

代码样例

import numpy as np
for it in xrange(10000):
    a = np.random.rand(5,6,2,4)
    b = np.random.rand(6,4,3)
    c = np.matmul(a,b)
    d = np.dot(a,b)
    #print 'c shape: ', c.shape,'d shape:', d.shape

    for i in range(5):
        for j in range(6):
            for k in range(2):
                for l in range(3):
                    if not c[i,j,k,l] == d[i,j,k,j,l]:
                        print it,i,j,k,l,c[i,j,k,l]==d[i,j,k,j,l] #you will not see them

In mathematics, I think the dot in numpy makes more sense

dot(a,b)_{i,j,k,a,b,c} =

since it gives the dot product when a and b are vectors, or the matrix multiplication when a and b are matrices


As for matmul operation in numpy, it consists of parts of dot result, and it can be defined as

>matmul(a,b)_{i,j,k,c} =

So, you can see that matmul(a,b) returns an array with a small shape, which has smaller memory consumption and make more sense in applications. In particular, combining with broadcasting, you can get

matmul(a,b)_{i,j,k,l} =

for example.


From the above two definitions, you can see the requirements to use those two operations. Assume a.shape=(s1,s2,s3,s4) and b.shape=(t1,t2,t3,t4)

  • To use dot(a,b) you need

    1. t3=s4;
  • To use matmul(a,b) you need

    1. t3=s4
    2. t2=s2, or one of t2 and s2 is 1
    3. t1=s1, or one of t1 and s1 is 1

Use the following piece of code to convince yourself.

Code sample

import numpy as np
for it in xrange(10000):
    a = np.random.rand(5,6,2,4)
    b = np.random.rand(6,4,3)
    c = np.matmul(a,b)
    d = np.dot(a,b)
    #print 'c shape: ', c.shape,'d shape:', d.shape

    for i in range(5):
        for j in range(6):
            for k in range(2):
                for l in range(3):
                    if not c[i,j,k,l] == d[i,j,k,j,l]:
                        print it,i,j,k,l,c[i,j,k,l]==d[i,j,k,j,l] #you will not see them

回答 4

这是与的比较,np.einsum以显示索引的投影方式

np.allclose(np.einsum('ijk,ijk->ijk', a,b), a*b)        # True 
np.allclose(np.einsum('ijk,ikl->ijl', a,b), a@b)        # True
np.allclose(np.einsum('ijk,lkm->ijlm',a,b), a.dot(b))   # True

Here is a comparison with np.einsum to show how the indices are projected

np.allclose(np.einsum('ijk,ijk->ijk', a,b), a*b)        # True 
np.allclose(np.einsum('ijk,ikl->ijl', a,b), a@b)        # True
np.allclose(np.einsum('ijk,lkm->ijlm',a,b), a.dot(b))   # True

回答 5

我对MATMUL和DOT的经验

尝试使用MATMUL时,我经常收到“ ValueError:传递的值的形状为(200,1),索引表示(200,3)”。我想要一个快速的解决方法,并发现DOT可以提供相同的功能。使用DOT我没有任何错误。我得到正确的答案

与MATMUL

X.shape
>>>(200, 3)

type(X)

>>>pandas.core.frame.DataFrame

w

>>>array([0.37454012, 0.95071431, 0.73199394])

YY = np.matmul(X,w)

>>>  ValueError: Shape of passed values is (200, 1), indices imply (200, 3)"

与DOT

YY = np.dot(X,w)
# no error message
YY
>>>array([ 2.59206877,  1.06842193,  2.18533396,  2.11366346,  0.28505879, 

YY.shape

>>> (200, )

My experience with MATMUL and DOT

I was constantly getting “ValueError: Shape of passed values is (200, 1), indices imply (200, 3)” when trying to use MATMUL. I wanted a quick workaround and found DOT to deliver the same functionality. I don’t get any error using DOT. I get the correct answer

with MATMUL

X.shape
>>>(200, 3)

type(X)

>>>pandas.core.frame.DataFrame

w

>>>array([0.37454012, 0.95071431, 0.73199394])

YY = np.matmul(X,w)

>>>  ValueError: Shape of passed values is (200, 1), indices imply (200, 3)"

with DOT

YY = np.dot(X,w)
# no error message
YY
>>>array([ 2.59206877,  1.06842193,  2.18533396,  2.11366346,  0.28505879, …

YY.shape

>>> (200, )

为什么在Python中“ 0,0 ==(0,0)”等于“(0,False)”?

问题:为什么在Python中“ 0,0 ==(0,0)”等于“(0,False)”?

在Python中(我仅使用Python 3.6进行过检查,但我相信它也适用于许多以前的版本):

(0, 0) == 0, 0   # results in a two element tuple: (False, 0)
0, 0 == (0, 0)   # results in a two element tuple: (0, False)
(0, 0) == (0, 0) # results in a boolean True

但:

a = 0, 0
b = (0, 0)
a == b # results in a boolean True

为什么两种方法的结果不同?相等运算符是否以不同的方式处理元组?

In Python (I checked only with Python 3.6 but I believe it should hold for many of the previous versions as well):

(0, 0) == 0, 0   # results in a two element tuple: (False, 0)
0, 0 == (0, 0)   # results in a two element tuple: (0, False)
(0, 0) == (0, 0) # results in a boolean True

But:

a = 0, 0
b = (0, 0)
a == b # results in a boolean True

Why does the result differ between the two approaches? Does the equality operator handle tuples differently?


回答 0

前两个表达式都解析为元组:

  1. (0, 0) == 0(即False),然后是0
  2. 0,其次是0 == (0, 0)(仍然False如此)。

由于逗号分隔符相对于相等运算符具有较高的优先级,因此将表达式进行拆分:Python看到一个元组包含两个表达式,其中一个恰好是一个相等测试,而不是两个元组之间的相等测试。

但是在第二组语句中,a = 0, 0 不能是元组。元组是值的集合,与相等性测试不同,赋值在Python中没有值。赋值不是表达式,而是语句。它没有可包含在元组或任何其他周围表达式中的值。如果您尝试执行类似(a = 0), 0的操作以强制将其解释为元组,则会出现语法错误。这样,将元组分配给变量(可以通过编写使其更明确)a = (0, 0)作为对的唯一有效解释a = 0, 0

因此,即使没有对的括号a,它和都b获得了赋值(0,0),因此a == b也是如此True

The first two expressions both parse as tuples:

  1. (0, 0) == 0 (which is False), followed by 0
  2. 0, followed by 0 == (0, 0) (which is still False that way around).

The expressions are split that way because of the relative precedence of the comma separator compared to the equality operator: Python sees a tuple containing two expressions, one of which happens to be an equality test, instead of an equality test between two tuples.

But in your second set of statements, a = 0, 0 cannot be a tuple. A tuple is a collection of values, and unlike an equality test, assignment has no value in Python. An assignment is not an expression, but a statement; it does not have a value that can be included into a tuple or any other surrounding expression. If you tried something like (a = 0), 0 in order to force interpretation as a tuple, you would get a syntax error. That leaves the assignment of a tuple to a variable – which could be made more explicit by writing it a = (0, 0) – as the only valid interpretation of a = 0, 0.

So even without the parentheses on the assignment to a, both it and b get assigned the value (0,0), so a == b is therefore True.


回答 1

您在所有3种情况下看到的都是语言语法规范的结果,以及如何解析源代码中遇到的标记以生成解析树。

看看这个低级代码应该可以帮助您了解幕后情况。我们可以使用以下python语句,将它们转换为字节码,然后使用以下dis模块对其进行反编译:

情况1: (0, 0) == 0, 0

>>> dis.dis(compile("(0, 0) == 0, 0", '', 'exec'))
  1           0 LOAD_CONST               2 ((0, 0))
              3 LOAD_CONST               0 (0)
              6 COMPARE_OP               2 (==)
              9 LOAD_CONST               0 (0)
             12 BUILD_TUPLE              2
             15 POP_TOP
             16 LOAD_CONST               1 (None)
             19 RETURN_VALUE

(0, 0)首先与0第一个进行比较,然后对进行评估False。然后,使用此结果构造一个元组,最后0得到结果(False, 0)

情况2: 0, 0 == (0, 0)

>>> dis.dis(compile("0, 0 == (0, 0)", '', 'exec'))
  1           0 LOAD_CONST               0 (0)
              3 LOAD_CONST               0 (0)
              6 LOAD_CONST               2 ((0, 0))
              9 COMPARE_OP               2 (==)
             12 BUILD_TUPLE              2
             15 POP_TOP
             16 LOAD_CONST               1 (None)
             19 RETURN_VALUE

将元组构造0为第一个元素。对于第二个元素,执行与第一种情况相同的检查,并求值为False,因此得到(0, False)

情况3: (0, 0) == (0, 0)

>>> dis.dis(compile("(0, 0) == (0, 0)", '', 'exec'))
  1           0 LOAD_CONST               2 ((0, 0))
              3 LOAD_CONST               3 ((0, 0))
              6 COMPARE_OP               2 (==)
              9 POP_TOP
             10 LOAD_CONST               1 (None)
             13 RETURN_VALUE

如您所见,在这里,您只是比较这两个(0, 0)元组并返回True

What you see in all 3 instances is a consequence of the grammar specification of the language, and how tokens encountered in the source code are parsed to generate the parse tree.

Taking a look at this low level code should help you understand what happens under the hood. We can take these python statements, convert them into byte code and then decompile them using the dis module:

Case 1: (0, 0) == 0, 0

>>> dis.dis(compile("(0, 0) == 0, 0", '', 'exec'))
  1           0 LOAD_CONST               2 ((0, 0))
              3 LOAD_CONST               0 (0)
              6 COMPARE_OP               2 (==)
              9 LOAD_CONST               0 (0)
             12 BUILD_TUPLE              2
             15 POP_TOP
             16 LOAD_CONST               1 (None)
             19 RETURN_VALUE

(0, 0) is first compared to 0 first and evaluated to False. A tuple is then constructed with this result and last 0, so you get (False, 0).

Case 2: 0, 0 == (0, 0)

>>> dis.dis(compile("0, 0 == (0, 0)", '', 'exec'))
  1           0 LOAD_CONST               0 (0)
              3 LOAD_CONST               0 (0)
              6 LOAD_CONST               2 ((0, 0))
              9 COMPARE_OP               2 (==)
             12 BUILD_TUPLE              2
             15 POP_TOP
             16 LOAD_CONST               1 (None)
             19 RETURN_VALUE

A tuple is constructed with 0 as the first element. For the second element, the same check is done as in the first case and evaluated to False, so you get (0, False).

Case 3: (0, 0) == (0, 0)

>>> dis.dis(compile("(0, 0) == (0, 0)", '', 'exec'))
  1           0 LOAD_CONST               2 ((0, 0))
              3 LOAD_CONST               3 ((0, 0))
              6 COMPARE_OP               2 (==)
              9 POP_TOP
             10 LOAD_CONST               1 (None)
             13 RETURN_VALUE

Here, as you see, you’re just comparing those two (0, 0) tuples and returning True.


回答 2

解释问题的另一种方法:您可能熟悉字典文字

{ "a": 1, "b": 2, "c": 3 }

和数组文字

[ "a", "b", "c" ]

和元组文字

( 1, 2, 3 )

但是您没有意识到的是,与字典和数组文字不同,您通常在元组文字周围看到的括号不是文字语法的一部分。元组的文字语法只是一系列用逗号分隔的表达式:

1, 2, 3

Python形式语法语言中的“ exprlist” )。

现在,您期望数组文字是什么

[ 0, 0 == (0, 0) ]

评价?看起来可能更像是应该

[ 0, (0 == (0, 0)) ]

当然是哪个[0, False]。类似地,使用显式带括号的元组文字

( 0, 0 == (0, 0) )

得到并不奇怪(0, False)。但是括号是可选的。

0, 0 == (0, 0)

是同一回事。这就是为什么你得到(0, False)


如果您想知道为什么元组文字周围的括号是可选的,那主要是因为必须以这种方式编写解构分配会很烦人:

(a, b) = (c, d) # meh
a, b = c, d     # better

Another way to explain the problem: You’re probably familiar with dictionary literals

{ "a": 1, "b": 2, "c": 3 }

and array literals

[ "a", "b", "c" ]

and tuple literals

( 1, 2, 3 )

but what you don’t realize is that, unlike dictionary and array literals, the parentheses you usually see around a tuple literal are not part of the literal syntax. The literal syntax for tuples is just a sequence of expressions separated by commas:

1, 2, 3

(an “exprlist” in the language of the formal grammar for Python).

Now, what do you expect the array literal

[ 0, 0 == (0, 0) ]

to evaluate to? That probably looks a lot more like it should be the same as

[ 0, (0 == (0, 0)) ]

which of course evaluates to [0, False]. Similarly, with an explicitly parenthesized tuple literal

( 0, 0 == (0, 0) )

it’s not surprising to get (0, False). But the parentheses are optional;

0, 0 == (0, 0)

is the same thing. And that’s why you get (0, False).


If you’re wondering why the parentheses around a tuple literal are optional, it is largely because it would be annoying to have to write destructuring assignments that way:

(a, b) = (c, d) # meh
a, b = c, d     # better

回答 3

在执行操作的顺序周围加上几个括号可以帮助您更好地理解结果:

# Build two element tuple comprising of 
# (0, 0) == 0 result and 0
>>> ((0, 0) == 0), 0
(False, 0)

# Build two element tuple comprising of
# 0 and result of (0, 0) == 0 
>>> 0, (0 == (0, 0))
(0, False)

# Create two tuples with elements (0, 0) 
# and compare them
>>> (0, 0) == (0, 0) 
True

逗号用于分隔表达式(使用括号,我们可以强制使用不同的行为)。查看您列出的代码段时,逗号,会将其分开并定义将对哪些表达式求值:

(0, 0) == 0 ,   0
#-----------|------
  expr 1      expr2

元组(0, 0)也可以类似的方式分解。逗号分隔两个包含文字的表达式0

Adding a couple of parentheses around the order in which actions are performed might help you understand the results better:

# Build two element tuple comprising of 
# (0, 0) == 0 result and 0
>>> ((0, 0) == 0), 0
(False, 0)

# Build two element tuple comprising of
# 0 and result of (0, 0) == 0 
>>> 0, (0 == (0, 0))
(0, False)

# Create two tuples with elements (0, 0) 
# and compare them
>>> (0, 0) == (0, 0) 
True

The comma is used to to separate expressions (using parentheses we can force different behavior, of course). When viewing the snippets you listed, the comma , will separate it and define what expressions will get evaluated:

(0, 0) == 0 ,   0
#-----------|------
  expr 1      expr2

The tuple (0, 0) can also be broken down in a similar way. The comma separates two expressions comprising of the literals 0.


回答 4

在第一个示例中,Python将两件事变成元组:

  1. 表达式(0, 0) == 0,其计算结果为False
  2. 常数 0

在第二个方面则相反。

In the first one Python is making a tuple of two things:

  1. The expression (0, 0) == 0, which evaluates to False
  2. The constant 0

In the second one it’s the other way around.


回答 5

看这个例子:

r = [1,0,1,0,1,1,0,0,0,1]
print(r==0,0,r,1,0)
print(r==r,0,1,0,1,0)

然后结果:

False 0 [1, 0, 1, 0, 1, 1, 0, 0, 0, 1] 1 0
True 0 1 0 1 0

然后比较就和示例中的第一个数字(0和r)进行比较。

look at this example:

r = [1,0,1,0,1,1,0,0,0,1]
print(r==0,0,r,1,0)
print(r==r,0,1,0,1,0)

then result:

False 0 [1, 0, 1, 0, 1, 1, 0, 0, 0, 1] 1 0
True 0 1 0 1 0

then comparison just does to the first number(0 and r) in the example.


Python结构的内存大小

问题:Python结构的内存大小

是否有关于32位和64位平台上Python数据结构的内存大小的参考?

如果没有,那么将其放在SO上会很好。越详尽越好!那么以下Python结构使用了多少字节(取决于相关时的len和内容类型)?

  • int
  • float
  • 参考
  • str
  • unicode字符串
  • tuple
  • list
  • dict
  • set
  • array.array
  • numpy.array
  • deque
  • 新型类对象
  • 旧式类对象
  • …以及我忘记的一切!

(对于仅保留对其他对象的引用的容器,我们显然不希望自己计算项目的大小,因为它可能是共享的。)

此外,是否有一种方法可以获取对象在运行时使用的内存(递归与否)?

Is there a reference for the memory size of Python data stucture on 32- and 64-bit platforms?

If not, this would be nice to have it on SO. The more exhaustive the better! So how many bytes are used by the following Python structures (depending on the len and the content type when relevant)?

  • int
  • float
  • reference
  • str
  • unicode string
  • tuple
  • list
  • dict
  • set
  • array.array
  • numpy.array
  • deque
  • new-style classes object
  • old-style classes object
  • … and everything I am forgetting!

(For containers that keep only references to other objects, we obviously do not want to count the size of the item themselves, since it might be shared.)

Furthermore, is there a way to get the memory used by an object at runtime (recursively or not)?


回答 0

对此先前的问题提出的建议是使用sys.getsizeof(),并引用:

>>> import sys
>>> x = 2
>>> sys.getsizeof(x)
14
>>> sys.getsizeof(sys.getsizeof)
32
>>> sys.getsizeof('this')
38
>>> sys.getsizeof('this also')
48

您可以采用这种方法:

>>> import sys
>>> import decimal
>>> 
>>> d = {
...     "int": 0,
...     "float": 0.0,
...     "dict": dict(),
...     "set": set(),
...     "tuple": tuple(),
...     "list": list(),
...     "str": "a",
...     "unicode": u"a",
...     "decimal": decimal.Decimal(0),
...     "object": object(),
... }
>>> for k, v in sorted(d.iteritems()):
...     print k, sys.getsizeof(v)
...
decimal 40
dict 140
float 16
int 12
list 36
object 8
set 116
str 25
tuple 28
unicode 28

2012-09-30

python 2.7(Linux,32位):

decimal 36
dict 136
float 16
int 12
list 32
object 8
set 112
str 22
tuple 24
unicode 32

python 3.3(Linux,32位)

decimal 52
dict 144
float 16
int 14
list 32
object 8
set 112
str 26
tuple 24
unicode 26

2016-08-01

OSX,Python 2.7.10(默认,2015年10月23日,19:19:21)[darwin上的[GCC 4.2.1兼容Apple LLVM 7.0.0(clang-700.0.59.5)]

decimal 80
dict 280
float 24
int 24
list 72
object 16
set 232
str 38
tuple 56
unicode 52

The recommendation from an earlier question on this was to use sys.getsizeof(), quoting:

>>> import sys
>>> x = 2
>>> sys.getsizeof(x)
14
>>> sys.getsizeof(sys.getsizeof)
32
>>> sys.getsizeof('this')
38
>>> sys.getsizeof('this also')
48

You could take this approach:

>>> import sys
>>> import decimal
>>> 
>>> d = {
...     "int": 0,
...     "float": 0.0,
...     "dict": dict(),
...     "set": set(),
...     "tuple": tuple(),
...     "list": list(),
...     "str": "a",
...     "unicode": u"a",
...     "decimal": decimal.Decimal(0),
...     "object": object(),
... }
>>> for k, v in sorted(d.iteritems()):
...     print k, sys.getsizeof(v)
...
decimal 40
dict 140
float 16
int 12
list 36
object 8
set 116
str 25
tuple 28
unicode 28

2012-09-30

python 2.7 (linux, 32-bit):

decimal 36
dict 136
float 16
int 12
list 32
object 8
set 112
str 22
tuple 24
unicode 32

python 3.3 (linux, 32-bit)

decimal 52
dict 144
float 16
int 14
list 32
object 8
set 112
str 26
tuple 24
unicode 26

2016-08-01

OSX, Python 2.7.10 (default, Oct 23 2015, 19:19:21) [GCC 4.2.1 Compatible Apple LLVM 7.0.0 (clang-700.0.59.5)] on darwin

decimal 80
dict 280
float 24
int 24
list 72
object 16
set 232
str 38
tuple 56
unicode 52

回答 1

我一直很高兴地将pympler用于此类任务。它与许多版本的Python兼容- asizeof特别是该模块可以追溯到2.2!

例如,使用hughdbrown的示例,但from pympler import asizeof在开头和print asizeof.asizeof(v)结尾处都看到了(MacOSX 10.5上的系统Python 2.5):

$ python pymp.py 
set 120
unicode 32
tuple 32
int 16
decimal 152
float 16
list 40
object 0
dict 144
str 32

显然这里有一些近似值,但是我发现它对于足迹分析和调整非常有用。

I’ve been happily using pympler for such tasks. It’s compatible with many versions of Python — the asizeof module in particular goes back to 2.2!

For example, using hughdbrown’s example but with from pympler import asizeof at the start and print asizeof.asizeof(v) at the end, I see (system Python 2.5 on MacOSX 10.5):

$ python pymp.py 
set 120
unicode 32
tuple 32
int 16
decimal 152
float 16
list 40
object 0
dict 144
str 32

Clearly there is some approximation here, but I’ve found it very useful for footprint analysis and tuning.


回答 2

这些答案都收集浅层尺寸信息。我怀疑访问此问题的访客最终将在这里回答以下问题:“此复杂对象在内存中有多大?”

这里有一个很好的答案:https : //goshippo.com/blog/measure-real-size-any-python-object/

重点:

import sys

def get_size(obj, seen=None):
    """Recursively finds size of objects"""
    size = sys.getsizeof(obj)
    if seen is None:
        seen = set()
    obj_id = id(obj)
    if obj_id in seen:
        return 0
    # Important mark as seen *before* entering recursion to gracefully handle
    # self-referential objects
    seen.add(obj_id)
    if isinstance(obj, dict):
        size += sum([get_size(v, seen) for v in obj.values()])
        size += sum([get_size(k, seen) for k in obj.keys()])
    elif hasattr(obj, '__dict__'):
        size += get_size(obj.__dict__, seen)
    elif hasattr(obj, '__iter__') and not isinstance(obj, (str, bytes, bytearray)):
        size += sum([get_size(i, seen) for i in obj])
    return size

像这样使用:

In [1]: get_size(1)
Out[1]: 24

In [2]: get_size([1])
Out[2]: 104

In [3]: get_size([[1]])
Out[3]: 184

如果您想更深入地了解Python的内存模型,这里有一篇很棒的文章,其中有类似的“总大小”代码段,作为较长说明的一部分:https : //code.tutsplus.com/tutorials/understand-how-您的Python记忆体大量使用–CMS-25609

These answers all collect shallow size information. I suspect that visitors to this question will end up here looking to answer the question, “How big is this complex object in memory?”

There’s a great answer here: https://goshippo.com/blog/measure-real-size-any-python-object/

The punchline:

import sys

def get_size(obj, seen=None):
    """Recursively finds size of objects"""
    size = sys.getsizeof(obj)
    if seen is None:
        seen = set()
    obj_id = id(obj)
    if obj_id in seen:
        return 0
    # Important mark as seen *before* entering recursion to gracefully handle
    # self-referential objects
    seen.add(obj_id)
    if isinstance(obj, dict):
        size += sum([get_size(v, seen) for v in obj.values()])
        size += sum([get_size(k, seen) for k in obj.keys()])
    elif hasattr(obj, '__dict__'):
        size += get_size(obj.__dict__, seen)
    elif hasattr(obj, '__iter__') and not isinstance(obj, (str, bytes, bytearray)):
        size += sum([get_size(i, seen) for i in obj])
    return size

Used like so:

In [1]: get_size(1)
Out[1]: 24

In [2]: get_size([1])
Out[2]: 104

In [3]: get_size([[1]])
Out[3]: 184

If you want to know Python’s memory model more deeply, there’s a great article here that has a similar “total size” snippet of code as part of a longer explanation: https://code.tutsplus.com/tutorials/understand-how-much-memory-your-python-objects-use–cms-25609


回答 3

尝试使用内存探查器。 内存分析器

Line #    Mem usage  Increment   Line Contents
==============================================
     3                           @profile
     4      5.97 MB    0.00 MB   def my_func():
     5     13.61 MB    7.64 MB       a = [1] * (10 ** 6)
     6    166.20 MB  152.59 MB       b = [2] * (2 * 10 ** 7)
     7     13.61 MB -152.59 MB       del b
     8     13.61 MB    0.00 MB       return a

Try memory profiler. memory profiler

Line #    Mem usage  Increment   Line Contents
==============================================
     3                           @profile
     4      5.97 MB    0.00 MB   def my_func():
     5     13.61 MB    7.64 MB       a = [1] * (10 ** 6)
     6    166.20 MB  152.59 MB       b = [2] * (2 * 10 ** 7)
     7     13.61 MB -152.59 MB       del b
     8     13.61 MB    0.00 MB       return a

回答 4

您也可以使用guppy模块。

>>> from guppy import hpy; hp=hpy()
>>> hp.heap()
Partition of a set of 25853 objects. Total size = 3320992 bytes.
 Index  Count   %     Size   % Cumulative  % Kind (class / dict of class)
     0  11731  45   929072  28    929072  28 str
     1   5832  23   469760  14   1398832  42 tuple
     2    324   1   277728   8   1676560  50 dict (no owner)
     3     70   0   216976   7   1893536  57 dict of module
     4    199   1   210856   6   2104392  63 dict of type
     5   1627   6   208256   6   2312648  70 types.CodeType
     6   1592   6   191040   6   2503688  75 function
     7    199   1   177008   5   2680696  81 type
     8    124   0   135328   4   2816024  85 dict of class
     9   1045   4    83600   3   2899624  87 __builtin__.wrapper_descriptor
<90 more rows. Type e.g. '_.more' to view.>

和:

>>> hp.iso(1, [1], "1", (1,), {1:1}, None)
Partition of a set of 6 objects. Total size = 560 bytes.
 Index  Count   %     Size   % Cumulative  % Kind (class / dict of class)
     0      1  17      280  50       280  50 dict (no owner)
     1      1  17      136  24       416  74 list
     2      1  17       64  11       480  86 tuple
     3      1  17       40   7       520  93 str
     4      1  17       24   4       544  97 int
     5      1  17       16   3       560 100 types.NoneType

Also you can use guppy module.

>>> from guppy import hpy; hp=hpy()
>>> hp.heap()
Partition of a set of 25853 objects. Total size = 3320992 bytes.
 Index  Count   %     Size   % Cumulative  % Kind (class / dict of class)
     0  11731  45   929072  28    929072  28 str
     1   5832  23   469760  14   1398832  42 tuple
     2    324   1   277728   8   1676560  50 dict (no owner)
     3     70   0   216976   7   1893536  57 dict of module
     4    199   1   210856   6   2104392  63 dict of type
     5   1627   6   208256   6   2312648  70 types.CodeType
     6   1592   6   191040   6   2503688  75 function
     7    199   1   177008   5   2680696  81 type
     8    124   0   135328   4   2816024  85 dict of class
     9   1045   4    83600   3   2899624  87 __builtin__.wrapper_descriptor
<90 more rows. Type e.g. '_.more' to view.>

And:

>>> hp.iso(1, [1], "1", (1,), {1:1}, None)
Partition of a set of 6 objects. Total size = 560 bytes.
 Index  Count   %     Size   % Cumulative  % Kind (class / dict of class)
     0      1  17      280  50       280  50 dict (no owner)
     1      1  17      136  24       416  74 list
     2      1  17       64  11       480  86 tuple
     3      1  17       40   7       520  93 str
     4      1  17       24   4       544  97 int
     5      1  17       16   3       560 100 types.NoneType

回答 5

也可以使用tracemallocPython标准库中的模块。对于类是用C实现的对象来说,它似乎工作得很好(例如,与Pympler不同)。

One can also make use of the tracemalloc module from the Python standard library. It seems to work well for objects whose class is implemented in C (unlike Pympler, for instance).


回答 6

使用dir([object])内置功能时,可以获得__sizeof__内置功能的。

>>> a = -1
>>> a.__sizeof__()
24

When you use the dir([object]) built-in function, you can get the __sizeof__ of the built-in function.

>>> a = -1
>>> a.__sizeof__()
24

使用SMTP从Python发送邮件

问题:使用SMTP从Python发送邮件

我正在使用以下方法使用SMTP从Python发送邮件。这是使用的正确方法,还是我想念的陷阱?

from smtplib import SMTP
import datetime

debuglevel = 0

smtp = SMTP()
smtp.set_debuglevel(debuglevel)
smtp.connect('YOUR.MAIL.SERVER', 26)
smtp.login('USERNAME@DOMAIN', 'PASSWORD')

from_addr = "John Doe <john@doe.net>"
to_addr = "foo@bar.com"

subj = "hello"
date = datetime.datetime.now().strftime( "%d/%m/%Y %H:%M" )

message_text = "Hello\nThis is a mail from your server\n\nBye\n"

msg = "From: %s\nTo: %s\nSubject: %s\nDate: %s\n\n%s" 
        % ( from_addr, to_addr, subj, date, message_text )

smtp.sendmail(from_addr, to_addr, msg)
smtp.quit()

I’m using the following method to send mail from Python using SMTP. Is it the right method to use or are there gotchas I’m missing ?

from smtplib import SMTP
import datetime

debuglevel = 0

smtp = SMTP()
smtp.set_debuglevel(debuglevel)
smtp.connect('YOUR.MAIL.SERVER', 26)
smtp.login('USERNAME@DOMAIN', 'PASSWORD')

from_addr = "John Doe <john@doe.net>"
to_addr = "foo@bar.com"

subj = "hello"
date = datetime.datetime.now().strftime( "%d/%m/%Y %H:%M" )

message_text = "Hello\nThis is a mail from your server\n\nBye\n"

msg = "From: %s\nTo: %s\nSubject: %s\nDate: %s\n\n%s" 
        % ( from_addr, to_addr, subj, date, message_text )

smtp.sendmail(from_addr, to_addr, msg)
smtp.quit()

回答 0

我使用的脚本非常相似。我将其张贴在此处,作为如何使用email。*模块生成MIME消息的示例。因此可以轻松修改此脚本以附加图片等。

我依靠我的ISP添加日期时间标头。

我的ISP要求我使用安全的smtp连接发送邮件,我依靠smtplib模块(可从http://www1.cs.columbia.edu/~db2501/ssmtplib.py下载)

就像在脚本中一样,用于在SMTP服务器上进行身份验证的用户名和密码(下面提供了伪值)在源中为纯文本格式。这是一个安全漏洞。但是最好的选择取决于您对这些保护有多认真(想要?)。

======================================

#! /usr/local/bin/python


SMTPserver = 'smtp.att.yahoo.com'
sender =     'me@my_email_domain.net'
destination = ['recipient@her_email_domain.com']

USERNAME = "USER_NAME_FOR_INTERNET_SERVICE_PROVIDER"
PASSWORD = "PASSWORD_INTERNET_SERVICE_PROVIDER"

# typical values for text_subtype are plain, html, xml
text_subtype = 'plain'


content="""\
Test message
"""

subject="Sent from Python"

import sys
import os
import re

from smtplib import SMTP_SSL as SMTP       # this invokes the secure SMTP protocol (port 465, uses SSL)
# from smtplib import SMTP                  # use this for standard SMTP protocol   (port 25, no encryption)

# old version
# from email.MIMEText import MIMEText
from email.mime.text import MIMEText

try:
    msg = MIMEText(content, text_subtype)
    msg['Subject']=       subject
    msg['From']   = sender # some SMTP servers will do this automatically, not all

    conn = SMTP(SMTPserver)
    conn.set_debuglevel(False)
    conn.login(USERNAME, PASSWORD)
    try:
        conn.sendmail(sender, destination, msg.as_string())
    finally:
        conn.quit()

except:
    sys.exit( "mail failed; %s" % "CUSTOM_ERROR" ) # give an error message

The script I use is quite similar; I post it here as an example of how to use the email.* modules to generate MIME messages; so this script can be easily modified to attach pictures, etc.

I rely on my ISP to add the date time header.

My ISP requires me to use a secure smtp connection to send mail, I rely on the smtplib module (downloadable at http://www1.cs.columbia.edu/~db2501/ssmtplib.py)

As in your script, the username and password, (given dummy values below), used to authenticate on the SMTP server, are in plain text in the source. This is a security weakness; but the best alternative depends on how careful you need (want?) to be about protecting these.

=======================================

#! /usr/local/bin/python


SMTPserver = 'smtp.att.yahoo.com'
sender =     'me@my_email_domain.net'
destination = ['recipient@her_email_domain.com']

USERNAME = "USER_NAME_FOR_INTERNET_SERVICE_PROVIDER"
PASSWORD = "PASSWORD_INTERNET_SERVICE_PROVIDER"

# typical values for text_subtype are plain, html, xml
text_subtype = 'plain'


content="""\
Test message
"""

subject="Sent from Python"

import sys
import os
import re

from smtplib import SMTP_SSL as SMTP       # this invokes the secure SMTP protocol (port 465, uses SSL)
# from smtplib import SMTP                  # use this for standard SMTP protocol   (port 25, no encryption)

# old version
# from email.MIMEText import MIMEText
from email.mime.text import MIMEText

try:
    msg = MIMEText(content, text_subtype)
    msg['Subject']=       subject
    msg['From']   = sender # some SMTP servers will do this automatically, not all

    conn = SMTP(SMTPserver)
    conn.set_debuglevel(False)
    conn.login(USERNAME, PASSWORD)
    try:
        conn.sendmail(sender, destination, msg.as_string())
    finally:
        conn.quit()

except:
    sys.exit( "mail failed; %s" % "CUSTOM_ERROR" ) # give an error message

回答 1

我通常使用的方法…没什么大的不同

import smtplib
from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText

msg = MIMEMultipart()
msg['From'] = 'me@gmail.com'
msg['To'] = 'you@gmail.com'
msg['Subject'] = 'simple email in python'
message = 'here is the email'
msg.attach(MIMEText(message))

mailserver = smtplib.SMTP('smtp.gmail.com',587)
# identify ourselves to smtp gmail client
mailserver.ehlo()
# secure our email with tls encryption
mailserver.starttls()
# re-identify ourselves as an encrypted connection
mailserver.ehlo()
mailserver.login('me@gmail.com', 'mypassword')

mailserver.sendmail('me@gmail.com','you@gmail.com',msg.as_string())

mailserver.quit()

而已

The method I commonly use…not much different but a little bit

import smtplib
from email.MIMEMultipart import MIMEMultipart
from email.MIMEText import MIMEText

msg = MIMEMultipart()
msg['From'] = 'me@gmail.com'
msg['To'] = 'you@gmail.com'
msg['Subject'] = 'simple email in python'
message = 'here is the email'
msg.attach(MIMEText(message))

mailserver = smtplib.SMTP('smtp.gmail.com',587)
# identify ourselves to smtp gmail client
mailserver.ehlo()
# secure our email with tls encryption
mailserver.starttls()
# re-identify ourselves as an encrypted connection
mailserver.ehlo()
mailserver.login('me@gmail.com', 'mypassword')

mailserver.sendmail('me@gmail.com','you@gmail.com',msg.as_string())

mailserver.quit()

That’s it


回答 2

另外,如果您想使用TLS(而不是SSL)执行smtp身份验证,则只需更改端口(使用587)并执行smtp.starttls()。这对我有用:

...
smtp.connect('YOUR.MAIL.SERVER', 587)
smtp.ehlo()
smtp.starttls()
smtp.ehlo()
smtp.login('USERNAME@DOMAIN', 'PASSWORD')
...

Also if you want to do smtp auth with TLS as opposed to SSL then you just have to change the port (use 587) and do smtp.starttls(). This worked for me:

...
smtp.connect('YOUR.MAIL.SERVER', 587)
smtp.ehlo()
smtp.starttls()
smtp.ehlo()
smtp.login('USERNAME@DOMAIN', 'PASSWORD')
...

回答 3

我看到的主要问题是,您没有处理任何错误:.login()和.sendmail()都记录了可以抛出的异常,而且看来.connect()必须有某种方式表明它是无法连接-可能是底层套接字代码引发的异常。

The main gotcha I see is that you’re not handling any errors: .login() and .sendmail() both have documented exceptions that they can throw, and it seems like .connect() must have some way to indicate that it was unable to connect – probably an exception thrown by the underlying socket code.


回答 4

确保您没有任何阻止SMTP的防火墙。我第一次尝试发送电子邮件时,它同时被Windows防火墙和McAfee阻止-花费了很多时间才能找到它们。

Make sure you don’t have any firewalls blocking SMTP. The first time I tried to send an email, it was blocked both by Windows Firewall and McAfee – took forever to find them both.


回答 5

那这个呢?

import smtplib

SERVER = "localhost"

FROM = "sender@example.com"
TO = ["user@example.com"] # must be a list

SUBJECT = "Hello!"

TEXT = "This message was sent with Python's smtplib."

# Prepare actual message

message = """\
From: %s
To: %s
Subject: %s

%s
""" % (FROM, ", ".join(TO), SUBJECT, TEXT)

# Send the mail

server = smtplib.SMTP(SERVER)
server.sendmail(FROM, TO, message)
server.quit()

What about this?

import smtplib

SERVER = "localhost"

FROM = "sender@example.com"
TO = ["user@example.com"] # must be a list

SUBJECT = "Hello!"

TEXT = "This message was sent with Python's smtplib."

# Prepare actual message

message = """\
From: %s
To: %s
Subject: %s

%s
""" % (FROM, ", ".join(TO), SUBJECT, TEXT)

# Send the mail

server = smtplib.SMTP(SERVER)
server.sendmail(FROM, TO, message)
server.quit()

回答 6

以下代码对我来说很好用:

import smtplib

to = 'mkyong2002@yahoo.com'
gmail_user = 'mkyong2002@gmail.com'
gmail_pwd = 'yourpassword'
smtpserver = smtplib.SMTP("smtp.gmail.com",587)
smtpserver.ehlo()
smtpserver.starttls()
smtpserver.ehlo() # extra characters to permit edit
smtpserver.login(gmail_user, gmail_pwd)
header = 'To:' + to + '\n' + 'From: ' + gmail_user + '\n' + 'Subject:testing \n'
print header
msg = header + '\n this is test msg from mkyong.com \n\n'
smtpserver.sendmail(gmail_user, to, msg)
print 'done!'
smtpserver.quit()

参考:http : //www.mkyong.com/python/how-do-send-email-in-python-via-smtplib/

following code is working fine for me:

import smtplib

to = 'mkyong2002@yahoo.com'
gmail_user = 'mkyong2002@gmail.com'
gmail_pwd = 'yourpassword'
smtpserver = smtplib.SMTP("smtp.gmail.com",587)
smtpserver.ehlo()
smtpserver.starttls()
smtpserver.ehlo() # extra characters to permit edit
smtpserver.login(gmail_user, gmail_pwd)
header = 'To:' + to + '\n' + 'From: ' + gmail_user + '\n' + 'Subject:testing \n'
print header
msg = header + '\n this is test msg from mkyong.com \n\n'
smtpserver.sendmail(gmail_user, to, msg)
print 'done!'
smtpserver.quit()

Ref: http://www.mkyong.com/python/how-do-send-email-in-python-via-smtplib/


回答 7

您应该确保以正确的格式RFC2822格式化日期。

You should make sure you format the date in the correct format – RFC2822.


回答 8

我使用SMTP发送邮件的示例代码。

import smtplib, ssl

smtp_server = "smtp.gmail.com"
port = 587  # For starttls
sender_email = "sender@email"
receiver_email = "receiver@email"
password = "<your password here>"
message = """ Subject: Hi there

This message is sent from Python."""


# Create a secure SSL context
context = ssl.create_default_context()

# Try to log in to server and send email
server = smtplib.SMTP(smtp_server,port)

try:
    server.ehlo() # Can be omitted
    server.starttls(context=context) # Secure the connection
    server.ehlo() # Can be omitted
    server.login(sender_email, password)
    server.sendmail(sender_email, receiver_email, message)
except Exception as e:
    # Print any error messages to stdout
    print(e)
finally:
    server.quit()

The example code which i did for send mail using SMTP.

import smtplib, ssl

smtp_server = "smtp.gmail.com"
port = 587  # For starttls
sender_email = "sender@email"
receiver_email = "receiver@email"
password = "<your password here>"
message = """ Subject: Hi there

This message is sent from Python."""


# Create a secure SSL context
context = ssl.create_default_context()

# Try to log in to server and send email
server = smtplib.SMTP(smtp_server,port)

try:
    server.ehlo() # Can be omitted
    server.starttls(context=context) # Secure the connection
    server.ehlo() # Can be omitted
    server.login(sender_email, password)
    server.sendmail(sender_email, receiver_email, message)
except Exception as e:
    # Print any error messages to stdout
    print(e)
finally:
    server.quit()

回答 9

查看所有这些宽大的答案?请允许我通过几行来自我推广。

导入并连接:

import yagmail
yag = yagmail.SMTP('john@doe.net', host = 'YOUR.MAIL.SERVER', port = 26)

然后,它只是一线:

yag.send('foo@bar.com', 'hello', 'Hello\nThis is a mail from your server\n\nBye\n')

当它超出范围时,它将实际上关闭(或可以手动关闭)。此外,它将允许您在密钥环中注册用户名,从而不必在脚本中写出密码(在写之前,这确实困扰了我yagmail!)

有关软件包/安装,提示和技巧,请查看gitpip,它们可用于Python 2和3。

See all those lenghty answers? Please allow me to self promote by doing it all in a couple of lines.

Import and Connect:

import yagmail
yag = yagmail.SMTP('john@doe.net', host = 'YOUR.MAIL.SERVER', port = 26)

Then it is just a one-liner:

yag.send('foo@bar.com', 'hello', 'Hello\nThis is a mail from your server\n\nBye\n')

It will actually close when it goes out of scope (or can be closed manually). Furthermore, it will allow you to register your username in your keyring such that you do not have to write out your password in your script (it really bothered me prior to writing yagmail!)

For the package/installation, tips and tricks please look at git or pip, available for both Python 2 and 3.


回答 10

你可以那样做

import smtplib
from email.mime.text import MIMEText
from email.header import Header


server = smtplib.SMTP('mail.servername.com', 25)
server.ehlo()
server.starttls()

server.login('username', 'password')
from = 'me@servername.com'
to = 'mygfriend@servername.com'
body = 'That A Message For My Girl Friend For tell Him If We will go to eat Something This Nigth'
subject = 'Invite to A Diner'
msg = MIMEText(body,'plain','utf-8')
msg['Subject'] = Header(subject, 'utf-8')
msg['From'] = Header(from, 'utf-8')
msg['To'] = Header(to, 'utf-8')
message = msg.as_string()
server.sendmail(from, to, message)

you can do like that

import smtplib
from email.mime.text import MIMEText
from email.header import Header


server = smtplib.SMTP('mail.servername.com', 25)
server.ehlo()
server.starttls()

server.login('username', 'password')
from = 'me@servername.com'
to = 'mygfriend@servername.com'
body = 'That A Message For My Girl Friend For tell Him If We will go to eat Something This Nigth'
subject = 'Invite to A Diner'
msg = MIMEText(body,'plain','utf-8')
msg['Subject'] = Header(subject, 'utf-8')
msg['From'] = Header(from, 'utf-8')
msg['To'] = Header(to, 'utf-8')
message = msg.as_string()
server.sendmail(from, to, message)

回答 11

这是Python 3.x的工作示例

#!/usr/bin/env python3

from email.message import EmailMessage
from getpass import getpass
from smtplib import SMTP_SSL
from sys import exit

smtp_server = 'smtp.gmail.com'
username = 'your_email_address@gmail.com'
password = getpass('Enter Gmail password: ')

sender = 'your_email_address@gmail.com'
destination = 'recipient_email_address@gmail.com'
subject = 'Sent from Python 3.x'
content = 'Hello! This was sent to you via Python 3.x!'

# Create a text/plain message
msg = EmailMessage()
msg.set_content(content)

msg['Subject'] = subject
msg['From'] = sender
msg['To'] = destination

try:
    s = SMTP_SSL(smtp_server)
    s.login(username, password)
    try:
        s.send_message(msg)
    finally:
        s.quit()

except Exception as E:
    exit('Mail failed: {}'.format(str(E)))

Here’s a working example for Python 3.x

#!/usr/bin/env python3

from email.message import EmailMessage
from getpass import getpass
from smtplib import SMTP_SSL
from sys import exit

smtp_server = 'smtp.gmail.com'
username = 'your_email_address@gmail.com'
password = getpass('Enter Gmail password: ')

sender = 'your_email_address@gmail.com'
destination = 'recipient_email_address@gmail.com'
subject = 'Sent from Python 3.x'
content = 'Hello! This was sent to you via Python 3.x!'

# Create a text/plain message
msg = EmailMessage()
msg.set_content(content)

msg['Subject'] = subject
msg['From'] = sender
msg['To'] = destination

try:
    s = SMTP_SSL(smtp_server)
    s.login(username, password)
    try:
        s.send_message(msg)
    finally:
        s.quit()

except Exception as E:
    exit('Mail failed: {}'.format(str(E)))

回答 12

基于此示例,我做了以下功能:

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

def send_email(host, port, user, pwd, recipients, subject, body, html=None, from_=None):
    """ copied and adapted from
        /programming/10147455/how-to-send-an-email-with-gmail-as-provider-using-python#12424439
    returns None if all ok, but if problem then returns exception object
    """

    PORT_LIST = (25, 587, 465)

    FROM = from_ if from_ else user 
    TO = recipients if isinstance(recipients, (list, tuple)) else [recipients]
    SUBJECT = subject
    TEXT = body.encode("utf8") if isinstance(body, unicode) else body
    HTML = html.encode("utf8") if isinstance(html, unicode) else html

    if not html:
        # Prepare actual message
        message = """From: %s\nTo: %s\nSubject: %s\n\n%s
        """ % (FROM, ", ".join(TO), SUBJECT, TEXT)
    else:
                # /programming/882712/sending-html-email-using-python#882770
        msg = MIMEMultipart('alternative')
        msg['Subject'] = SUBJECT
        msg['From'] = FROM
        msg['To'] = ", ".join(TO)

        # Record the MIME types of both parts - text/plain and text/html.
        # utf-8 -> /programming/5910104/python-how-to-send-utf-8-e-mail#5910530
        part1 = MIMEText(TEXT, 'plain', "utf-8")
        part2 = MIMEText(HTML, 'html', "utf-8")

        # Attach parts into message container.
        # According to RFC 2046, the last part of a multipart message, in this case
        # the HTML message, is best and preferred.
        msg.attach(part1)
        msg.attach(part2)

        message = msg.as_string()


    try:
        if port not in PORT_LIST: 
            raise Exception("Port %s not one of %s" % (port, PORT_LIST))

        if port in (465,):
            server = smtplib.SMTP_SSL(host, port)
        else:
            server = smtplib.SMTP(host, port)

        # optional
        server.ehlo()

        if port in (587,): 
            server.starttls()

        server.login(user, pwd)
        server.sendmail(FROM, TO, message)
        server.close()
        # logger.info("SENT_EMAIL to %s: %s" % (recipients, subject))
    except Exception, ex:
        return ex

    return None

如果仅通过,body则将发送纯文本邮件,但是,如果将html论点与论点一起传递body,则将发送html电子邮件(对于不支持html / mime类型的电子邮件客户端,将回退到文本内容)。

用法示例:

ex = send_email(
      host        = 'smtp.gmail.com'
   #, port        = 465 # OK
    , port        = 587  #OK
    , user        = "xxx@gmail.com"
    , pwd         = "xxx"
    , from_       = 'xxx@gmail.com'
    , recipients  = ['yyy@gmail.com']
    , subject     = "Test from python"
    , body        = "Test from python - body"
    )
if ex: 
    print("Mail sending failed: %s" % ex)
else:
    print("OK - mail sent"

顺便说一句。如果要将gmail用作测试或生产SMTP服务器,请启用对安全性较低的应用程序的临时访问或永久访问:

Based on this example I made following function:

import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

def send_email(host, port, user, pwd, recipients, subject, body, html=None, from_=None):
    """ copied and adapted from
        https://stackoverflow.com/questions/10147455/how-to-send-an-email-with-gmail-as-provider-using-python#12424439
    returns None if all ok, but if problem then returns exception object
    """

    PORT_LIST = (25, 587, 465)

    FROM = from_ if from_ else user 
    TO = recipients if isinstance(recipients, (list, tuple)) else [recipients]
    SUBJECT = subject
    TEXT = body.encode("utf8") if isinstance(body, unicode) else body
    HTML = html.encode("utf8") if isinstance(html, unicode) else html

    if not html:
        # Prepare actual message
        message = """From: %s\nTo: %s\nSubject: %s\n\n%s
        """ % (FROM, ", ".join(TO), SUBJECT, TEXT)
    else:
                # https://stackoverflow.com/questions/882712/sending-html-email-using-python#882770
        msg = MIMEMultipart('alternative')
        msg['Subject'] = SUBJECT
        msg['From'] = FROM
        msg['To'] = ", ".join(TO)

        # Record the MIME types of both parts - text/plain and text/html.
        # utf-8 -> https://stackoverflow.com/questions/5910104/python-how-to-send-utf-8-e-mail#5910530
        part1 = MIMEText(TEXT, 'plain', "utf-8")
        part2 = MIMEText(HTML, 'html', "utf-8")

        # Attach parts into message container.
        # According to RFC 2046, the last part of a multipart message, in this case
        # the HTML message, is best and preferred.
        msg.attach(part1)
        msg.attach(part2)

        message = msg.as_string()


    try:
        if port not in PORT_LIST: 
            raise Exception("Port %s not one of %s" % (port, PORT_LIST))

        if port in (465,):
            server = smtplib.SMTP_SSL(host, port)
        else:
            server = smtplib.SMTP(host, port)

        # optional
        server.ehlo()

        if port in (587,): 
            server.starttls()

        server.login(user, pwd)
        server.sendmail(FROM, TO, message)
        server.close()
        # logger.info("SENT_EMAIL to %s: %s" % (recipients, subject))
    except Exception, ex:
        return ex

    return None

if you pass only body then plain text mail will be sent, but if you pass html argument along with body argument, html email will be sent (with fallback to text content for email clients that don’t support html/mime types).

Example usage:

ex = send_email(
      host        = 'smtp.gmail.com'
   #, port        = 465 # OK
    , port        = 587  #OK
    , user        = "xxx@gmail.com"
    , pwd         = "xxx"
    , from_       = 'xxx@gmail.com'
    , recipients  = ['yyy@gmail.com']
    , subject     = "Test from python"
    , body        = "Test from python - body"
    )
if ex: 
    print("Mail sending failed: %s" % ex)
else:
    print("OK - mail sent"

Btw. If you want to use gmail as testing or production SMTP server, enable temp or permanent access to less secured apps:


用于多个测试的Unittest setUp / tearDown

问题:用于多个测试的Unittest setUp / tearDown

在测试场景的开头/结尾是否触发了某个功能?在每次测试之前/之后都会触发setUp和tearDown函数。

我通常希望拥有:

class TestSequenceFunctions(unittest.TestCase):

    def setUpScenario(self):
        start() #launched at the beginning, once

    def test_choice(self):
        element = random.choice(self.seq)
        self.assertTrue(element in self.seq)

    def test_sample(self):
        with self.assertRaises(ValueError):
            random.sample(self.seq, 20)
        for element in random.sample(self.seq, 5):
            self.assertTrue(element in self.seq)

    def tearDownScenario(self):
        end() #launched at the end, once

目前,这些setUp和tearDown是单元测试,并且分散在我所有的场景中(包含许多测试),一个是第一个测试,另一个是最后一个测试。

Is there a function that is fired at the beginning/end of a scenario of tests? The functions setUp and tearDown are fired before/after every single test.

I typically would like to have this:

class TestSequenceFunctions(unittest.TestCase):

    def setUpScenario(self):
        start() #launched at the beginning, once

    def test_choice(self):
        element = random.choice(self.seq)
        self.assertTrue(element in self.seq)

    def test_sample(self):
        with self.assertRaises(ValueError):
            random.sample(self.seq, 20)
        for element in random.sample(self.seq, 5):
            self.assertTrue(element in self.seq)

    def tearDownScenario(self):
        end() #launched at the end, once

For now, these setUp and tearDown are unit tests and spread in all my scenarios (containing many tests), one is the first test, the other is the last test.


回答 0

从2.7开始(根据文档),分别在给定类中的测试运行之前和之后获得,setUpClasstearDownClass分别执行。或者,如果您在一个文件中有一组,则可以使用setUpModuletearDownModule文档)。

否则,最好的选择是创建自己的派生TestSuite并重写run()。所有其他调用将由父级处理,运行会在父级run方法的调用前后调用您的设置和拆卸代码。

As of 2.7 (per the documentation) you get setUpClass and tearDownClass which execute before and after the tests in a given class are run, respectively. Alternatively, if you have a group of them in one file, you can use setUpModule and tearDownModule (documentation).

Otherwise your best bet is probably going to be to create your own derived TestSuite and override run(). All other calls would be handled by the parent, and run would call your setup and teardown code around a call up to the parent’s run method.


回答 1

我有同样的情况,对我来说setUpClass和tearDownClass方法可以完美地工作

import unittest

class Test(unittest.TestCase):
    @classmethod
    def setUpClass(cls):
        cls._connection = createExpensiveConnectionObject()

    @classmethod
    def tearDownClass(cls):
        cls._connection.destroy()

I have the same scenario, for me setUpClass and tearDownClass methods works perfectly

import unittest

class Test(unittest.TestCase):
    @classmethod
    def setUpClass(cls):
        cls._connection = createExpensiveConnectionObject()

    @classmethod
    def tearDownClass(cls):
        cls._connection.destroy()

回答 2

对于python 2.5,以及在使用pydev时,这有点困难。似乎pydev并没有使用测试套件,而是找到所有单独的测试用例并分别运行它们。

我的解决方案是使用这样的类变量:

class TestCase(unittest.TestCase):
    runCount = 0

    def setUpClass(self):
        pass # overridden in actual testcases

    def run(self, result=None):
        if type(self).runCount == 0:
            self.setUpClass()

        super(TestCase, self).run(result)
        type(self).runCount += 1

使用此技巧,从此继承TestCase(而不是从原始unittest.TestCase)继承时,您还将继承runCount0。然后在run方法中,runCount检查并增加子测试用例的。此类的runCount变量保留为0。

这意味着setUpClass遗嘱每个类只能运行一次,而不是每个实例一次。

我还没有tearDownClass方法,但是我想使用该计数器可能会有所作为。

For python 2.5, and when working with pydev, it’s a bit hard. It appears that pydev doesn’t use the test suite, but finds all individual test cases and runs them all separately.

My solution for this was using a class variable like this:

class TestCase(unittest.TestCase):
    runCount = 0

    def setUpClass(self):
        pass # overridden in actual testcases

    def run(self, result=None):
        if type(self).runCount == 0:
            self.setUpClass()

        super(TestCase, self).run(result)
        type(self).runCount += 1

With this trick, when you inherit from this TestCase (instead of from the original unittest.TestCase), you’ll also inherit the runCount of 0. Then in the run method, the runCount of the child testcase is checked and incremented. This leaves the runCount variable for this class at 0.

This means the setUpClass will only be ran once per class and not once per instance.

I don’t have a tearDownClass method yet, but I guess something could be made with using that counter.


回答 3

这是一个示例:3种测试方法访问一个共享资源,该资源一次创建,而不是每个测试一次。

import unittest
import random

class TestSimulateLogistics(unittest.TestCase):

    shared_resource = None

    @classmethod
    def setUpClass(cls):
        cls.shared_resource = random.randint(1, 100)

    @classmethod
    def tearDownClass(cls):
        cls.shared_resource = None

    def test_1(self):
        print('test 1:', self.shared_resource)

    def test_2(self):
        print('test 2:', self.shared_resource)

    def test_3(self):
        print('test 3:', self.shared_resource)

Here is an example: 3 test methods access a shared resource, which is created once, not per test.

import unittest
import random

class TestSimulateLogistics(unittest.TestCase):

    shared_resource = None

    @classmethod
    def setUpClass(cls):
        cls.shared_resource = random.randint(1, 100)

    @classmethod
    def tearDownClass(cls):
        cls.shared_resource = None

    def test_1(self):
        print('test 1:', self.shared_resource)

    def test_2(self):
        print('test 2:', self.shared_resource)

    def test_3(self):
        print('test 3:', self.shared_resource)


如何将python脚本编译为二进制可执行文件

问题:如何将python脚本编译为二进制可执行文件

我需要将Python脚本转换为Windows可执行文件。

我已将Python 2.6安装到python26

我创建了一个脚本并将其保存在中C:\pythonscript。在此文件夹中有两个文件

Setup.pyoldlogs.py(此文件需要掩盖)

setup.py 代码是

from distutils.core import setup
import py2exe

setup(console=['oldlogs.py'])

如何将oldlogs.py转换为exe文件?

I need to convert a Python script to a Windows executable.

I have Python 2.6 installed to python26.

I have created one script and kept it in C:\pythonscript. Inside this folder there are two files

Setup.py and oldlogs.py (this file need coversion)

setup.py code is

from distutils.core import setup
import py2exe

setup(console=['oldlogs.py'])

How can I convert oldlogs.py to an exe file?


回答 0

或者使用PyInstaller替代py2exe。这是一个很好的起点。PyInstaller还可以让您为Linux和Mac创建可执行文件。

这是一个相当容易使用PyInstaller解决当前问题的方法:

pyinstaller oldlogs.py

从该工具的文档中:

PyInstaller分析myscript.py并执行以下操作:

  • 在脚本所在的文件夹中写入myscript.spec。
  • 如果脚本不存在,则在与脚本相同的文件夹中创建文件夹。
  • 在构建文件夹中写入一些日志文件和工作文件。
  • 如果脚本不存在,则在与脚本相同的文件夹中创建文件夹dist。
  • 将myscript可执行文件文件夹写入dist文件夹。

在dist文件夹中,您找到分发给用户的捆绑应用程序。

Or use PyInstaller as an alternative to py2exe. Here is a good starting point. PyInstaller also lets you create executables for linux and mac…

Here is how one could fairly easily use PyInstaller to solve the issue at hand:

pyinstaller oldlogs.py

From the tool’s documentation:

PyInstaller analyzes myscript.py and:

  • Writes myscript.spec in the same folder as the script.
  • Creates a folder build in the same folder as the script if it does not exist.
  • Writes some log files and working files in the build folder.
  • Creates a folder dist in the same folder as the script if it does not exist.
  • Writes the myscript executable folder in the dist folder.

In the dist folder you find the bundled app you distribute to your users.


回答 1

我建议使用PyInstaller,可以使用以下命令将简单的python脚本转换为exe:

utils/Makespec.py [--onefile] oldlogs.py

这将创建一个yourprogram.spec文件,该文件是用于构建最终exe的配置。下一条命令从配置文件生成exe:

utils/Build.py oldlogs.spec

这里可以找到更多

I recommend PyInstaller, a simple python script can be converted to an exe with the following commands:

utils/Makespec.py [--onefile] oldlogs.py

which creates a yourprogram.spec file which is a configuration for building the final exe. Next command builds the exe from the configuration file:

utils/Build.py oldlogs.spec

More can be found here


回答 2

# -*- mode: python -*-

block_cipher = None

a = Analysis(['SCRIPT.py'],
             pathex=[
                 'folder path',
                 'C:\\Windows\\WinSxS\\x86_microsoft-windows-m..namespace-downlevel_31bf3856ad364e35_10.0.17134.1_none_50c6cb8431e7428f',
                 'C:\\Windows\\WinSxS\\x86_microsoft-windows-m..namespace-downlevel_31bf3856ad364e35_10.0.17134.1_none_c4f50889467f081d'
             ],
             binaries=[(''C:\\Users\\chromedriver.exe'')],
             datas=[],
             hiddenimports=[],
             hookspath=[],
             runtime_hooks=[],
             excludes=[],
             win_no_prefer_redirects=False,
             win_private_assemblies=False,
             cipher=block_cipher)
pyz = PYZ(a.pure, a.zipped_data,
             cipher=block_cipher)
exe = EXE(pyz,
          a.scripts,
          a.binaries,
          a.zipfiles,
          a.datas,
          name='NAME OF YOUR EXE',
          debug=False,
          strip=False,
          upx=True,
          runtime_tmpdir=None,
          console=True )
# -*- mode: python -*-

block_cipher = None

a = Analysis(['SCRIPT.py'],
             pathex=[
                 'folder path',
                 'C:\\Windows\\WinSxS\\x86_microsoft-windows-m..namespace-downlevel_31bf3856ad364e35_10.0.17134.1_none_50c6cb8431e7428f',
                 'C:\\Windows\\WinSxS\\x86_microsoft-windows-m..namespace-downlevel_31bf3856ad364e35_10.0.17134.1_none_c4f50889467f081d'
             ],
             binaries=[(''C:\\Users\\chromedriver.exe'')],
             datas=[],
             hiddenimports=[],
             hookspath=[],
             runtime_hooks=[],
             excludes=[],
             win_no_prefer_redirects=False,
             win_private_assemblies=False,
             cipher=block_cipher)
pyz = PYZ(a.pure, a.zipped_data,
             cipher=block_cipher)
exe = EXE(pyz,
          a.scripts,
          a.binaries,
          a.zipfiles,
          a.datas,
          name='NAME OF YOUR EXE',
          debug=False,
          strip=False,
          upx=True,
          runtime_tmpdir=None,
          console=True )

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