在Python列表的第一位置插入[关闭]

问题:在Python列表的第一位置插入[关闭]

如何在列表的第一个索引处插入元素?如果我使用list.insert(0,elem),elem是否会修改第一个索引的内容?还是我必须使用第一个元素创建一个新列表,然后将旧列表复制到这个新列表中?

How can I insert an element at the first index of a list ? If I use list.insert(0,elem), do elem modify the content of the first index? Or do I have to create a new list with the first elem and then copy the old list inside this new one?


回答 0

用途insert

In [1]: ls = [1,2,3]

In [2]: ls.insert(0, "new")

In [3]: ls
Out[3]: ['new', 1, 2, 3]

Use insert:

In [1]: ls = [1,2,3]

In [2]: ls.insert(0, "new")

In [3]: ls
Out[3]: ['new', 1, 2, 3]

回答 1

从文档中:

list.insert(i,x)
在给定位置插入项目。第一个参数是要在其之前插入的元素的索引,因此a.insert(0, x)将其插入 到列表的开头,并且a.insert(len(a),x)等效于a.append(x)

http://docs.python.org/2/tutorial/datastructures.html#more-on-lists

From the documentation:

list.insert(i, x)
Insert an item at a given position. The first argument is the index of the element before which to insert, so a.insert(0, x) inserts at the front of the list, and a.insert(len(a),x) is equivalent to a.append(x)

http://docs.python.org/2/tutorial/datastructures.html#more-on-lists


Tensorflow 2.0-AttributeError:模块’tensorflow’没有属性’Session’

问题:Tensorflow 2.0-AttributeError:模块’tensorflow’没有属性’Session’

sess = tf.Session()在Tensorflow 2.0环境中执行命令时,出现如下错误消息:

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: module 'tensorflow' has no attribute 'Session'

系统信息:

  • 操作系统平台和发行版:Windows 10
  • python版本:3.7.1
  • Tensorflow版本:2.0.0-alpha0(随pip一起安装)

重现步骤:

安装:

  1. 点安装-升级点
  2. pip install tensorflow == 2.0.0-alpha0
  3. 点安装keras
  4. 点安装numpy == 1.16.2

执行:

  1. 执行命令:将tensorflow导入为tf
  2. 执行命令:sess = tf.Session()

When I am executing the command sess = tf.Session() in Tensorflow 2.0 environment, I am getting an error message as below:

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: module 'tensorflow' has no attribute 'Session'

System Information:

  • OS Platform and Distribution: Windows 10
  • Python Version: 3.7.1
  • Tensorflow Version: 2.0.0-alpha0 (installed with pip)

Steps to reproduce:

Installation:

  1. pip install –upgrade pip
  2. pip install tensorflow==2.0.0-alpha0
  3. pip install keras
  4. pip install numpy==1.16.2

Execution:

  1. Execute command: import tensorflow as tf
  2. Execute command: sess = tf.Session()

回答 0

根据TF 1:1 Symbols Map,在TF 2.0中,您应该使用tf.compat.v1.Session()而不是tf.Session()

https://docs.google.com/spreadsheets/d/1FLFJLzg7WNP6JHODX5q8BDgptKafq_slHpnHVbJIteQ/edit#gid=0

要获得TF 2.0中类似TF 1.x的行为,可以运行

import tensorflow.compat.v1 as tf
tf.disable_v2_behavior()

但后来人们无法受益于TF 2.0所做的许多改进。有关更多详细信息,请参阅迁移指南 https://www.tensorflow.org/guide/migrate

According to TF 1:1 Symbols Map, in TF 2.0 you should use tf.compat.v1.Session() instead of tf.Session()

https://docs.google.com/spreadsheets/d/1FLFJLzg7WNP6JHODX5q8BDgptKafq_slHpnHVbJIteQ/edit#gid=0

To get TF 1.x like behaviour in TF 2.0 one can run

import tensorflow.compat.v1 as tf
tf.disable_v2_behavior()

but then one cannot benefit of many improvements made in TF 2.0. For more details please refer to the migration guide https://www.tensorflow.org/guide/migrate


回答 1

TF2默认情况下运行急切执行,因此无需会话。如果要运行静态图,则更正确的方法是tf.function()在TF2中使用。虽然仍然可以通过tf.compat.v1.Session()TF2访问Session ,但我不建议使用它。通过比较问候世界中的差异来证明这种差异可能会有所帮助:

TF1.x你好世界:

import tensorflow as tf
msg = tf.constant('Hello, TensorFlow!')
sess = tf.Session()
print(sess.run(msg))

TF2.x你好世界:

import tensorflow as tf
msg = tf.constant('Hello, TensorFlow!')
tf.print(msg)

有关更多信息,请参见Effective TensorFlow 2

TF2 runs Eager Execution by default, thus removing the need for Sessions. If you want to run static graphs, the more proper way is to use tf.function() in TF2. While Session can still be accessed via tf.compat.v1.Session() in TF2, I would discourage using it. It may be helpful to demonstrate this difference by comparing the difference in hello worlds:

TF1.x hello world:

import tensorflow as tf
msg = tf.constant('Hello, TensorFlow!')
sess = tf.Session()
print(sess.run(msg))

TF2.x hello world:

import tensorflow as tf
msg = tf.constant('Hello, TensorFlow!')
tf.print(msg)

For more info, see Effective TensorFlow 2


回答 2

安装后第一次尝试python时遇到了这个问题 windows10 + python3.7(64bit) + anacconda3 + jupyter notebook.

我通过参考“ https://vispud.blogspot.com/2019/05/tensorflow200a0-attributeerror-module.html ”解决了此问题

我同意

我相信TF 2.0已删除了“ Session()”。

我插入了两行。一个是tf.compat.v1.disable_eager_execution(),另一个是sess = tf.compat.v1.Session()

我的Hello.py如下:

import tensorflow as tf

tf.compat.v1.disable_eager_execution()

hello = tf.constant('Hello, TensorFlow!')

sess = tf.compat.v1.Session()

print(sess.run(hello))

I faced this problem when I first tried python after installing windows10 + python3.7(64bit) + anacconda3 + jupyter notebook.

I solved this problem by refering to “https://vispud.blogspot.com/2019/05/tensorflow200a0-attributeerror-module.html

I agree with

I believe “Session()” has been removed with TF 2.0.

I inserted two lines. One is tf.compat.v1.disable_eager_execution() and the other is sess = tf.compat.v1.Session()

My Hello.py is as follows:

import tensorflow as tf

tf.compat.v1.disable_eager_execution()

hello = tf.constant('Hello, TensorFlow!')

sess = tf.compat.v1.Session()

print(sess.run(hello))

回答 3

对于TF2.x,您可以这样做。

import tensorflow as tf
with tf.compat.v1.Session() as sess:
    hello = tf.constant('hello world')
    print(sess.run(hello))

>>> b'hello world

For TF2.x, you can do like this.

import tensorflow as tf
with tf.compat.v1.Session() as sess:
    hello = tf.constant('hello world')
    print(sess.run(hello))

>>> b'hello world


回答 4

尝试这个

import tensorflow as tf

tf.compat.v1.disable_eager_execution()

hello = tf.constant('Hello, TensorFlow!')

sess = tf.compat.v1.Session()

print(sess.run(hello))

try this

import tensorflow as tf

tf.compat.v1.disable_eager_execution()

hello = tf.constant('Hello, TensorFlow!')

sess = tf.compat.v1.Session()

print(sess.run(hello))

回答 5

如果这是您的代码,则正确的解决方案是将其重写为不使用Session(),因为在TensorFlow 2中不再需要

如果这只是您正在运行的代码,则可以通过运行降级到TensorFlow 1

pip3 install --upgrade --force-reinstall tensorflow-gpu==1.15.0 

(或TensorFlow 1最新版本

If this is your code, the correct solution is to rewrite it to not use Session(), since that’s no longer necessary in TensorFlow 2

If this is just code you’re running, you can downgrade to TensorFlow 1 by running

pip3 install --upgrade --force-reinstall tensorflow-gpu==1.15.0 

(or whatever the latest version of TensorFlow 1 is)


回答 6

Tensorflow 2.x支持默认执行Eager Execution,因此不支持Session。

Tensorflow 2.x support’s Eager Execution by default hence Session is not supported.


回答 7

使用Anaconda + Spyder(Python 3.7)

[码]

import tensorflow as tf
valor1 = tf.constant(2)
valor2 = tf.constant(3)
type(valor1)
print(valor1)
soma=valor1+valor2
type(soma)
print(soma)
sess = tf.compat.v1.Session()
with sess:
    print(sess.run(soma))

[安慰]

import tensorflow as tf
valor1 = tf.constant(2)
valor2 = tf.constant(3)
type(valor1)
print(valor1)
soma=valor1+valor2
type(soma)
Tensor("Const_8:0", shape=(), dtype=int32)
Out[18]: tensorflow.python.framework.ops.Tensor

print(soma)
Tensor("add_4:0", shape=(), dtype=int32)

sess = tf.compat.v1.Session()

with sess:
    print(sess.run(soma))
5

Using Anaconda + Spyder (Python 3.7)

[code]

import tensorflow as tf
valor1 = tf.constant(2)
valor2 = tf.constant(3)
type(valor1)
print(valor1)
soma=valor1+valor2
type(soma)
print(soma)
sess = tf.compat.v1.Session()
with sess:
    print(sess.run(soma))

[console]

import tensorflow as tf
valor1 = tf.constant(2)
valor2 = tf.constant(3)
type(valor1)
print(valor1)
soma=valor1+valor2
type(soma)
Tensor("Const_8:0", shape=(), dtype=int32)
Out[18]: tensorflow.python.framework.ops.Tensor

print(soma)
Tensor("add_4:0", shape=(), dtype=int32)

sess = tf.compat.v1.Session()

with sess:
    print(sess.run(soma))
5

回答 8

TF v2.0支持Eager模式和v1.0的Graph模式。因此,v2.0不支持tf.session()。因此,建议您重写代码以在Eager模式下工作。

TF v2.0 supports Eager mode vis-a-vis Graph mode of v1.0. Hence, tf.session() is not supported on v2.0. Hence, would suggest you to rewrite your code to work in Eager mode.


回答 9

import tensorflow as tf
sess = tf.Session()

此代码将在版本2.x上显示属性错误

在版本2.x中使用版本1.x代码

尝试这个

import tensorflow.compat.v1 as tf
sess = tf.Session()
import tensorflow as tf
sess = tf.Session()

this code will show an Attribute error on version 2.x

to use version 1.x code in version 2.x

try this

import tensorflow.compat.v1 as tf
sess = tf.Session()

在运行时确定带有upload_to的Django FileField

问题:在运行时确定带有upload_to的Django FileField

我正在尝试设置我的上传文件,以便如果用户joe上传文件,则文件将转到MEDIA_ROOT / joe,而不是让每个人的文件都转到MEDIA_ROOT。问题是我不知道如何在模型中定义它。这是当前的外观:

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to='.')

所以我想要的不是“。” 作为upload_to,将其作为用户名。

我知道从Django 1.0开始,您可以定义自己的函数来处理upload_to,但是该函数也不知道谁将成为谁,所以我有点迷失了。

谢谢您的帮助!

I’m trying to set up my uploads so that if user joe uploads a file it goes to MEDIA_ROOT/joe as opposed to having everyone’s files go to MEDIA_ROOT. The problem is I don’t know how to define this in the model. Here is how it currently looks:

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to='.')

So what I want is instead of ‘.’ as the upload_to, have it be the user’s name.

I understand that as of Django 1.0 you can define your own function to handle the upload_to but that function has no idea of who the user will be either so I’m a bit lost.

Thanks for the help!


回答 0

您可能已经阅读了文档,所以这里有一个简单的示例可以使之有意义:

def content_file_name(instance, filename):
    return '/'.join(['content', instance.user.username, filename])

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to=content_file_name)

如您所见,您甚至不需要使用给定的文件名-如果愿意,您也可以覆盖您可调用的upload_to中的文件名。

You’ve probably read the documentation, so here’s an easy example to make it make sense:

def content_file_name(instance, filename):
    return '/'.join(['content', instance.user.username, filename])

class Content(models.Model):
    name = models.CharField(max_length=200)
    user = models.ForeignKey(User)
    file = models.FileField(upload_to=content_file_name)

As you can see, you don’t even need to use the filename given – you could override that in your upload_to callable too if you liked.


回答 1

这确实有帮助。为了简洁起见,决定在我的情况下使用lambda:

file = models.FileField(
    upload_to=lambda instance, filename: '/'.join(['mymodel', str(instance.pk), filename]),
)

This really helped. For a bit more brevity’s sake, decided to use lambda in my case:

file = models.FileField(
    upload_to=lambda instance, filename: '/'.join(['mymodel', str(instance.pk), filename]),
)

回答 2

关于使用“实例”对象的pk值的注释。根据文档:

在大多数情况下,此对象尚未保存到数据库,因此,如果使用默认的AutoField,则它的主键字段可能尚未具有值。

因此,使用pk的有效性取决于特定模型的定义。

A note on using the ‘instance’ object’s pk value. According to the documentation:

In most cases, this object will not have been saved to the database yet, so if it uses the default AutoField, it might not yet have a value for its primary key field.

Therefore the validity of using pk depends on how your particular model is defined.


回答 3

如果您在迁移时遇到问题,则可能应该使用@deconstructible装饰器。

import datetime
import os
import unicodedata

from django.core.files.storage import default_storage
from django.utils.deconstruct import deconstructible
from django.utils.encoding import force_text, force_str


@deconstructible
class UploadToPath(object):
    def __init__(self, upload_to):
        self.upload_to = upload_to

    def __call__(self, instance, filename):
        return self.generate_filename(filename)

    def get_directory_name(self):
        return os.path.normpath(force_text(datetime.datetime.now().strftime(force_str(self.upload_to))))

    def get_filename(self, filename):
        filename = default_storage.get_valid_name(os.path.basename(filename))
        filename = force_text(filename)
        filename = unicodedata.normalize('NFKD', filename).encode('ascii', 'ignore').decode('ascii')
        return os.path.normpath(filename)

    def generate_filename(self, filename):
        return os.path.join(self.get_directory_name(), self.get_filename(filename))

用法:

class MyModel(models.Model):
    file = models.FileField(upload_to=UploadToPath('files/%Y/%m/%d'), max_length=255)

If you have problems with migrations you probably should be using @deconstructible decorator.

import datetime
import os
import unicodedata

from django.core.files.storage import default_storage
from django.utils.deconstruct import deconstructible
from django.utils.encoding import force_text, force_str


@deconstructible
class UploadToPath(object):
    def __init__(self, upload_to):
        self.upload_to = upload_to

    def __call__(self, instance, filename):
        return self.generate_filename(filename)

    def get_directory_name(self):
        return os.path.normpath(force_text(datetime.datetime.now().strftime(force_str(self.upload_to))))

    def get_filename(self, filename):
        filename = default_storage.get_valid_name(os.path.basename(filename))
        filename = force_text(filename)
        filename = unicodedata.normalize('NFKD', filename).encode('ascii', 'ignore').decode('ascii')
        return os.path.normpath(filename)

    def generate_filename(self, filename):
        return os.path.join(self.get_directory_name(), self.get_filename(filename))

Usage:

class MyModel(models.Model):
    file = models.FileField(upload_to=UploadToPath('files/%Y/%m/%d'), max_length=255)

通过架构以部署用户身份激活virtualenv

问题:通过架构以部署用户身份激活virtualenv

我想在本地运行我的结构脚本,这将依次登录到我的服务器,切换用户以进行部署,激活项目.virtualenv,这将把dir更改为项目并发出git pull。

def git_pull():
    sudo('su deploy')
    # here i need to switch to the virtualenv
    run('git pull')

我通常使用来自virtualenvwrapper的workon命令,该命令提供激活文件,后激活文件会将我放在项目文件夹中。在这种情况下,似乎因为结构是在shell中运行的,所以控制权移交给了结构,所以我不能将bash的源内置到’$ source〜/ .virtualenv / myvenv / bin / activate’中。

有人举一个例子,并解释他们如何做到这一点吗?

I want to run my fabric script locally, which will in turn, log into my server, switch user to deploy, activate the projects .virtualenv, which will change dir to the project and issue a git pull.

def git_pull():
    sudo('su deploy')
    # here i need to switch to the virtualenv
    run('git pull')

I typically use the workon command from virtualenvwrapper which sources the activate file and the postactivate file will put me in the project folder. In this case, it seems that because fabric runs from within shell, control is give over to fabric, so I can’t use bash’s source built-in to ‘$source ~/.virtualenv/myvenv/bin/activate’

Anybody have an example and explanation of how they have done this?


回答 0

现在,您可以做我所要做的事情,这很笨拙,但效果很好*(此用法假设您正在使用virtualenvwrapper-应该如此-但您可以轻松地替换为您提到的更长的“源”调用, 如果不):

def task():
    workon = 'workon myvenv && '
    run(workon + 'git pull')
    run(workon + 'do other stuff, etc')

从1.0版开始,Fabric具有使用此技术的prefix上下文管理器,因此您可以例如:

def task():
    with prefix('workon myvenv'):
        run('git pull')
        run('do other stuff, etc')

*在某些情况下,使用这种command1 && command2方法可能会炸毁您,例如command1失败时(command2永远不会运行)或command1无法正确转义并且包含特殊的shell字符等。

Right now, you can do what I do, which is kludgy but works perfectly well* (this usage assumes you’re using virtualenvwrapper — which you should be — but you can easily substitute in the rather longer ‘source’ call you mentioned, if not):

def task():
    workon = 'workon myvenv && '
    run(workon + 'git pull')
    run(workon + 'do other stuff, etc')

Since version 1.0, Fabric has a prefix context manager which uses this technique so you can for example:

def task():
    with prefix('workon myvenv'):
        run('git pull')
        run('do other stuff, etc')

* There are bound to be cases where using the command1 && command2 approach may blow up on you, such as when command1 fails (command2 will never run) or if command1 isn’t properly escaped and contains special shell characters, and so forth.


回答 1

作为对bitprophet预测的更新:使用Fabric 1.0,您可以使用prefix()和您自己的上下文管理器。

from __future__ import with_statement
from fabric.api import *
from contextlib import contextmanager as _contextmanager

env.hosts = ['servername']
env.user = 'deploy'
env.keyfile = ['$HOME/.ssh/deploy_rsa']
env.directory = '/path/to/virtualenvs/project'
env.activate = 'source /path/to/virtualenvs/project/bin/activate'

@_contextmanager
def virtualenv():
    with cd(env.directory):
        with prefix(env.activate):
            yield

def deploy():
    with virtualenv():
        run('pip freeze')

As an update to bitprophet’s forecast: With Fabric 1.0 you can make use of prefix() and your own context managers.

from __future__ import with_statement
from fabric.api import *
from contextlib import contextmanager as _contextmanager

env.hosts = ['servername']
env.user = 'deploy'
env.keyfile = ['$HOME/.ssh/deploy_rsa']
env.directory = '/path/to/virtualenvs/project'
env.activate = 'source /path/to/virtualenvs/project/bin/activate'

@_contextmanager
def virtualenv():
    with cd(env.directory):
        with prefix(env.activate):
            yield

def deploy():
    with virtualenv():
        run('pip freeze')

回答 2

我只是使用一个简单的包装函数virtualenv()而不是run()即可调用。它不使用cd上下文管理器,因此可以使用相对路径。

def virtualenv(command):
    """
    Run a command in the virtualenv. This prefixes the command with the source
    command.
    Usage:
        virtualenv('pip install django')
    """
    source = 'source %(project_directory)s/bin/activate && ' % env
    run(source + command)

I’m just using a simple wrapper function virtualenv() that can be called instead of run(). It doesn’t use the cd context manager, so relative paths can be used.

def virtualenv(command):
    """
    Run a command in the virtualenv. This prefixes the command with the source
    command.
    Usage:
        virtualenv('pip install django')
    """
    source = 'source %(project_directory)s/bin/activate && ' % env
    run(source + command)

回答 3

virtualenvwrapper 可以简化一点

  1. 使用@ nh2的方法(该方法在使用时也适用local,但仅适用workon$PATHin中的virtualenvwrapper安装程序,换句话说-Windows)

    from contextlib import contextmanager
    from fabric.api import prefix
    
    @contextmanager
    def virtualenv():
        with prefix("workon env1"):
            yield
    
    def deploy():
        with virtualenv():
            run("pip freeze > requirements.txt")
  2. 或部署fab文件并在本地运行。通过此设置,您可以为本地或远程命令激活virtualenv。这种方法功能强大,因为它可以解决local无法使用bash -l以下命令运行.bashrc的问题:

    @contextmanager
    def local_prefix(shell, prefix):
        def local_call(command):
            return local("%(sh)s \"%(pre)s && %(cmd)s\"" % 
                {"sh": shell, "pre": prefix, "cmd": command})
        yield local_prefix
    
    def write_requirements(shell="/bin/bash -lic", env="env1"):
        with local_prefix(shell, "workon %s" % env) as local:
            local("pip freeze > requirements.txt")
    
    write_requirements()  # locally
    run("fab write_requirements")

virtualenvwrapper can make this a little simpler

  1. Using @nh2’s approach (this approach also works when using local, but only for virtualenvwrapper installations where workon is in $PATH, in other words — Windows)

    from contextlib import contextmanager
    from fabric.api import prefix
    
    @contextmanager
    def virtualenv():
        with prefix("workon env1"):
            yield
    
    def deploy():
        with virtualenv():
            run("pip freeze > requirements.txt")
    
  2. Or deploy your fab file and run this locally. This setup lets you activate the virtualenv for local or remote commands. This approach is powerful because it works around local‘s inability to run .bashrc using bash -l:

    @contextmanager
    def local_prefix(shell, prefix):
        def local_call(command):
            return local("%(sh)s \"%(pre)s && %(cmd)s\"" % 
                {"sh": shell, "pre": prefix, "cmd": command})
        yield local_prefix
    
    def write_requirements(shell="/bin/bash -lic", env="env1"):
        with local_prefix(shell, "workon %s" % env) as local:
            local("pip freeze > requirements.txt")
    
    write_requirements()  # locally
    run("fab write_requirements")
    

回答 4

这是我在virtualenv本地部署中使用的方法。

使用fabric的path()上下文管理器,您可以运行virtualenv pippython使用virtualenv中的二进制文件。

from fabric.api import lcd, local, path

project_dir = '/www/my_project/sms/'
env_bin_dir = project_dir + '../env/bin/'

def deploy():
    with lcd(project_dir):
        local('git pull origin')
        local('git checkout -f')
        with path(env_bin_dir, behavior='prepend'):
            local('pip freeze')
            local('pip install -r requirements/staging.txt')
            local('./manage.py migrate') # Django related

            # Note: previous line is the same as:
            local('python manage.py migrate')

            # Using next line, you can make sure that python 
            # from virtualenv directory is used:
            local('which python')

This is my approach on using virtualenv with local deployments.

Using fabric’s path() context manager you can run pip or python with binaries from virtualenv.

from fabric.api import lcd, local, path

project_dir = '/www/my_project/sms/'
env_bin_dir = project_dir + '../env/bin/'

def deploy():
    with lcd(project_dir):
        local('git pull origin')
        local('git checkout -f')
        with path(env_bin_dir, behavior='prepend'):
            local('pip freeze')
            local('pip install -r requirements/staging.txt')
            local('./manage.py migrate') # Django related

            # Note: previous line is the same as:
            local('python manage.py migrate')

            # Using next line, you can make sure that python 
            # from virtualenv directory is used:
            local('which python')

回答 5

感谢发布的所有答案,我想为此添加另一种替代方法。有一个模块fabric-virtualenv,可以提供与相同代码相同的功能:

>>> from fabvenv import virtualenv
>>> with virtualenv('/home/me/venv/'):
...     run('python foo')

fabric-virtualenv使用fabric.context_managers.prefix,这可能是一个好方法:)

Thanks to all answers posted and I would like to add one more alternative for this. There is an module, fabric-virtualenv, which can provide the function as the same code:

>>> from fabvenv import virtualenv
>>> with virtualenv('/home/me/venv/'):
...     run('python foo')

fabric-virtualenv makes use of fabric.context_managers.prefix, which might be a good way :)


回答 6

如果您想将软件包安装到环境中,或者要根据环境中的软件包运行命令,我发现此技巧可以解决我的问题,而不是编写复杂的Fabric方法或安装新的OS软件包:

/path/to/virtualenv/bin/python manage.py migrate/runserver/makemigrations  # for running commands under virtualenv

local("/home/user/env/bin/python manage.py migrate")    # fabric command


/path/to/virtualenv/bin/pip install -r requirements.txt   # installing/upgrading virtualenv

local("/home/user/env/bin/pip install -r requirements.txt")  #  fabric command

这样,您可能不需要激活环境,但是可以在该环境下执行命令。

If you want to install the packages to environment or want to run commands according to the packages you have in environment, I have found this hack to solve my problem, instead of writing complex methods of fabric or installing new OS packages:

/path/to/virtualenv/bin/python manage.py migrate/runserver/makemigrations  # for running commands under virtualenv

local("/home/user/env/bin/python manage.py migrate")    # fabric command


/path/to/virtualenv/bin/pip install -r requirements.txt   # installing/upgrading virtualenv

local("/home/user/env/bin/pip install -r requirements.txt")  #  fabric command

This way you might not need to activate the environment, but you can execute commands under the environment.


回答 7

以下是装饰器的代码,该代码将导致对任何运行/ sudo调用使用虚拟环境:

# This is the bash code to update the $PATH as activate does
UPDATE_PYTHON_PATH = r'PATH="{}:$PATH"'.format(VIRTUAL_ENV_BIN_DIR)

def with_venv(func, *args, **kwargs):
  "Use Virtual Environment for the command"

  def wrapped(*args, **kwargs):
    with prefix(UPDATE_PYTHON_PATH):
      return func(*args, **kwargs)

  wrapped.__name__ = func.__name__
  wrapped.__doc__ = func.__doc__
  return wrapped

然后要使用装饰器,请注意装饰器的顺序很重要:

@task
@with_venv
def which_python():
  "Gets which python is being used"
  run("which python")

Here is code for a decorator that will result in the use of Virtual Environment for any run/sudo calls:

# This is the bash code to update the $PATH as activate does
UPDATE_PYTHON_PATH = r'PATH="{}:$PATH"'.format(VIRTUAL_ENV_BIN_DIR)

def with_venv(func, *args, **kwargs):
  "Use Virtual Environment for the command"

  def wrapped(*args, **kwargs):
    with prefix(UPDATE_PYTHON_PATH):
      return func(*args, **kwargs)

  wrapped.__name__ = func.__name__
  wrapped.__doc__ = func.__doc__
  return wrapped

and then to use the decorator, note the order of the decorators is important:

@task
@with_venv
def which_python():
  "Gets which python is being used"
  run("which python")

回答 8

这种方法对我有用,您也可以应用。

from fabric.api import run 
# ... other code...
def install_pip_requirements():
    run("/bin/bash -l -c 'source venv/bin/activate' "
        "&& pip install -r requirements.txt "
        "&& /bin/bash -l -c 'deactivate'")

假定venv您的虚拟环境目录,并在适当的地方添加此方法。

This approach worked for me, you can apply this too.

from fabric.api import run 
# ... other code...
def install_pip_requirements():
    run("/bin/bash -l -c 'source venv/bin/activate' "
        "&& pip install -r requirements.txt "
        "&& /bin/bash -l -c 'deactivate'")

Assuming venv is your virtual env directory and add this method wherever appropriate.


NumPy矩阵与数组类的乘法有何不同?

问题:NumPy矩阵与数组类的乘法有何不同?

numpy文档建议使用数组而不是矩阵来处理矩阵。但是,与八度(我直到最近才使用)不同,*不执行矩阵乘法,您需要使用函数matrixmultipy()。我觉得这使代码非常不可读。

是否有人分享我的观点并找到了解决方案?

The numpy docs recommend using array instead of matrix for working with matrices. However, unlike octave (which I was using till recently), * doesn’t perform matrix multiplication, you need to use the function matrixmultipy(). I feel this makes the code very unreadable.

Does anybody share my views, and has found a solution?


回答 0

避免使用的主要原因 matrix该类的是:a)本质上是二维的,并且b)与“常规” numpy数组相比,存在额外的开销。如果您要做的只是线性代数,那么请务必使用矩阵类…就我个人而言,我发现它比它值得的麻烦更多。

对于数组(Python 3.5之前的版本),请使用dot代替matrixmultiply

例如

import numpy as np
x = np.arange(9).reshape((3,3))
y = np.arange(3)

print np.dot(x,y)

或在新版本的numpy中,只需使用 x.dot(y)

就个人而言,我发现它比*表示矩阵乘法的运算符更具可读性…

对于Python 3.5中的数组,请使用x @ y

The main reason to avoid using the matrix class is that a) it’s inherently 2-dimensional, and b) there’s additional overhead compared to a “normal” numpy array. If all you’re doing is linear algebra, then by all means, feel free to use the matrix class… Personally I find it more trouble than it’s worth, though.

For arrays (prior to Python 3.5), use dot instead of matrixmultiply.

E.g.

import numpy as np
x = np.arange(9).reshape((3,3))
y = np.arange(3)

print np.dot(x,y)

Or in newer versions of numpy, simply use x.dot(y)

Personally, I find it much more readable than the * operator implying matrix multiplication…

For arrays in Python 3.5, use x @ y.


回答 1

与在NumPy 矩阵上进行操作相比,在NumPy 数组上进行操作要了解的关键事项是:

  • NumPy矩阵是NumPy数组的子类

  • NumPy 数组操作是基于元素的(一旦考虑了广播)

  • NumPy 矩阵运算遵循线性代数的一般规则

一些代码片段来说明:

>>> from numpy import linalg as LA
>>> import numpy as NP

>>> a1 = NP.matrix("4 3 5; 6 7 8; 1 3 13; 7 21 9")
>>> a1
matrix([[ 4,  3,  5],
        [ 6,  7,  8],
        [ 1,  3, 13],
        [ 7, 21,  9]])

>>> a2 = NP.matrix("7 8 15; 5 3 11; 7 4 9; 6 15 4")
>>> a2
matrix([[ 7,  8, 15],
        [ 5,  3, 11],
        [ 7,  4,  9],
        [ 6, 15,  4]])

>>> a1.shape
(4, 3)

>>> a2.shape
(4, 3)

>>> a2t = a2.T
>>> a2t.shape
(3, 4)

>>> a1 * a2t         # same as NP.dot(a1, a2t) 
matrix([[127,  84,  85,  89],
        [218, 139, 142, 173],
        [226, 157, 136, 103],
        [352, 197, 214, 393]])

但是如果将以下两个NumPy矩阵转换为数组,则此操作将失败:

>>> a1 = NP.array(a1)
>>> a2t = NP.array(a2t)

>>> a1 * a2t
Traceback (most recent call last):
   File "<pyshell#277>", line 1, in <module>
   a1 * a2t
   ValueError: operands could not be broadcast together with shapes (4,3) (3,4) 

尽管使用NP.dot语法可以处理数组 ; 该操作类似于矩阵乘法:

>> NP.dot(a1, a2t)
array([[127,  84,  85,  89],
       [218, 139, 142, 173],
       [226, 157, 136, 103],
       [352, 197, 214, 393]])

那么您是否需要NumPy矩阵?即,NumPy数组是否足以进行线性代数计算(前提是您知道正确的语法,即NP.dot)?

规则似乎是,如果参数(数组)的形状(mxn)与给定的线性代数运算兼容,那么您就可以了,否则,NumPy抛出。

我遇到的唯一exceptions(可能还有其他exceptions)是计算矩阵逆

下面是我称为纯线性代数运算(实际上是从Numpy的线性代数模块)并传递给NumPy数组的代码片段

数组的行列式

>>> m = NP.random.randint(0, 10, 16).reshape(4, 4)
>>> m
array([[6, 2, 5, 2],
       [8, 5, 1, 6],
       [5, 9, 7, 5],
       [0, 5, 6, 7]])

>>> type(m)
<type 'numpy.ndarray'>

>>> md = LA.det(m)
>>> md
1772.9999999999995

特征向量/特征值对:

>>> LA.eig(m)
(array([ 19.703+0.j   ,   0.097+4.198j,   0.097-4.198j,   5.103+0.j   ]), 
array([[-0.374+0.j   , -0.091+0.278j, -0.091-0.278j, -0.574+0.j   ],
       [-0.446+0.j   ,  0.671+0.j   ,  0.671+0.j   , -0.084+0.j   ],
       [-0.654+0.j   , -0.239-0.476j, -0.239+0.476j, -0.181+0.j   ],
       [-0.484+0.j   , -0.387+0.178j, -0.387-0.178j,  0.794+0.j   ]]))

矩阵范数

>>>> LA.norm(m)
22.0227

qr因式分解

>>> LA.qr(a1)
(array([[ 0.5,  0.5,  0.5],
        [ 0.5,  0.5, -0.5],
        [ 0.5, -0.5,  0.5],
        [ 0.5, -0.5, -0.5]]), 
 array([[ 6.,  6.,  6.],
        [ 0.,  0.,  0.],
        [ 0.,  0.,  0.]]))

矩阵等级

>>> m = NP.random.rand(40).reshape(8, 5)
>>> m
array([[ 0.545,  0.459,  0.601,  0.34 ,  0.778],
       [ 0.799,  0.047,  0.699,  0.907,  0.381],
       [ 0.004,  0.136,  0.819,  0.647,  0.892],
       [ 0.062,  0.389,  0.183,  0.289,  0.809],
       [ 0.539,  0.213,  0.805,  0.61 ,  0.677],
       [ 0.269,  0.071,  0.377,  0.25 ,  0.692],
       [ 0.274,  0.206,  0.655,  0.062,  0.229],
       [ 0.397,  0.115,  0.083,  0.19 ,  0.701]])
>>> LA.matrix_rank(m)
5

矩阵条件

>>> a1 = NP.random.randint(1, 10, 12).reshape(4, 3)
>>> LA.cond(a1)
5.7093446189400954

反演需要一个NumPy矩阵

>>> a1 = NP.matrix(a1)
>>> type(a1)
<class 'numpy.matrixlib.defmatrix.matrix'>

>>> a1.I
matrix([[ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028]])
>>> a1 = NP.array(a1)
>>> a1.I

Traceback (most recent call last):
   File "<pyshell#230>", line 1, in <module>
   a1.I
   AttributeError: 'numpy.ndarray' object has no attribute 'I'

但是Moore-Penrose伪逆似乎工作得很好

>>> LA.pinv(m)
matrix([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
        [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
        [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
        [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
        [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])

>>> m = NP.array(m)

>>> LA.pinv(m)
array([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
       [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
       [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
       [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
       [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])

the key things to know for operations on NumPy arrays versus operations on NumPy matrices are:

  • NumPy matrix is a subclass of NumPy array

  • NumPy array operations are element-wise (once broadcasting is accounted for)

  • NumPy matrix operations follow the ordinary rules of linear algebra

some code snippets to illustrate:

>>> from numpy import linalg as LA
>>> import numpy as NP

>>> a1 = NP.matrix("4 3 5; 6 7 8; 1 3 13; 7 21 9")
>>> a1
matrix([[ 4,  3,  5],
        [ 6,  7,  8],
        [ 1,  3, 13],
        [ 7, 21,  9]])

>>> a2 = NP.matrix("7 8 15; 5 3 11; 7 4 9; 6 15 4")
>>> a2
matrix([[ 7,  8, 15],
        [ 5,  3, 11],
        [ 7,  4,  9],
        [ 6, 15,  4]])

>>> a1.shape
(4, 3)

>>> a2.shape
(4, 3)

>>> a2t = a2.T
>>> a2t.shape
(3, 4)

>>> a1 * a2t         # same as NP.dot(a1, a2t) 
matrix([[127,  84,  85,  89],
        [218, 139, 142, 173],
        [226, 157, 136, 103],
        [352, 197, 214, 393]])

but this operations fails if these two NumPy matrices are converted to arrays:

>>> a1 = NP.array(a1)
>>> a2t = NP.array(a2t)

>>> a1 * a2t
Traceback (most recent call last):
   File "<pyshell#277>", line 1, in <module>
   a1 * a2t
   ValueError: operands could not be broadcast together with shapes (4,3) (3,4) 

though using the NP.dot syntax works with arrays; this operations works like matrix multiplication:

>> NP.dot(a1, a2t)
array([[127,  84,  85,  89],
       [218, 139, 142, 173],
       [226, 157, 136, 103],
       [352, 197, 214, 393]])

so do you ever need a NumPy matrix? ie, will a NumPy array suffice for linear algebra computation (provided you know the correct syntax, ie, NP.dot)?

the rule seems to be that if the arguments (arrays) have shapes (m x n) compatible with the a given linear algebra operation, then you are ok, otherwise, NumPy throws.

the only exception i have come across (there are likely others) is calculating matrix inverse.

below are snippets in which i have called a pure linear algebra operation (in fact, from Numpy’s Linear Algebra module) and passed in a NumPy array

determinant of an array:

>>> m = NP.random.randint(0, 10, 16).reshape(4, 4)
>>> m
array([[6, 2, 5, 2],
       [8, 5, 1, 6],
       [5, 9, 7, 5],
       [0, 5, 6, 7]])

>>> type(m)
<type 'numpy.ndarray'>

>>> md = LA.det(m)
>>> md
1772.9999999999995

eigenvectors/eigenvalue pairs:

>>> LA.eig(m)
(array([ 19.703+0.j   ,   0.097+4.198j,   0.097-4.198j,   5.103+0.j   ]), 
array([[-0.374+0.j   , -0.091+0.278j, -0.091-0.278j, -0.574+0.j   ],
       [-0.446+0.j   ,  0.671+0.j   ,  0.671+0.j   , -0.084+0.j   ],
       [-0.654+0.j   , -0.239-0.476j, -0.239+0.476j, -0.181+0.j   ],
       [-0.484+0.j   , -0.387+0.178j, -0.387-0.178j,  0.794+0.j   ]]))

matrix norm:

>>>> LA.norm(m)
22.0227

qr factorization:

>>> LA.qr(a1)
(array([[ 0.5,  0.5,  0.5],
        [ 0.5,  0.5, -0.5],
        [ 0.5, -0.5,  0.5],
        [ 0.5, -0.5, -0.5]]), 
 array([[ 6.,  6.,  6.],
        [ 0.,  0.,  0.],
        [ 0.,  0.,  0.]]))

matrix rank:

>>> m = NP.random.rand(40).reshape(8, 5)
>>> m
array([[ 0.545,  0.459,  0.601,  0.34 ,  0.778],
       [ 0.799,  0.047,  0.699,  0.907,  0.381],
       [ 0.004,  0.136,  0.819,  0.647,  0.892],
       [ 0.062,  0.389,  0.183,  0.289,  0.809],
       [ 0.539,  0.213,  0.805,  0.61 ,  0.677],
       [ 0.269,  0.071,  0.377,  0.25 ,  0.692],
       [ 0.274,  0.206,  0.655,  0.062,  0.229],
       [ 0.397,  0.115,  0.083,  0.19 ,  0.701]])
>>> LA.matrix_rank(m)
5

matrix condition:

>>> a1 = NP.random.randint(1, 10, 12).reshape(4, 3)
>>> LA.cond(a1)
5.7093446189400954

inversion requires a NumPy matrix though:

>>> a1 = NP.matrix(a1)
>>> type(a1)
<class 'numpy.matrixlib.defmatrix.matrix'>

>>> a1.I
matrix([[ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028]])
>>> a1 = NP.array(a1)
>>> a1.I

Traceback (most recent call last):
   File "<pyshell#230>", line 1, in <module>
   a1.I
   AttributeError: 'numpy.ndarray' object has no attribute 'I'

but the Moore-Penrose pseudoinverse seems to works just fine

>>> LA.pinv(m)
matrix([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
        [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
        [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
        [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
        [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])

>>> m = NP.array(m)

>>> LA.pinv(m)
array([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
       [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
       [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
       [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
       [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])

回答 2

在3.5中,Python终于有了一个矩阵乘法运算符。语法为a @ b

In 3.5, Python finally got a matrix multiplication operator. The syntax is a @ b.


回答 3

在处理数组和处理矩阵时,点运算符会给出不同的答案。例如,假设以下内容:

>>> a=numpy.array([1, 2, 3])
>>> b=numpy.array([1, 2, 3])

让我们将它们转换成矩阵:

>>> am=numpy.mat(a)
>>> bm=numpy.mat(b)

现在,我们可以看到两种情况的不同输出:

>>> print numpy.dot(a.T, b)
14
>>> print am.T*bm
[[1.  2.  3.]
 [2.  4.  6.]
 [3.  6.  9.]]

There is a situation where the dot operator will give different answers when dealing with arrays as with dealing with matrices. For example, suppose the following:

>>> a=numpy.array([1, 2, 3])
>>> b=numpy.array([1, 2, 3])

Lets convert them into matrices:

>>> am=numpy.mat(a)
>>> bm=numpy.mat(b)

Now, we can see a different output for the two cases:

>>> print numpy.dot(a.T, b)
14
>>> print am.T*bm
[[1.  2.  3.]
 [2.  4.  6.]
 [3.  6.  9.]]

回答 4

来自http://docs.scipy.org/doc/scipy/reference/tutorial/linalg.html的参考

…,使用的numpy.matrix气馁,因为它增加了什么,无法与2D来完成numpy.ndarray对象,并可能导致混乱,其中正在使用的类。例如,

>>> import numpy as np
>>> from scipy import linalg
>>> A = np.array([[1,2],[3,4]])
>>> A
    array([[1, 2],
           [3, 4]])
>>> linalg.inv(A)
array([[-2. ,  1. ],
      [ 1.5, -0.5]])
>>> b = np.array([[5,6]]) #2D array
>>> b
array([[5, 6]])
>>> b.T
array([[5],
      [6]])
>>> A*b #not matrix multiplication!
array([[ 5, 12],
      [15, 24]])
>>> A.dot(b.T) #matrix multiplication
array([[17],
      [39]])
>>> b = np.array([5,6]) #1D array
>>> b
array([5, 6])
>>> b.T  #not matrix transpose!
array([5, 6])
>>> A.dot(b)  #does not matter for multiplication
array([17, 39])

scipy.linalg操作可以同等地应用于numpy.matrix或2D numpy.ndarray对象。

Reference from http://docs.scipy.org/doc/scipy/reference/tutorial/linalg.html

…, the use of the numpy.matrix class is discouraged, since it adds nothing that cannot be accomplished with 2D numpy.ndarray objects, and may lead to a confusion of which class is being used. For example,

>>> import numpy as np
>>> from scipy import linalg
>>> A = np.array([[1,2],[3,4]])
>>> A
    array([[1, 2],
           [3, 4]])
>>> linalg.inv(A)
array([[-2. ,  1. ],
      [ 1.5, -0.5]])
>>> b = np.array([[5,6]]) #2D array
>>> b
array([[5, 6]])
>>> b.T
array([[5],
      [6]])
>>> A*b #not matrix multiplication!
array([[ 5, 12],
      [15, 24]])
>>> A.dot(b.T) #matrix multiplication
array([[17],
      [39]])
>>> b = np.array([5,6]) #1D array
>>> b
array([5, 6])
>>> b.T  #not matrix transpose!
array([5, 6])
>>> A.dot(b)  #does not matter for multiplication
array([17, 39])

scipy.linalg operations can be applied equally to numpy.matrix or to 2D numpy.ndarray objects.


回答 5

这个技巧可能就是您想要的。这是一种简单的运算符重载。

然后,您可以使用类似建议的Infix类的东西:

a = np.random.rand(3,4)
b = np.random.rand(4,3)
x = Infix(lambda x,y: np.dot(x,y))
c = a |x| b

This trick could be what you are looking for. It is a kind of simple operator overload.

You can then use something like the suggested Infix class like this:

a = np.random.rand(3,4)
b = np.random.rand(4,3)
x = Infix(lambda x,y: np.dot(x,y))
c = a |x| b

回答 6

来自PEP 465的相关报价 @ petr-viktorin提到的用于矩阵乘法的专用中缀运算符,阐明了OP遇到的问题:

numpy提供了两种使用不同__mul__方法的不同类型。对于numpy.ndarray对象,*执行元素乘法,矩阵乘法必须使用函数调用(numpy.dot)。对于numpy.matrix对象,*执行矩阵乘法,而元素乘法则需要函数语法。使用编写代码numpy.ndarray效果很好。使用编写代码numpy.matrix也可以。但是,一旦我们尝试将这两段代码集成在一起,麻烦就会开始。预期为ndarray并得到matrix或相反的代码可能会崩溃或返回错误的结果

@infix运算符的引入应有助于统一和简化python矩阵代码。

A pertinent quote from PEP 465 – A dedicated infix operator for matrix multiplication , as mentioned by @petr-viktorin, clarifies the problem the OP was getting at:

[…] numpy provides two different types with different __mul__ methods. For numpy.ndarray objects, * performs elementwise multiplication, and matrix multiplication must use a function call (numpy.dot). For numpy.matrix objects, * performs matrix multiplication, and elementwise multiplication requires function syntax. Writing code using numpy.ndarray works fine. Writing code using numpy.matrix also works fine. But trouble begins as soon as we try to integrate these two pieces of code together. Code that expects an ndarray and gets a matrix, or vice-versa, may crash or return incorrect results

The introduction of the @ infix operator should help to unify and simplify python matrix code.


回答 7

函数matmul(自numpy 1.10.1起)对两种类型均适用,并以numpy矩阵类返回结果:

import numpy as np

A = np.mat('1 2 3; 4 5 6; 7 8 9; 10 11 12')
B = np.array(np.mat('1 1 1 1; 1 1 1 1; 1 1 1 1'))
print (A, type(A))
print (B, type(B))

C = np.matmul(A, B)
print (C, type(C))

输出:

(matrix([[ 1,  2,  3],
        [ 4,  5,  6],
        [ 7,  8,  9],
        [10, 11, 12]]), <class 'numpy.matrixlib.defmatrix.matrix'>)
(array([[1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]]), <type 'numpy.ndarray'>)
(matrix([[ 6,  6,  6,  6],
        [15, 15, 15, 15],
        [24, 24, 24, 24],
        [33, 33, 33, 33]]), <class 'numpy.matrixlib.defmatrix.matrix'>)

由于python 3.5 如前所述,您还可以使用新的矩阵乘法运算符,@例如

C = A @ B

并获得与上述相同的结果。

Function matmul (since numpy 1.10.1) works fine for both types and return result as a numpy matrix class:

import numpy as np

A = np.mat('1 2 3; 4 5 6; 7 8 9; 10 11 12')
B = np.array(np.mat('1 1 1 1; 1 1 1 1; 1 1 1 1'))
print (A, type(A))
print (B, type(B))

C = np.matmul(A, B)
print (C, type(C))

Output:

(matrix([[ 1,  2,  3],
        [ 4,  5,  6],
        [ 7,  8,  9],
        [10, 11, 12]]), <class 'numpy.matrixlib.defmatrix.matrix'>)
(array([[1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]]), <type 'numpy.ndarray'>)
(matrix([[ 6,  6,  6,  6],
        [15, 15, 15, 15],
        [24, 24, 24, 24],
        [33, 33, 33, 33]]), <class 'numpy.matrixlib.defmatrix.matrix'>)

Since python 3.5 as mentioned early you also can use a new matrix multiplication operator @ like

C = A @ B

and get the same result as above.


Python中漂亮的图形和图表[关闭]

问题:Python中漂亮的图形和图表[关闭]

有哪些可用的库在Python应用程序中创建漂亮的图表和图形?

What are the available libraries for creating pretty charts and graphs in a Python application?


回答 0

我是支持CairoPlot的人,对此感到非常自豪。matplotlib当然很棒,但是我相信CairoPlot会更好看。因此,对于演示文稿和网站,这是一个很好的选择。

今天,我发布了1.1版。如果有兴趣,请在CairoPlot v1.1中进行检查

编辑:经过漫长而寒冷的冬天后,CairoPlot再次被开发。在GitHub上检查新版本。

I’m the one supporting CairoPlot and I’m very proud it came up here. Surely matplotlib is great, but I believe CairoPlot is better looking. So, for presentations and websites, it’s a very good choice.

Today I released version 1.1. If interested, check it out at CairoPlot v1.1

EDIT: After a long and cold winter, CairoPlot is being developed again. Check out the new version on GitHub.


回答 1

对于交互工作,Matplotlib是成熟的标准。它提供了OO风格的API以及Matlab风格的交互式API。

Chaco是Enthought的人们提供的更现代的绘图库。它使用Enthought的Kiva矢量绘图库,目前仅在OpenGL的情况下可用于Wx和Qt(Matplotlib具有Tk,Qt,Wx,Cocoa的后端,以及许多图像类型,如PDF,EPS,PNG等)。Chaco的主要优点是其相对于Matplotlib的速度以及与Enthought的用于交互应用程序的Traits API的集成。

For interactive work, Matplotlib is the mature standard. It provides an OO-style API as well as a Matlab-style interactive API.

Chaco is a more modern plotting library from the folks at Enthought. It uses Enthought’s Kiva vector drawing library and currently works only with Wx and Qt with OpenGL on the way (Matplotlib has backends for Tk, Qt, Wx, Cocoa, and many image types such as PDF, EPS, PNG, etc.). The main advantages of Chaco are its speed relative to Matplotlib and its integration with Enthought’s Traits API for interactive applications.


回答 2

您也可以使用使用Google Chart API的pygooglechart。这不是您始终想要使用的东西,但是如果您想要少量的简单,良好的图表并且始终在线,尤其是如果您仍在浏览器中显示,则这是一个不错的选择。

You can also use pygooglechart, which uses the Google Chart API. This isn’t something you’d always want to use, but if you want a small number of good, simple, charts, and are always online, and especially if you’re displaying in a browser anyway, it’s a good choice.


回答 3

您没有提到所需的输出格式,但是reportlab擅长以pdf和位图(例如png)格式创建图表。

这是png和pdf格式的条形图的简单示例:

from reportlab.graphics.shapes import Drawing
from reportlab.graphics.charts.barcharts import VerticalBarChart

d = Drawing(300, 200)

chart = VerticalBarChart()
chart.width = 260
chart.height = 160
chart.x = 20
chart.y = 20
chart.data = [[1,2], [3,4]]
chart.categoryAxis.categoryNames = ['foo', 'bar']
chart.valueAxis.valueMin = 0

d.add(chart)
d.save(fnRoot='test', formats=['png', 'pdf'])

替代文字http://i40.tinypic.com/2j677tl.jpg

注意:图像主机已将图像转换为jpg。

You didn’t mention what output format you need but reportlab is good at creating charts both in pdf and bitmap (e.g. png) format.

Here is a simple example of a barchart in png and pdf format:

from reportlab.graphics.shapes import Drawing
from reportlab.graphics.charts.barcharts import VerticalBarChart

d = Drawing(300, 200)

chart = VerticalBarChart()
chart.width = 260
chart.height = 160
chart.x = 20
chart.y = 20
chart.data = [[1,2], [3,4]]
chart.categoryAxis.categoryNames = ['foo', 'bar']
chart.valueAxis.valueMin = 0

d.add(chart)
d.save(fnRoot='test', formats=['png', 'pdf'])

alt text http://i40.tinypic.com/2j677tl.jpg

Note: the image has been converted to jpg by the image host.


回答 4


回答 5

我使用pychart,并认为它非常简单。

http://home.gna.org/pychart/

它全部是本机python,没有大量的依赖项。我确定matplotlib很可爱,但是我要下载和安装几天,我只想要一张measley条形图!

它似乎几年没有更新,但是嘿,它起作用了!

I used pychart and thought it was very straightforward.

http://home.gna.org/pychart/

It’s all native python and does not have a busload of dependencies. I’m sure matplotlib is lovely but I’d be downloading and installing for days and I just want one measley bar chart!

It doesn’t seem to have been updated in a few years but hey it works!


回答 6

您是否研究过适用于Python的ChartDirector

我不能谈论这个,但是我已经将ChartDirector用于PHP,这非常好。

Have you looked into ChartDirector for Python?

I can’t speak about this one, but I’ve used ChartDirector for PHP and it’s pretty good.


回答 7

NodeBox非常适合创建原始图形。

NodeBox is awesome for raw graphics creation.


回答 8

如果您想使用gnuplot进行绘图,则应考虑使用Gnuplot.py。它为gnuplot提供了一个面向对象的界面,还允许您将命令直接传递给gnuplot。不幸的是,它不再被积极开发。

If you like to use gnuplot for plotting, you should consider Gnuplot.py. It provides an object-oriented interface to gnuplot, and also allows you to pass commands directly to gnuplot. Unfortunately, it is no longer being actively developed.


回答 9

查科enthought是另一种选择

Chaco from enthought is another option


回答 10

您还应该考虑使用PyCha http://www.lorenzogil.com/projects/pycha/


回答 11

我是PyOFC2的粉丝: http

它只是一个软件包,可轻松生成非常漂亮的Open Flash Charts 2所需的JSON数据。在上面的链接上查看示例。

I am a fan on PyOFC2 : http://btbytes.github.com/pyofc2/

It just just a package that makes it easy to generate the JSON data needed for Open Flash Charts 2, which are very beautiful. Check out the examples on the link above.


回答 12

请查看嵌入WHIFF的Open Flash Chart http://aaron.oirt.rutgers.edu/myapp/docs/W1100_1600.openFlashCharts 和嵌入WHIFF的amCharts http://aaron.oirt.rutgers.edu/myapp/ amcharts / doc。谢谢。

Please look at the Open Flash Chart embedding for WHIFF http://aaron.oirt.rutgers.edu/myapp/docs/W1100_1600.openFlashCharts and the amCharts embedding for WHIFF too http://aaron.oirt.rutgers.edu/myapp/amcharts/doc. Thanks.


回答 13

您也可以考虑使用Google图表

从技术上讲,这不是python API,但是您可以从python中使用它,它的编码速度相当快,并且结果看起来不错。如果您碰巧正在在线使用地块,那么这将是一个更好的解决方案。

You could also consider google charts.

Not technically a python API, but you can use it from python, it’s reasonably fast to code for, and the results tend to look nice. If you happen to be using your plots online, then this would be an even better solution.


回答 14

PLplot是用于创建科学图的跨平台软件包。它们不是很漂亮(引人注目),但看起来足够好。看一些例子(源代码和图片)。

PLplot核心库可用于创建标准的xy图,半对数图,对数对数图,轮廓图,3D表面图,网格图,条形图和饼图。它可以在Windows(2000,XP和Vista),Linux,Mac OS X和其他Unices上运行。

PLplot is a cross-platform software package for creating scientific plots. They aren’t very pretty (eye catching), but they look good enough. Have a look at some examples (both source code and pictures).

The PLplot core library can be used to create standard x-y plots, semi-log plots, log-log plots, contour plots, 3D surface plots, mesh plots, bar charts and pie charts. It runs on Windows (2000, XP and Vista), Linux, Mac OS X, and other Unices.


泡泡排序作业

问题:泡泡排序作业

在课堂上,我们正在做排序算法,尽管我在谈论它们并编写伪代码时理解得很好,但是我在为它们编写实际代码时遇到了问题。

这是我在Python中的尝试:

mylist = [12, 5, 13, 8, 9, 65]

def bubble(badList):
    length = len(badList) - 1
    unsorted = True

    while unsorted:
        for element in range(0,length):
            unsorted = False
            if badList[element] > badList[element + 1]:
                hold = badList[element + 1]
                badList[element + 1] = badList[element]
                badList[element] = hold
                print badList
            else:
                unsorted = True

print bubble(mylist)

现在,(据我所知)该排序正确,但是一旦完成,它就会无限循环。

如何修复此代码,以便函数正确完成并正确排序任何(合理)大小的列表?

PS:我知道我不应该在函数中真正打印输出,而应该得到返回值,但是我还没有这样做,因为我的代码还没有真正起作用。

In class we are doing sorting algorithms and, although I understand them fine when talking about them and writing pseudocode, I am having problems writing actual code for them.

This is my attempt in Python:

mylist = [12, 5, 13, 8, 9, 65]

def bubble(badList):
    length = len(badList) - 1
    unsorted = True

    while unsorted:
        for element in range(0,length):
            unsorted = False
            if badList[element] > badList[element + 1]:
                hold = badList[element + 1]
                badList[element + 1] = badList[element]
                badList[element] = hold
                print badList
            else:
                unsorted = True

print bubble(mylist)

Now, this (as far as I can tell) sorts correctly, but once it finishes it just loops indefinitely.

How can this code be fixed so the function finishes properly and correctly sorts a list of any (reasonable) size?

P.S. I know I should not really have prints in a function and I should have a return, but I just have not done that yet as my code does not really work yet.


回答 0

为了解释为什么您的脚本现在无法正常工作,我将变量重命名unsortedsorted

首先,您的列表尚未排序。当然,我们设置sortedFalse

一旦开始while循环,我们就假定该列表已经排序。想法是这样的:一旦找到两个顺序不正确的元素,便将其设置sortedFalse只有在没有顺序错误的元素时sorted才会保留。True

sorted = False  # We haven't started sorting yet

while not sorted:
    sorted = True  # Assume the list is now sorted
    for element in range(0, length):
        if badList[element] > badList[element + 1]:
            sorted = False  # We found two elements in the wrong order
            hold = badList[element + 1]
            badList[element + 1] = badList[element]
            badList[element] = hold
    # We went through the whole list. At this point, if there were no elements
    # in the wrong order, sorted is still True. Otherwise, it's false, and the
    # while loop executes again.

还有一些小问题,可以帮助提高代码的效率或可读性。

  • for循环中,使用变量element。从技术上讲,element不是元素;它是代表列表索引的数字。而且,它很长。在这些情况下,只需使用一个临时变量名即可,例如i“ index”。

    for i in range(0, length):
  • range命令还可以仅接受一个参数(名为stop)。在这种情况下,您将获得从0到该参数的所有整数的列表。

    for i in range(length):
  • Python样式指南》建议使用下划线将变量命名为小写。对于这样的小脚本,这是一个很小的缺点。它会让您习惯于Python代码最常类似于的东西。

    def bubble(bad_list):
  • 要交换两个变量的值,请将它们写为元组分配。右侧被评估为元组(例如(badList[i+1], badList[i])is (3, 5)),然后被分配给左侧的两个变量((badList[i], badList[i+1]))。

    bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]

放在一起,你会得到:

my_list = [12, 5, 13, 8, 9, 65]

def bubble(bad_list):
    length = len(bad_list) - 1
    sorted = False

    while not sorted:
        sorted = True
        for i in range(length):
            if bad_list[i] > bad_list[i+1]:
                sorted = False
                bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]

bubble(my_list)
print my_list

(顺便说一句,我也删除了您的打印声明。)

To explain why your script isn’t working right now, I’ll rename the variable unsorted to sorted.

At first, your list isn’t yet sorted. Of course, we set sorted to False.

As soon as we start the while loop, we assume that the list is already sorted. The idea is this: as soon as we find two elements that are not in the right order, we set sorted back to False. sorted will remain True only if there were no elements in the wrong order.

sorted = False  # We haven't started sorting yet

while not sorted:
    sorted = True  # Assume the list is now sorted
    for element in range(0, length):
        if badList[element] > badList[element + 1]:
            sorted = False  # We found two elements in the wrong order
            hold = badList[element + 1]
            badList[element + 1] = badList[element]
            badList[element] = hold
    # We went through the whole list. At this point, if there were no elements
    # in the wrong order, sorted is still True. Otherwise, it's false, and the
    # while loop executes again.

There are also minor little issues that would help the code be more efficient or readable.

  • In the for loop, you use the variable element. Technically, element is not an element; it’s a number representing a list index. Also, it’s quite long. In these cases, just use a temporary variable name, like i for “index”.

    for i in range(0, length):
    
  • The range command can also take just one argument (named stop). In that case, you get a list of all the integers from 0 to that argument.

    for i in range(length):
    
  • The Python Style Guide recommends that variables be named in lowercase with underscores. This is a very minor nitpick for a little script like this; it’s more to get you accustomed to what Python code most often resembles.

    def bubble(bad_list):
    
  • To swap the values of two variables, write them as a tuple assignment. The right hand side gets evaluated as a tuple (say, (badList[i+1], badList[i]) is (3, 5)) and then gets assigned to the two variables on the left hand side ((badList[i], badList[i+1])).

    bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]
    

Put it all together, and you get this:

my_list = [12, 5, 13, 8, 9, 65]

def bubble(bad_list):
    length = len(bad_list) - 1
    sorted = False

    while not sorted:
        sorted = True
        for i in range(length):
            if bad_list[i] > bad_list[i+1]:
                sorted = False
                bad_list[i], bad_list[i+1] = bad_list[i+1], bad_list[i]

bubble(my_list)
print my_list

(I removed your print statement too, by the way.)


回答 1

气泡分类的目的是在每轮底部移动较重的项目,同时向上移动较轻的项目。在内部循环中,您可以比较元素,而不必每次都迭代整个列表。将最重的已经排在最后。该交换变量是一个额外的检查,所以我们可以标记列表现在可以正确排序,并避免不必要的计算仍在继续。

def bubble(badList):
    length = len(badList)
    for i in range(0,length):
        swapped = False
        for element in range(0, length-i-1):
            if badList[element] > badList[element + 1]:
                hold = badList[element + 1]
                badList[element + 1] = badList[element]
                badList[element] = hold
                swapped = True
        if not swapped: break

    return badList

您的版本1已更正:

def bubble(badList):
    length = len(badList) - 1
    unsorted = True
    while unsorted:
        unsorted = False
        for element in range(0,length):
            #unsorted = False
            if badList[element] > badList[element + 1]:
                 hold = badList[element + 1]
                 badList[element + 1] = badList[element]
                 badList[element] = hold
                 unsorted = True
                 #print badList
             #else:
                 #unsorted = True

     return badList

The goal of bubble sort is to move the heavier items at the bottom in each round, while moving the lighter items up. In the inner loop, where you compare the elements, you don’t have to iterate the whole list in each turn. The heaviest is already placed last. The swapped variable is an extra check so we can mark that the list is now sorted and avoid continuing with unnecessary calculations.

def bubble(badList):
    length = len(badList)
    for i in range(0,length):
        swapped = False
        for element in range(0, length-i-1):
            if badList[element] > badList[element + 1]:
                hold = badList[element + 1]
                badList[element + 1] = badList[element]
                badList[element] = hold
                swapped = True
        if not swapped: break

    return badList

Your version 1, corrected:

def bubble(badList):
    length = len(badList) - 1
    unsorted = True
    while unsorted:
        unsorted = False
        for element in range(0,length):
            #unsorted = False
            if badList[element] > badList[element + 1]:
                 hold = badList[element + 1]
                 badList[element + 1] = badList[element]
                 badList[element] = hold
                 unsorted = True
                 #print badList
             #else:
                 #unsorted = True

     return badList

回答 2

当您使用负含义的变量名时,您需要反转它们的值,这就是发生的情况。以下内容将更容易理解:

sorted = False
while not sorted:
    ...

另一方面,算法的逻辑有点偏离。您需要检查在for循环期间是否交换了两个元素。这是我的写法:

def bubble(values):
    length = len(values) - 1
    sorted = False
    while not sorted:
        sorted = True
        for element in range(0,length):
            if values[element] > values[element + 1]:
                 hold = values[element + 1]
                 values[element + 1] = values[element]
                 values[element] = hold
                 sorted = False
    return values

This is what happens when you use variable name of negative meaning, you need to invert their values. The following would be easier to understand:

sorted = False
while not sorted:
    ...

On the other hand, the logic of the algorithm is a little bit off. You need to check whether two elements swapped during the for loop. Here’s how I would write it:

def bubble(values):
    length = len(values) - 1
    sorted = False
    while not sorted:
        sorted = True
        for element in range(0,length):
            if values[element] > values[element + 1]:
                 hold = values[element + 1]
                 values[element + 1] = values[element]
                 values[element] = hold
                 sorted = False
    return values

回答 3

您对Unsorted变量的使用是错误的;您想要一个变量来告诉您是否交换了两个元素;如果这样做,则可以退出循环,否则,需要再次循环。要解决您在此处遇到的问题,只需在if情况的正文中输入“ unsorted = false”;删除其他情况;并在for循环前添加“ unsorted = true” 。

Your use of the Unsorted variable is wrong; you want to have a variable that tells you if you have swapped two elements; if you have done that, you can exit your loop, otherwise, you need to loop again. To fix what you’ve got here, just put the “unsorted = false” in the body of your if case; remove your else case; and put “unsorted = true before your for loop.


回答 4

def bubble_sort(l):
    for passes_left in range(len(l)-1, 0, -1):
        for index in range(passes_left):
            if l[index] < l[index + 1]:
               l[index], l[index + 1] = l[index + 1], l[index]
    return l
def bubble_sort(l):
    for passes_left in range(len(l)-1, 0, -1):
        for index in range(passes_left):
            if l[index] < l[index + 1]:
               l[index], l[index + 1] = l[index + 1], l[index]
    return l

回答 5

#一个非常简单的函数,可以通过减少第二个数组的问题空间来优化(显然)。但是相同的O(n ^ 2)复杂度。

def bubble(arr):
    l = len(arr)        
    for a in range(l):
        for b in range(l-1):
            if (arr[a] < arr[b]):
            arr[a], arr[b] = arr[b], arr[a]
    return arr 

#A very simple function, can be optimized (obviously) by decreasing the problem space of the 2nd array. But same O(n^2) complexity.

def bubble(arr):
    l = len(arr)        
    for a in range(l):
        for b in range(l-1):
            if (arr[a] < arr[b]):
            arr[a], arr[b] = arr[b], arr[a]
    return arr 

回答 6

您那里有几个错误。第一个是长度,第二个是您未排序的使用(如McWafflestix所述)。如果要打印列表,您可能还想返回该列表:

mylist = [12, 5, 13, 8, 9, 65]

def bubble(badList):
    length = len(badList) - 2
    unsorted = True

    while unsorted:
        for element in range(0,length):
            unsorted = False

            if badList[element] > badList[element + 1]:
                hold = badList[element + 1]
                badList[element + 1] = badList[element]
                badList[element] = hold
                print badList
                unsorted = True

    return badList

print bubble(mylist)

埃塔:你是对的,上面的小车就像地狱一样。我的缺点是不通过更多示例进行测试。

def bubble2(badList):
    swapped = True
    length = len(badList) - 2

    while swapped:
        swapped = False
        for i in range(0, length):
            if badList[i] > badList[i + 1]:

                # swap
                hold = badList[i + 1]
                badList[i + 1] = badList[i]
                badList[i] = hold

                swapped = True

    return badList

You’ve got a couple of errors in there. The first is in length, and the second is in your use of unsorted (as stated by McWafflestix). You probably also want to return the list if you’re going to print it:

mylist = [12, 5, 13, 8, 9, 65]

def bubble(badList):
    length = len(badList) - 2
    unsorted = True

    while unsorted:
        for element in range(0,length):
            unsorted = False

            if badList[element] > badList[element + 1]:
                hold = badList[element + 1]
                badList[element + 1] = badList[element]
                badList[element] = hold
                print badList
                unsorted = True

    return badList

print bubble(mylist)

eta: You’re right, the above is buggy as hell. My bad for not testing through some more examples.

def bubble2(badList):
    swapped = True
    length = len(badList) - 2

    while swapped:
        swapped = False
        for i in range(0, length):
            if badList[i] > badList[i + 1]:

                # swap
                hold = badList[i + 1]
                badList[i + 1] = badList[i]
                badList[i] = hold

                swapped = True

    return badList

回答 7

我是一个新鲜的初学者,昨天开始阅读有关Python的文章。受到您的榜样的启发,我创作了一些也许具有80年代风格的作品,但仍然可以

lista1 = [12, 5, 13, 8, 9, 65]

i=0
while i < len(lista1)-1:
    if lista1[i] > lista1[i+1]:
        x = lista1[i]
        lista1[i] = lista1[i+1]
        lista1[i+1] = x
        i=0
        continue
    else:
        i+=1

print(lista1)

I am a fresh fresh beginner, started to read about Python yesterday. Inspired by your example I created something maybe more in the 80-ties style, but nevertheless it kinda works

lista1 = [12, 5, 13, 8, 9, 65]

i=0
while i < len(lista1)-1:
    if lista1[i] > lista1[i+1]:
        x = lista1[i]
        lista1[i] = lista1[i+1]
        lista1[i+1] = x
        i=0
        continue
    else:
        i+=1

print(lista1)

回答 8

原始算法的问题在于,如果列表中的数字更小,它将无法将其带到正确的排序位置。程序每次都需要从头开始,以确保数字始终排序。

我简化了代码,它现在适用于任何数字列表,而不管列表如何,即使有重复的数字也是如此。这是代码

mylist = [9, 8, 5, 4, 12, 1, 7, 5, 2]
print mylist

def bubble(badList):
    length = len(badList) - 1
    element = 0
    while element < length:
        if badList[element] > badList[element + 1]:
            hold = badList[element + 1]
            badList[element + 1] = badList[element]
            badList[element] = hold
            element = 0
            print badList
        else:
            element = element + 1

print bubble(mylist)

The problem with the original algorithm is that if you had a lower number further in the list, it would not bring it to the correct sorted position. The program needs to go back the the beginning each time to ensure that the numbers sort all the way through.

I simplified the code and it will now work for any list of numbers regardless of the list and even if there are repeating numbers. Here’s the code

mylist = [9, 8, 5, 4, 12, 1, 7, 5, 2]
print mylist

def bubble(badList):
    length = len(badList) - 1
    element = 0
    while element < length:
        if badList[element] > badList[element + 1]:
            hold = badList[element + 1]
            badList[element + 1] = badList[element]
            badList[element] = hold
            element = 0
            print badList
        else:
            element = element + 1

print bubble(mylist)

回答 9

def bubble_sort(l):
    exchanged = True
    iteration = 0
    n = len(l)

    while(exchanged):
        iteration += 1
        exchanged = False

        # Move the largest element to the end of the list
        for i in range(n-1):
            if l[i] > l[i+1]:
                exchanged = True
                l[i], l[i+1] = l[i+1], l[i]
        n -= 1   # Largest element already towards the end

    print 'Iterations: %s' %(iteration)
    return l
def bubble_sort(l):
    exchanged = True
    iteration = 0
    n = len(l)

    while(exchanged):
        iteration += 1
        exchanged = False

        # Move the largest element to the end of the list
        for i in range(n-1):
            if l[i] > l[i+1]:
                exchanged = True
                l[i], l[i+1] = l[i+1], l[i]
        n -= 1   # Largest element already towards the end

    print 'Iterations: %s' %(iteration)
    return l

回答 10

def bubbleSort(alist):
if len(alist) <= 1:
    return alist
for i in range(0,len(alist)):
   print "i is :%d",i
   for j in range(0,i):
      print "j is:%d",j
      print "alist[i] is :%d, alist[j] is :%d"%(alist[i],alist[j])
      if alist[i] > alist[j]:
         alist[i],alist[j] = alist[j],alist[i]
return alist

alist = [54,26,93,17,77,31,44,55,20,-23,-34,16,11,11,11]

打印bubbleSort(alist)

def bubbleSort(alist):
if len(alist) <= 1:
    return alist
for i in range(0,len(alist)):
   print "i is :%d",i
   for j in range(0,i):
      print "j is:%d",j
      print "alist[i] is :%d, alist[j] is :%d"%(alist[i],alist[j])
      if alist[i] > alist[j]:
         alist[i],alist[j] = alist[j],alist[i]
return alist

alist = [54,26,93,17,77,31,44,55,20,-23,-34,16,11,11,11]

print bubbleSort(alist)


回答 11

def bubble_sort(a):
    t = 0
    sorted = False # sorted = False because we have not began to sort
    while not sorted:
    sorted = True # Assume sorted = True first, it will switch only there is any change
        for key in range(1,len(a)):
            if a[key-1] > a[key]:
                sorted = False
                t = a[key-1]; a[key-1] = a[key]; a[key] = t;
    print a
def bubble_sort(a):
    t = 0
    sorted = False # sorted = False because we have not began to sort
    while not sorted:
    sorted = True # Assume sorted = True first, it will switch only there is any change
        for key in range(1,len(a)):
            if a[key-1] > a[key]:
                sorted = False
                t = a[key-1]; a[key-1] = a[key]; a[key] = t;
    print a

回答 12

一个简单的例子:

a = len(alist)-1
while a > 0:
    for b in range(0,a):
        #compare with the adjacent element
        if alist[b]>=alist[b+1]:
            #swap both elements
            alist[b], alist[b+1] = alist[b+1], alist[b]
    a-=1

这只是简单地将元素从0到a(基本上是该回合中所有未排序的元素),并将其与相邻元素进行比较,如果大于相邻元素,则进行交换。在回合结束时,对最后一个元素进行排序,然后在没有该元素的情况下再次运行该过程,直到对所有元素进行了排序。

不需要条件是否sort成立。

请注意,此算法仅在交换时才考虑数字的位置,因此重复的数字不会对其产生影响。

PS。我知道这个问题发布已经很长时间了,但是我只想分享这个想法。

A simpler example:

a = len(alist)-1
while a > 0:
    for b in range(0,a):
        #compare with the adjacent element
        if alist[b]>=alist[b+1]:
            #swap both elements
            alist[b], alist[b+1] = alist[b+1], alist[b]
    a-=1

This simply takes the elements from 0 to a(basically, all the unsorted elements in that round) and compares it with its adjacent element, and making a swap if it is greater than its adjacent element. At the end the round, the last element is sorted, and the process runs again without it, until all elements have been sorted.

There is no need for a condition whether sort is true or not.

Note that this algorithm takes into consideration the position of the numbers only when swapping, so repeated numbers will not affect it.

PS. I know it has been very long since this question was posted, but I just wanted to share this idea.


回答 13

arr = [5,4,3,1,6,8,10,9] # array not sorted

for i in range(len(arr)):
    for j in range(i, len(arr)):
        if(arr[i] > arr[j]):
            arr[i], arr[j] = arr[j], arr[i]

            print (arr)
arr = [5,4,3,1,6,8,10,9] # array not sorted

for i in range(len(arr)):
    for j in range(i, len(arr)):
        if(arr[i] > arr[j]):
            arr[i], arr[j] = arr[j], arr[i]

            print (arr)

回答 14

def bubble_sort(li):
    l = len(li)
    tmp = None
    sorted_l = sorted(li)
    while (li != sorted_l):
        for ele in range(0,l-1):
            if li[ele] > li[ele+1]:
                tmp = li[ele+1]
                li[ele+1] = li [ele]
                li[ele] = tmp
    return li
def bubble_sort(li):
    l = len(li)
    tmp = None
    sorted_l = sorted(li)
    while (li != sorted_l):
        for ele in range(0,l-1):
            if li[ele] > li[ele+1]:
                tmp = li[ele+1]
                li[ele+1] = li [ele]
                li[ele] = tmp
    return li

回答 15

def bubbleSort ( arr ):
    swapped = True 
    length = len ( arr )
    j = 0

    while swapped:
        swapped = False
        j += 1 
        for i in range ( length  - j ):
            if arr [ i ] > arr [ i + 1 ]:
                # swap
                tmp = arr [ i ]
                arr [ i ] = arr [ i + 1]
                arr [ i + 1 ] = tmp 

                swapped = True

if __name__ == '__main__':
    # test list
    a = [ 67, 45, 39, -1, -5, -44 ];

    print ( a )
    bubbleSort ( a )
    print ( a )
def bubbleSort ( arr ):
    swapped = True 
    length = len ( arr )
    j = 0

    while swapped:
        swapped = False
        j += 1 
        for i in range ( length  - j ):
            if arr [ i ] > arr [ i + 1 ]:
                # swap
                tmp = arr [ i ]
                arr [ i ] = arr [ i + 1]
                arr [ i + 1 ] = tmp 

                swapped = True

if __name__ == '__main__':
    # test list
    a = [ 67, 45, 39, -1, -5, -44 ];

    print ( a )
    bubbleSort ( a )
    print ( a )

回答 16

def bubblesort(array):
    for i in range(len(array)-1):
        for j in range(len(array)-1-i):
            if array[j] > array[j+1]:
                array[j], array[j+1] = array[j+1], array[j]
    return(array)

print(bubblesort([3,1,6,2,5,4]))
def bubblesort(array):
    for i in range(len(array)-1):
        for j in range(len(array)-1-i):
            if array[j] > array[j+1]:
                array[j], array[j+1] = array[j+1], array[j]
    return(array)

print(bubblesort([3,1,6,2,5,4]))

回答 17

我考虑添加我的解决方案,因为这里曾经有解决方案

  1. 更长的时间
  2. 更大的空间复杂度
  3. 或进行过多的操作

那应该是

所以,这是我的解决方案:


def countInversions(arr):
    count = 0
    n = len(arr)
    for i in range(n):
        _count = count
        for j in range(0, n - i - 1):
            if arr[j] > arr[j + 1]:
                count += 1
                arr[j], arr[j + 1] = arr[j + 1], arr[j]
        if _count == count:
            break
    return count

I consider adding my solution because ever solution here is having

  1. greater time
  2. greater space complexity
  3. or doing too much operations

then is should be

So, here is my solution:


def countInversions(arr):
    count = 0
    n = len(arr)
    for i in range(n):
        _count = count
        for j in range(0, n - i - 1):
            if arr[j] > arr[j + 1]:
                count += 1
                arr[j], arr[j + 1] = arr[j + 1], arr[j]
        if _count == count:
            break
    return count

回答 18

如果有人对使用列表理解的较短实现感兴趣:

def bubble_sort(lst: list) -> None:
    [swap_items(lst, i, i+1) for left in range(len(lst)-1, 0, -1) for i in range(left) if lst[i] > lst[i+1]]


def swap_items(lst: list, pos1: int, pos2: int) -> None:
    lst[pos1], lst[pos2] = lst[pos2], lst[pos1]

If anyone is interested in a shorter implementation using a list comprehension:

def bubble_sort(lst: list) -> None:
    [swap_items(lst, i, i+1) for left in range(len(lst)-1, 0, -1) for i in range(left) if lst[i] > lst[i+1]]


def swap_items(lst: list, pos1: int, pos2: int) -> None:
    lst[pos1], lst[pos2] = lst[pos2], lst[pos1]

回答 19

这是不带for循环的气泡排序的另一种形式。基本上,您正在考虑的lastIndexarray然后慢慢进行decrementing,直到它成为数组的第一个索引为止。

algorithm将继续通过这样的阵列移动,直到整个通而没有任何由swaps发生。

泡沫基本上Quadratic Time: O(n²)是关于性能的。

class BubbleSort: 
  def __init__(self, arr):
    self.arr = arr;

  def bubbleSort(self):
    count = 0;
    lastIndex = len(self.arr) - 1;
    
    while(count < lastIndex):
      if(self.arr[count] > self.arr[count + 1]):
        self.swap(count)  
      count = count + 1;

      if(count == lastIndex):
        count = 0;
        lastIndex = lastIndex - 1;   

  def swap(self, count):
    temp = self.arr[count];
    self.arr[count] = self.arr[count + 1];
    self.arr[count + 1] = temp;
    
arr = [9, 1, 5, 3, 8, 2]
p1 = BubbleSort(arr)

print(p1.bubbleSort())

Here is a different variation of bubble sort without for loop. Basically you are considering the lastIndex of the array and slowly decrementing it until it first index of the array.

The algorithm will continue to move through the array like this until an entire pass is made without any swaps occurring.

The bubble is sort is basically Quadratic Time: O(n²) when it comes to performance.

class BubbleSort: 
  def __init__(self, arr):
    self.arr = arr;

  def bubbleSort(self):
    count = 0;
    lastIndex = len(self.arr) - 1;
    
    while(count < lastIndex):
      if(self.arr[count] > self.arr[count + 1]):
        self.swap(count)  
      count = count + 1;

      if(count == lastIndex):
        count = 0;
        lastIndex = lastIndex - 1;   

  def swap(self, count):
    temp = self.arr[count];
    self.arr[count] = self.arr[count + 1];
    self.arr[count + 1] = temp;
    
arr = [9, 1, 5, 3, 8, 2]
p1 = BubbleSort(arr)

print(p1.bubbleSort())

回答 20

the-fury和Martin Cote提供的答案解决了无限循环的问题,但是我的代码仍然无法正常工作(对于较大的列表,它将无法正确排序)。我最终放弃了unsorted变量,而是使用了计数器。

def bubble(badList):
    length = len(badList) - 1
    n = 0
    while n < len(badList):
        for element in range(0,length):
            if badList[element] > badList[element + 1]:
                hold = badList[element + 1]
                badList[element + 1] = badList[element]
                badList[element] = hold
                n = 0
            else:
                n += 1
    return badList

if __name__ == '__main__':
    mylist = [90, 10, 2, 76, 17, 66, 57, 23, 57, 99]
    print bubble(mylist)

如果有人可以在注释中提供任何有关如何改进我的代码的指针,将不胜感激。

Answers provided by the-fury and Martin Cote fixed the problem of the infinite loop, but my code would still not work correctly (for a larger list, it would not sort correctly.). I ended up ditching the unsorted variable and used a counter instead.

def bubble(badList):
    length = len(badList) - 1
    n = 0
    while n < len(badList):
        for element in range(0,length):
            if badList[element] > badList[element + 1]:
                hold = badList[element + 1]
                badList[element + 1] = badList[element]
                badList[element] = hold
                n = 0
            else:
                n += 1
    return badList

if __name__ == '__main__':
    mylist = [90, 10, 2, 76, 17, 66, 57, 23, 57, 99]
    print bubble(mylist)

If anyone could provide any pointers on how to improve my code in the comments, it would be much appreciated.


回答 21

试试这个

a = int(input("Enter Limit"))


val = []

for z in range(0,a):
    b = int(input("Enter Number in List"))
    val.append(b)


for y in range(0,len(val)):
   for x in range(0,len(val)-1):
       if val[x]>val[x+1]:
           t = val[x]
           val[x] = val[x+1]
           val[x+1] = t

print(val)

Try this

a = int(input("Enter Limit"))


val = []

for z in range(0,a):
    b = int(input("Enter Number in List"))
    val.append(b)


for y in range(0,len(val)):
   for x in range(0,len(val)-1):
       if val[x]>val[x+1]:
           t = val[x]
           val[x] = val[x+1]
           val[x+1] = t

print(val)

回答 22

idk如果这可能会在9年后对您有所帮助…它是一个简单的气泡排序程序

    l=[1,6,3,7,5,9,8,2,4,10]

    for i in range(1,len(l)):
        for j in range (i+1,len(l)):
            if l[i]>l[j]:
                l[i],l[j]=l[j],l[i]

idk if this might help you after 9 years… its a simple bubble sort program

    l=[1,6,3,7,5,9,8,2,4,10]

    for i in range(1,len(l)):
        for j in range (i+1,len(l)):
            if l[i]>l[j]:
                l[i],l[j]=l[j],l[i]

回答 23

def merge_bubble(arr):
    k = len(arr)
    while k>2:
        for i in range(0,k-1):
            for j in range(0,k-1):
                if arr[j] > arr[j+1]:
                    arr[j],arr[j+1] = arr[j+1],arr[j]

        return arr
        break
    else:
        if arr[0] > arr[1]:
            arr[0],arr[1] = arr[1],arr[0]
        return arr 
def merge_bubble(arr):
    k = len(arr)
    while k>2:
        for i in range(0,k-1):
            for j in range(0,k-1):
                if arr[j] > arr[j+1]:
                    arr[j],arr[j+1] = arr[j+1],arr[j]

        return arr
        break
    else:
        if arr[0] > arr[1]:
            arr[0],arr[1] = arr[1],arr[0]
        return arr 

回答 24

def bubble_sort(l):
    for i in range(len(l) -1):
        for j in range(len(l)-i-1):
            if l[j] > l[j+1]:
                l[j],l[j+1] = l[j+1], l[j]
    return l
def bubble_sort(l):
    for i in range(len(l) -1):
        for j in range(len(l)-i-1):
            if l[j] > l[j+1]:
                l[j],l[j+1] = l[j+1], l[j]
    return l

回答 25

def bubble_sorted(arr:list):
    while True:
        for i in range(0,len(arr)-1):
            count = 0
            if arr[i] > arr[i+1]:
                count += 1
                arr[i], arr[i+1] = arr[i+1], arr[i]
        if count == 0:
            break
    return arr
arr = [30,20,80,40,50,10,60,70,90]
print(bubble_sorted(arr))
#[20, 30, 40, 50, 10, 60, 70, 80, 90]
def bubble_sorted(arr:list):
    while True:
        for i in range(0,len(arr)-1):
            count = 0
            if arr[i] > arr[i+1]:
                count += 1
                arr[i], arr[i+1] = arr[i+1], arr[i]
        if count == 0:
            break
    return arr
arr = [30,20,80,40,50,10,60,70,90]
print(bubble_sorted(arr))
#[20, 30, 40, 50, 10, 60, 70, 80, 90]

回答 26

def bubbleSort(a): def swap(x, y): temp = a[x] a[x] = a[y] a[y] = temp #outer loop for j in range(len(a)): #slicing to the center, inner loop, python style for i in range(j, len(a) - j):
#find the min index and swap if a[i] < a[j]: swap(j, i) #find the max index and swap if a[i] > a[len(a) - j - 1]: swap(len(a) - j - 1, i) return a

def bubbleSort(a): def swap(x, y): temp = a[x] a[x] = a[y] a[y] = temp #outer loop for j in range(len(a)): #slicing to the center, inner loop, python style for i in range(j, len(a) - j):
#find the min index and swap if a[i] < a[j]: swap(j, i) #find the max index and swap if a[i] > a[len(a) - j - 1]: swap(len(a) - j - 1, i) return a


使用matplotlib绘制水平线

问题:使用matplotlib绘制水平线

我使用样条插值法来平滑时间序列,并且还想在绘图中添加一条水平线。但是似乎有一个我无法控制的问题。任何帮助都会非常有帮助。这是我所拥有的:

annual = np.arange(1,21,1)
l = np.array(value_list) # a list with 20 values
spl = UnivariateSpline(annual,l)
xs = np.linspace(1,21,200)
plt.plot(xs,spl(xs),'b')

plt.plot([0,len(xs)],[40,40],'r--',lw=2)
pylab.ylim([0,200])
plt.show()

问题似乎与我[0,len(xs)]对水平线图的使用有关。

I have used spline interpolation to smooth a time series and would also like to add a horizontal line to the plot. But there seems to be an issue that is out of my grips. Any assistance would be really helpful. Here is what I have:

annual = np.arange(1,21,1)
l = np.array(value_list) # a list with 20 values
spl = UnivariateSpline(annual,l)
xs = np.linspace(1,21,200)
plt.plot(xs,spl(xs),'b')

plt.plot([0,len(xs)],[40,40],'r--',lw=2)
pylab.ylim([0,200])
plt.show()

problem seems to be with my use of [0,len(xs)] for horizontal line plotting.


回答 0

你是对的,我认为这使[0,len(xs)]你失望了。您将要重用原始的x轴变量,xs并使用另一个包含变量的相同长度的numpy数组对其进行绘制。

annual = np.arange(1,21,1)
l = np.array(value_list) # a list with 20 values
spl = UnivariateSpline(annual,l)
xs = np.linspace(1,21,200)
plt.plot(xs,spl(xs),'b')

#####horizontal line
horiz_line_data = np.array([40 for i in xrange(len(xs))])
plt.plot(xs, horiz_line_data, 'r--') 
###########plt.plot([0,len(xs)],[40,40],'r--',lw=2)
pylab.ylim([0,200])
plt.show()

希望可以解决问题!

You are correct, I think the [0,len(xs)] is throwing you off. You’ll want to reuse the original x-axis variable xs and plot that with another numpy array of the same length that has your variable in it.

annual = np.arange(1,21,1)
l = np.array(value_list) # a list with 20 values
spl = UnivariateSpline(annual,l)
xs = np.linspace(1,21,200)
plt.plot(xs,spl(xs),'b')

#####horizontal line
horiz_line_data = np.array([40 for i in xrange(len(xs))])
plt.plot(xs, horiz_line_data, 'r--') 
###########plt.plot([0,len(xs)],[40,40],'r--',lw=2)
pylab.ylim([0,200])
plt.show()

Hopefully that fixes the problem!


回答 1

您正在寻找axhline(水平轴线)。例如,以下代码将为您提供一条水平线y = 0.5

import matplotlib.pyplot as plt
plt.axhline(y=0.5, color='r', linestyle='-')
plt.show()

You’re looking for axhline (a horizontal axis line). For example, the following will give you a horizontal line at y = 0.5:

import matplotlib.pyplot as plt
plt.axhline(y=0.5, color='r', linestyle='-')
plt.show()


回答 2

如果要在轴上绘制一条水平线,也可以尝试ax.hlines()方法。您需要在数据坐标中指定y位置和xminxmax(即,您在x轴上的实际数据范围)。示例代码段为:

import matplotlib.pyplot as plt
import numpy as np

x = np.linspace(1, 21, 200)
y = np.exp(-x)

fig, ax = plt.subplots()
ax.plot(x, y)
ax.hlines(y=0.2, xmin=4, xmax=20, linewidth=2, color='r')

plt.show()

上面的代码段将在处的轴上绘制一条水平线y=0.2。水平线的起点是x=4,终点为x=20。生成的图像是:

If you want to draw a horizontal line in the axes, you might also try ax.hlines() method. You need to specify y position and xmin and xmax in the data coordinate (i.e, your actual data range in the x-axis). A sample code snippet is:

import matplotlib.pyplot as plt
import numpy as np

x = np.linspace(1, 21, 200)
y = np.exp(-x)

fig, ax = plt.subplots()
ax.plot(x, y)
ax.hlines(y=0.2, xmin=4, xmax=20, linewidth=2, color='r')

plt.show()

The snippet above will plot a horizontal line in the axes at y=0.2. The horizontal line starts at x=4 and ends at x=20. The generated image is:


回答 3

用途matplotlib.pyplot.hlines

  • y 可以作为单个位置传递: y=40
  • y 可以作为多个位置传递: y=[39, 40, 41]
  • 如果您在绘制的东西,如一个数字fig, ax = plt.subplots(),然后更换plt.hlinesplt.axhlineax.hlinesax.axhline分别。
  • matplotlib.pyplot.axhline只能绘制一个位置(例如y=40
import numpy as np
import matplotlib.pyplot as plt

xs = np.linspace(1, 21, 200)
plt.hlines(y=40, xmin=0, xmax=len(xs), colors='r', linestyles='--', lw=2)
plt.show()

Use matplotlib.pyplot.hlines:

  • Can plot multiple horizontal lines by passing a list to the y parameter.
  • y can be passed as a single location: y=40
  • y can be passed as multiple locations: y=[39, 40, 41]
  • If you’re a plotting a figure with something like fig, ax = plt.subplots(), then replace plt.hlines or plt.axhline with ax.hlines or ax.axhline, respectively.
  • matplotlib.pyplot.axhline can only plot a single location (e.g. y=40)

plt.plot

import numpy as np
import matplotlib.pyplot as plt

xs = np.linspace(1, 21, 200)

plt.figure(figsize=(6, 3))
plt.hlines(y=39.5, xmin=100, xmax=175, colors='aqua', linestyles='-', lw=2, label='Single Short Line')
plt.hlines(y=[39, 40, 41], xmin=[0, 25, 50], xmax=[len(xs)], colors='purple', linestyles='--', lw=2, label='Multiple Lines')
plt.legend(bbox_to_anchor=(1.04,0.5), loc="center left", borderaxespad=0)

ax.plot

import numpy as np
import matplotlib.pyplot as plt

xs = np.linspace(1, 21, 200)
fig, (ax1, ax2) = plt.subplots(2, 1, figsize=(6, 6))

ax1.hlines(y=40, xmin=0, xmax=len(xs), colors='r', linestyles='--', lw=2)
ax1.set_title('One Line')

ax2.hlines(y=[39, 40, 41], xmin=0, xmax=len(xs), colors='purple', linestyles='--', lw=2)
ax2.set_title('Multiple Lines')

plt.tight_layout()
plt.show()

Time Series Axis

  • xmin and xmax will accept a date like '2020-09-10' or datetime(2020, 9, 10)
    • xmin=datetime(2020, 9, 10), xmax=datetime(2020, 9, 10) + timedelta(days=3)
    • Given date = df.index[9], xmin=date, xmax=date + pd.Timedelta(days=3), where the index is a DatetimeIndex.
import pandas_datareader as web  # conda or pip install this; not part of pandas
import pandas as pd
import matplotlib.pyplot as plt

# get test data
df = web.DataReader('^gspc', data_source='yahoo', start='2020-09-01', end='2020-09-28').iloc[:, :2]

# plot dataframe
ax = df.plot(figsize=(9, 6), title='S&P 500', ylabel='Price')

# add horizontal line
ax.hlines(y=3450, xmin='2020-09-10', xmax='2020-09-17', color='purple', label='test')

ax.legend()
plt.show()

  • Sample time series data if web.DataReader doesn’t work.
data = {pd.Timestamp('2020-09-01 00:00:00'): {'High': 3528.03, 'Low': 3494.6}, pd.Timestamp('2020-09-02 00:00:00'): {'High': 3588.11, 'Low': 3535.23}, pd.Timestamp('2020-09-03 00:00:00'): {'High': 3564.85, 'Low': 3427.41}, pd.Timestamp('2020-09-04 00:00:00'): {'High': 3479.15, 'Low': 3349.63}, pd.Timestamp('2020-09-08 00:00:00'): {'High': 3379.97, 'Low': 3329.27}, pd.Timestamp('2020-09-09 00:00:00'): {'High': 3424.77, 'Low': 3366.84}, pd.Timestamp('2020-09-10 00:00:00'): {'High': 3425.55, 'Low': 3329.25}, pd.Timestamp('2020-09-11 00:00:00'): {'High': 3368.95, 'Low': 3310.47}, pd.Timestamp('2020-09-14 00:00:00'): {'High': 3402.93, 'Low': 3363.56}, pd.Timestamp('2020-09-15 00:00:00'): {'High': 3419.48, 'Low': 3389.25}, pd.Timestamp('2020-09-16 00:00:00'): {'High': 3428.92, 'Low': 3384.45}, pd.Timestamp('2020-09-17 00:00:00'): {'High': 3375.17, 'Low': 3328.82}, pd.Timestamp('2020-09-18 00:00:00'): {'High': 3362.27, 'Low': 3292.4}, pd.Timestamp('2020-09-21 00:00:00'): {'High': 3285.57, 'Low': 3229.1}, pd.Timestamp('2020-09-22 00:00:00'): {'High': 3320.31, 'Low': 3270.95}, pd.Timestamp('2020-09-23 00:00:00'): {'High': 3323.35, 'Low': 3232.57}, pd.Timestamp('2020-09-24 00:00:00'): {'High': 3278.7, 'Low': 3209.45}, pd.Timestamp('2020-09-25 00:00:00'): {'High': 3306.88, 'Low': 3228.44}, pd.Timestamp('2020-09-28 00:00:00'): {'High': 3360.74, 'Low': 3332.91}}

df = pd.DataFrame.from_dict(data, 'index')

回答 4

除了最upvoted答案在这里,你也可以使用链axhline打完电话后plotpandasDataFrame

import pandas as pd

(pd.DataFrame([1, 2, 3])
   .plot(kind='bar', color='orange')
   .axhline(y=1.5));

In addition to the most upvoted answer here, one can also chain axhline after calling plot on a pandas‘s DataFrame.

import pandas as pd

(pd.DataFrame([1, 2, 3])
   .plot(kind='bar', color='orange')
   .axhline(y=1.5));


回答 5

对于那些总是忘记命令的人来说,一个不错的简便方法axhline

plt.plot(x, [y]*len(x))

你的情况xs = xy = 40。如果len(x)大,则效率低下,您应该真正使用axhline

A nice and easy way for those people who always forget the command axhline is the following

plt.plot(x, [y]*len(x))

In your case xs = x and y = 40. If len(x) is large, then this becomes inefficient and you should really use axhline.


回答 6

您可以使用plt.grid绘制水平线。

import numpy as np
from matplotlib import pyplot as plt
from scipy.interpolate import UnivariateSpline
from matplotlib.ticker import LinearLocator

# your data here
annual = np.arange(1,21,1)
l = np.random.random(20)
spl = UnivariateSpline(annual,l)
xs = np.linspace(1,21,200)

# plot your data
plt.plot(xs,spl(xs),'b')

# horizental line?
ax = plt.axes()
# three ticks:
ax.yaxis.set_major_locator(LinearLocator(3))
# plot grids only on y axis on major locations
plt.grid(True, which='major', axis='y')

# show
plt.show()

You can use plt.grid to draw a horizontal line.

import numpy as np
from matplotlib import pyplot as plt
from scipy.interpolate import UnivariateSpline
from matplotlib.ticker import LinearLocator

# your data here
annual = np.arange(1,21,1)
l = np.random.random(20)
spl = UnivariateSpline(annual,l)
xs = np.linspace(1,21,200)

# plot your data
plt.plot(xs,spl(xs),'b')

# horizental line?
ax = plt.axes()
# three ticks:
ax.yaxis.set_major_locator(LinearLocator(3))
# plot grids only on y axis on major locations
plt.grid(True, which='major', axis='y')

# show
plt.show()


“如果a或b或c但不是全部”的Python语法

问题:“如果a或b或c但不是全部”的Python语法

我有一个可以接收零个或三个命令行参数的python脚本。(要么以默认行为运行,要么需要指定所有三个值。)

诸如此类的理想语法是什么:

if a and (not b or not c) or b and (not a or not c) or c and (not b or not a):

I have a python script that can receive either zero or three command line arguments. (Either it runs on default behavior or needs all three values specified.)

What’s the ideal syntax for something like:

if a and (not b or not c) or b and (not a or not c) or c and (not b or not a):

?


回答 0

如果您的意思是最小形式,请执行以下操作:

if (not a or not b or not c) and (a or b or c):

这将翻译您的问题的标题。

更新:正如Volatility和Supr正确地说的那样,您可以应用De Morgan的定律并获得等效项:

if (a or b or c) and not (a and b and c):

我的建议是使用对您和其他程序员更重要的形式。第一个意思是“有一些错误,但也有一些真实”,第二个意思是“有一些真实,但不是一切”。如果要在硬件上进行优化或执行此操作,则选择第二个,这里只选择最易读的代码(还要考虑要测试的条件及其名称)。我选了第一个。

If you mean a minimal form, go with this:

if (not a or not b or not c) and (a or b or c):

Which translates the title of your question.

UPDATE: as correctly said by Volatility and Supr, you can apply De Morgan’s law and obtain equivalent:

if (a or b or c) and not (a and b and c):

My advice is to use whichever form is more significant to you and to other programmers. The first means “there is something false, but also something true”, the second “There is something true, but not everything”. If I were to optimize or do this in hardware, I would choose the second, here just choose the most readable (also taking in consideration the conditions you will be testing and their names). I picked the first.


回答 1

怎么样:

conditions = [a, b, c]
if any(conditions) and not all(conditions):
   ...

其他变体:

if 1 <= sum(map(bool, conditions)) <= 2:
   ...

How about:

conditions = [a, b, c]
if any(conditions) and not all(conditions):
   ...

Other variant:

if 1 <= sum(map(bool, conditions)) <= 2:
   ...

回答 2

这个问题已经有很多被高度认可的答案和一个被接受的答案,但是到目前为止,所有这些都被表达布尔问题的各种方法分散了注意力,并且错过了关键点:

我有一个可以接收零或三个命令行参数的python脚本。(要么以默认行为运行,要么需要指定所有三个值)

首先,此逻辑不应由您的代码负责,而应由argparse模块处理。不用费心编写复杂的if语句,而喜欢设置参数解析器,如下所示:

#!/usr/bin/env python
import argparse as ap
parser = ap.ArgumentParser()
parser.add_argument('--foo', nargs=3, default=['x', 'y', 'z'])
args = parser.parse_args()
print(args.foo)

是的,它应该是一个选项而不是位置参数,因为它毕竟是optional


编辑: 为了解决LarsH在注释中的问题,下面是一个示例示例,如果您确定要使用3个或0个位置 args的接口,可以如何编写它。我认为以前的接口是更好的样式,因为可选参数应该是 options,但是出于完整性考虑,这是一种替代方法。usage在创建解析器时,请注意最重要的kwarg,argparse否则会自动生成误导性的使用消息!

#!/usr/bin/env python
import argparse as ap
parser = ap.ArgumentParser(usage='%(prog)s [-h] [a b c]\n')
parser.add_argument('abc', nargs='*', help='specify 3 or 0 items', default=['x', 'y', 'z'])
args = parser.parse_args()
if len(args.abc) != 3:
  parser.error('expected 3 arguments')
print(args.abc)

以下是一些用法示例:

# default case
wim@wim-zenbook:/tmp$ ./three_or_none.py 
['x', 'y', 'z']

# explicit case
wim@wim-zenbook:/tmp$ ./three_or_none.py 1 2 3
['1', '2', '3']

# example failure mode
wim@wim-zenbook:/tmp$ ./three_or_none.py 1 2 
usage: three_or_none.py [-h] [a b c]
three_or_none.py: error: expected 3 arguments

This question already had many highly upvoted answers and an accepted answer, but all of them so far were distracted by various ways to express the boolean problem and missed a crucial point:

I have a python script that can receive either zero or three command line arguments. (Either it runs on default behavior or needs all three values specified)

This logic should not be the responsibility of your code in the first place, rather it should be handled by argparse module. Don’t bother writing a complex if statement, instead prefer to setup your argument parser something like this:

#!/usr/bin/env python
import argparse as ap
parser = ap.ArgumentParser()
parser.add_argument('--foo', nargs=3, default=['x', 'y', 'z'])
args = parser.parse_args()
print(args.foo)

And yes, it should be an option not a positional argument, because it is after all optional.


edited: To address the concern of LarsH in the comments, below is an example of how you could write it if you were certain you wanted the interface with either 3 or 0 positional args. I am of the opinion that the previous interface is better style, because optional arguments should be options, but here’s an alternative approach for the sake of completeness. Note the overriding kwarg usage when creating your parser, because argparse will auto-generate a misleading usage message otherwise!

#!/usr/bin/env python
import argparse as ap
parser = ap.ArgumentParser(usage='%(prog)s [-h] [a b c]\n')
parser.add_argument('abc', nargs='*', help='specify 3 or 0 items', default=['x', 'y', 'z'])
args = parser.parse_args()
if len(args.abc) != 3:
  parser.error('expected 3 arguments')
print(args.abc)

Here are some usage examples:

# default case
wim@wim-zenbook:/tmp$ ./three_or_none.py 
['x', 'y', 'z']

# explicit case
wim@wim-zenbook:/tmp$ ./three_or_none.py 1 2 3
['1', '2', '3']

# example failure mode
wim@wim-zenbook:/tmp$ ./three_or_none.py 1 2 
usage: three_or_none.py [-h] [a b c]
three_or_none.py: error: expected 3 arguments

回答 3

我会去:

conds = iter([a, b, c])
if any(conds) and not any(conds):
    # okay...

我认为这应该有效地短路

说明

通过做conds一个迭代器,any如果有任何一项为true ,则第一次使用will将使迭代器短路并使迭代器指向下一个元素。否则,它将消耗整个列表,并且为False。下一个any将迭代器中的其余项取为零,并确保没有其他任何真值…如果有,则整个语句不可能为真,因此就没有一个唯一元素(因此短路再次)。最后一个any将返回False或将耗尽iterable和be True

注意:以上检查是否仅设置了一个条件


如果要检查是否设置了一个或多个项目,但没有设置每个项目,则可以使用:

not all(conds) and any(conds)

I’d go for:

conds = iter([a, b, c])
if any(conds) and not any(conds):
    # okay...

I think this should short-circuit fairly efficiently

Explanation

By making conds an iterator, the first use of any will short circuit and leave the iterator pointing to the next element if any item is true; otherwise, it will consume the entire list and be False. The next any takes the remaining items in the iterable, and makes sure than there aren’t any other true values… If there are, the whole statement can’t be true, thus there isn’t one unique element (so short circuits again). The last any will either return False or will exhaust the iterable and be True.

note: the above checks if only a single condition is set


If you want to check if one or more items, but not every item is set, then you can use:

not all(conds) and any(conds)

回答 4

英文句子:

“如果是a或b或c,但不是全部”

转换为以下逻辑:

(a or b or c) and not (a and b and c)

单词“但是”通常表示连词,即“和”。此外,“所有这些”转换为的条件结合:这种情况下,该条件下,其它条件。“ not”反转整个连接。

我不同意接受的答案。作者忽略了将最直接的解释应用于规范,而忽略了应用戴摩根定律(De Morgan’s Law)将表达式简化为更少的运算符:

 not a or not b or not c  ->  not (a and b and c)

同时声称答案是“最小形式”。

The English sentence:

“if a or b or c but not all of them”

Translates to this logic:

(a or b or c) and not (a and b and c)

The word “but” usually implies a conjunction, in other words “and”. Furthermore, “all of them” translates to a conjunction of conditions: this condition, and that condition, and other condition. The “not” inverts that entire conjunction.

I do not agree that the accepted answer. The author neglected to apply the most straightforward interpretation to the specification, and neglected to apply De Morgan’s Law to simplify the expression to fewer operators:

 not a or not b or not c  ->  not (a and b and c)

while claiming that the answer is a “minimal form”.


回答 5

True如果三个条件之一并且只有一个是,则返回此值True。可能是您在示例代码中想要的。

if sum(1 for x in (a,b,c) if x) == 1:

This returns True if one and only one of the three conditions is True. Probably what you wanted in your example code.

if sum(1 for x in (a,b,c) if x) == 1:

回答 6

关于:(唯一条件)

if (bool(a) + bool(b) + bool(c) == 1):

注意,如果您同时允许两个条件,也可以这样做

if (bool(a) + bool(b) + bool(c) in [1,2]):

What about: (unique condition)

if (bool(a) + bool(b) + bool(c) == 1):

Notice, if you allow two conditions too you could do that

if (bool(a) + bool(b) + bool(c) in [1,2]):

回答 7

需要明确的是,您要基于多少个参数为逻辑TRUE(在使用字符串参数的情况下-不为空)做出决定?

argsne = (1 if a else 0) + (1 if b else 0) + (1 if c else 0)

然后您做出了决定:

if ( 0 < argsne < 3 ):
 doSth() 

现在的逻辑更加清晰了。

To be clear, you want to made your decision based on how much of the parameters are logical TRUE (in case of string arguments – not empty)?

argsne = (1 if a else 0) + (1 if b else 0) + (1 if c else 0)

Then you made a decision:

if ( 0 < argsne < 3 ):
 doSth() 

Now the logic is more clear.


回答 8

为何不仅仅计算它们呢?

import sys
a = sys.argv
if len(a) = 1 :  
    # No arguments were given, the program name count as one
elif len(a) = 4 :
    # Three arguments were given
else :
    # another amount of arguments was given

And why not just count them ?

import sys
a = sys.argv
if len(a) = 1 :  
    # No arguments were given, the program name count as one
elif len(a) = 4 :
    # Three arguments were given
else :
    # another amount of arguments was given

回答 9

如果您不介意有点晦涩难懂,则可以简单地滚动一下0 < (a + b + c) < 3true如果您有一个和两个真实的陈述,则返回,如果全部为假或没有为假,则返回false。

这也简化了您是否使用函数来评估布尔值,因为您只评估了一次变量,这意味着您可以内联编写函数,而无需临时存储变量。(例如:0 < ( a(x) + b(x) + c(x) ) < 3。)

If you don’t mind being a bit cryptic you can simly roll with 0 < (a + b + c) < 3 which will return true if you have between one and two true statements and false if all are false or none is false.

This also simplifies if you use functions to evaluate the bools as you only evaluate the variables once and which means you can write the functions inline and do not need to temporarily store the variables. (Example: 0 < ( a(x) + b(x) + c(x) ) < 3.)


回答 10

问题表明您需要全部三个参数(a和b以及c),或者都不需要(而不是(a或b或c))

这给出:

(a and b and c)否(a或b或c)

The question states that you need either all three arguments (a and b and c) or none of them (not (a or b or c))

This gives:

(a and b and c) or not (a or b or c)


回答 11

据我了解,您有一个接收3个参数的函数,但是如果不这样做,它将以默认行为运行。由于您尚未说明提供1或2个参数时应该发生的情况,因此我假设它应该只是执行默认行为。在这种情况下,我认为您会发现以下答案非常有利:

def method(a=None, b=None, c=None):
    if all([a, b, c]):
        # received 3 arguments
    else:
        # default behavior

但是,如果要对1或2个参数进行不同的处理:

def method(a=None, b=None, c=None):
    args = [a, b, c]
    if all(args):
        # received 3 arguments
    elif not any(args):
        # default behavior
    else:
        # some args (raise exception?)

注意:假设“ False”值不会传递到此方法中。

As I understand it, you have a function that receives 3 arguments, but if it does not it will run on default behavior. Since you have not explained what should happen when 1 or 2 arguments are supplied I will assume it should simply do the default behavior. In which case, I think you will find the following answer very advantageous:

def method(a=None, b=None, c=None):
    if all([a, b, c]):
        # received 3 arguments
    else:
        # default behavior

However, if you want 1 or 2 arguments to be handled differently:

def method(a=None, b=None, c=None):
    args = [a, b, c]
    if all(args):
        # received 3 arguments
    elif not any(args):
        # default behavior
    else:
        # some args (raise exception?)

note: This assumes that “False” values will not be passed into this method.


回答 12

如果使用条件迭代器,则访问速度可能很慢。但是,您不需要一次访问每个元素,也不总是需要读取所有元素。这是可与无限生成器一起使用的解决方案:

#!/usr/bin/env python3
from random import randint
from itertools import tee

def generate_random():
    while True:
        yield bool(randint(0,1))

def any_but_not_all2(s): # elegant
    t1, t2 = tee(s)
    return False in t1 and True in t2 # could also use "not all(...) and any(...)"

def any_but_not_all(s): # simple
    hadFalse = False
    hadTrue = False
    for i in s:
        if i:
            hadTrue = True
        else:
            hadFalse = True
        if hadTrue and hadFalse:
            return True
    return False


r1, r2 = tee(generate_random())
assert any_but_not_all(r1)
assert any_but_not_all2(r2)

assert not any_but_not_all([True, True])
assert not any_but_not_all2([True, True])

assert not any_but_not_all([])
assert not any_but_not_all2([])

assert any_but_not_all([True, False])
assert any_but_not_all2([True, False])

If you work with an iterator of conditions, it could be slow to access. But you don’t need to access each element more than once, and you don’t always need to read all of it. Here’s a solution that will work with infinite generators:

#!/usr/bin/env python3
from random import randint
from itertools import tee

def generate_random():
    while True:
        yield bool(randint(0,1))

def any_but_not_all2(s): # elegant
    t1, t2 = tee(s)
    return False in t1 and True in t2 # could also use "not all(...) and any(...)"

def any_but_not_all(s): # simple
    hadFalse = False
    hadTrue = False
    for i in s:
        if i:
            hadTrue = True
        else:
            hadFalse = True
        if hadTrue and hadFalse:
            return True
    return False


r1, r2 = tee(generate_random())
assert any_but_not_all(r1)
assert any_but_not_all2(r2)

assert not any_but_not_all([True, True])
assert not any_but_not_all2([True, True])

assert not any_but_not_all([])
assert not any_but_not_all2([])

assert any_but_not_all([True, False])
assert any_but_not_all2([True, False])

回答 13

当每一个给定boolTrue,或者当每一个给定boolFalse……
他们都是彼此相等!

所以,我们只需要找到两个元素的计算结果为不同的这bool小号
知道至少有一个True和至少一个False

我的简短解决方案:

not bool(a)==bool(b)==bool(c)

我相信它会短路,导致AFAIK a==b==c等于a==b and b==c

我的广义解决方案:

def _any_but_not_all(first, iterable): #doing dirty work
    bool_first=bool(first)
    for x in iterable:
        if bool(x) is not bool_first:
            return True
    return False

def any_but_not_all(arg, *args): #takes any amount of args convertable to bool
    return _any_but_not_all(arg, args)

def v_any_but_not_all(iterable): #takes iterable or iterator
    iterator=iter(iterable)
    return _any_but_not_all(next(iterator), iterator)

我也写了一些处理多个可迭代对象的代码,但是我从这里删除了它,因为我认为这没有意义。但是,这里仍然可用。

When every given bool is True, or when every given bool is False
they all are equal to each other!

So, we just need to find two elements which evaluates to different bools
to know that there is at least one True and at least one False.

My short solution:

not bool(a)==bool(b)==bool(c)

I belive it short-circuits, cause AFAIK a==b==c equals a==b and b==c.

My generalized solution:

def _any_but_not_all(first, iterable): #doing dirty work
    bool_first=bool(first)
    for x in iterable:
        if bool(x) is not bool_first:
            return True
    return False

def any_but_not_all(arg, *args): #takes any amount of args convertable to bool
    return _any_but_not_all(arg, args)

def v_any_but_not_all(iterable): #takes iterable or iterator
    iterator=iter(iterable)
    return _any_but_not_all(next(iterator), iterator)

I wrote also some code dealing with multiple iterables, but I deleted it from here because I think it’s pointless. It’s however still available here.


回答 14

基本上,这是“一些(但不是全部)”功能(与any()all()内置功能相比)。

这意味着结果之间应该有Falses True s。因此,您可以执行以下操作:

some = lambda ii: frozenset(bool(i) for i in ii).issuperset((True, False))

# one way to test this is...
test = lambda iterable: (any(iterable) and (not all(iterable))) # see also http://stackoverflow.com/a/16522290/541412

# Some test cases...
assert(some(()) == False)       # all() is true, and any() is false
assert(some((False,)) == False) # any() is false
assert(some((True,)) == False)  # any() and all() are true

assert(some((False,False)) == False)
assert(some((True,True)) == False)
assert(some((True,False)) == True)
assert(some((False,True)) == True)

此代码的优点之一是,您只需要对结果(布尔值)项目进行一次迭代。

缺点之一是,所有这些真值表达式总是被求值,并且不像/ 运算符那样发生短路orand

This is basically a “some (but not all)” functionality (when contrasted with the any() and all() builtin functions).

This implies that there should be Falses and Trues among the results. Therefore, you can do the following:

some = lambda ii: frozenset(bool(i) for i in ii).issuperset((True, False))

# one way to test this is...
test = lambda iterable: (any(iterable) and (not all(iterable))) # see also http://stackoverflow.com/a/16522290/541412

# Some test cases...
assert(some(()) == False)       # all() is true, and any() is false
assert(some((False,)) == False) # any() is false
assert(some((True,)) == False)  # any() and all() are true

assert(some((False,False)) == False)
assert(some((True,True)) == False)
assert(some((True,False)) == True)
assert(some((False,True)) == True)

One advantage of this code is that you only need to iterate once through the resulting (booleans) items.

One disadvantage is that all these truth-expressions are always evaluated, and do not do short-circuiting like the or/and operators.


如何从C#运行Python脚本?

问题:如何从C#运行Python脚本?

之前曾在不同程度上提出过这样的问题,但我觉得还没有以简明的方式回答,因此我再次提出。

我想在Python中运行脚本。可以说是这样的:

if __name__ == '__main__':
    with open(sys.argv[1], 'r') as f:
        s = f.read()
    print s

它获取文件位置,读取它,然后打印其内容。没那么复杂。

好吧,那我该如何在C#中运行它呢?

这就是我现在所拥有的:

    private void run_cmd(string cmd, string args)
    {
        ProcessStartInfo start = new ProcessStartInfo();
        start.FileName = cmd;
        start.Arguments = args;
        start.UseShellExecute = false;
        start.RedirectStandardOutput = true;
        using (Process process = Process.Start(start))
        {
            using (StreamReader reader = process.StandardOutput)
            {
                string result = reader.ReadToEnd();
                Console.Write(result);
            }
        }
    }

当我传递code.py位置as cmdfilename位置as args无效时。有人告诉我,我应该通过python.execmd,然后code.py filename作为args

我已经寻找了一段时间,只能找到建议使用IronPython之类的人。但是必须有一种从C#调用Python脚本的方法。

一些澄清:

我需要从C#运行它,我需要捕获输出,并且不能使用IronPython或其他任何东西。无论您有什么技巧,都可以。

PS:我正在运行的实际Python代码要复杂得多,并且它返回我在C#中需要的输出,并且C#代码将不断调用Python代码。

假设这是我的代码:

    private void get_vals()
    {
        for (int i = 0; i < 100; i++)
        {
            run_cmd("code.py", i);
        }
    }

This sort of question has been asked before in varying degrees, but I feel it has not been answered in a concise way and so I ask it again.

I want to run a script in Python. Let’s say it’s this:

if __name__ == '__main__':
    with open(sys.argv[1], 'r') as f:
        s = f.read()
    print s

Which gets a file location, reads it, then prints its contents. Not so complicated.

Okay, so how do I run this in C#?

This is what I have now:

    private void run_cmd(string cmd, string args)
    {
        ProcessStartInfo start = new ProcessStartInfo();
        start.FileName = cmd;
        start.Arguments = args;
        start.UseShellExecute = false;
        start.RedirectStandardOutput = true;
        using (Process process = Process.Start(start))
        {
            using (StreamReader reader = process.StandardOutput)
            {
                string result = reader.ReadToEnd();
                Console.Write(result);
            }
        }
    }

When I pass the code.py location as cmd and the filename location as args it doesn’t work. I was told I should pass python.exe as the cmd, and then code.py filename as the args.

I have been looking for a while now and can only find people suggesting to use IronPython or such. But there must be a way to call a Python script from C#.

Some clarification:

I need to run it from C#, I need to capture the output, and I can’t use IronPython or anything else. Whatever hack you have will be fine.

P.S.: The actual Python code I’m running is much more complex than this, and it returns output which I need in C#, and the C# code will be constantly calling the Python code.

Pretend this is my code:

    private void get_vals()
    {
        for (int i = 0; i < 100; i++)
        {
            run_cmd("code.py", i);
        }
    }

回答 0

它不起作用的原因是因为您有UseShellExecute = false

如果不使用外壳,则必须以方式提供python可执行文件的完整路径FileName,并构建Arguments字符串以提供脚本和要读取的文件。

另请注意,RedirectStandardOutput除非您不能这样做UseShellExecute = false

我不太确定应如何为python格式化参数字符串,但是您将需要以下内容:

private void run_cmd(string cmd, string args)
{
     ProcessStartInfo start = new ProcessStartInfo();
     start.FileName = "my/full/path/to/python.exe";
     start.Arguments = string.Format("{0} {1}", cmd, args);
     start.UseShellExecute = false;
     start.RedirectStandardOutput = true;
     using(Process process = Process.Start(start))
     {
         using(StreamReader reader = process.StandardOutput)
         {
             string result = reader.ReadToEnd();
             Console.Write(result);
         }
     }
}

The reason it isn’t working is because you have UseShellExecute = false.

If you don’t use the shell, you will have to supply the complete path to the python executable as FileName, and build the Arguments string to supply both your script and the file you want to read.

Also note, that you can’t RedirectStandardOutput unless UseShellExecute = false.

I’m not quite sure how the argument string should be formatted for python, but you will need something like this:

private void run_cmd(string cmd, string args)
{
     ProcessStartInfo start = new ProcessStartInfo();
     start.FileName = "my/full/path/to/python.exe";
     start.Arguments = string.Format("{0} {1}", cmd, args);
     start.UseShellExecute = false;
     start.RedirectStandardOutput = true;
     using(Process process = Process.Start(start))
     {
         using(StreamReader reader = process.StandardOutput)
         {
             string result = reader.ReadToEnd();
             Console.Write(result);
         }
     }
}

回答 1

如果您愿意使用IronPython,则可以直接在C#中执行脚本:

using IronPython.Hosting;
using Microsoft.Scripting.Hosting;

private static void doPython()
{
    ScriptEngine engine = Python.CreateEngine();
    engine.ExecuteFile(@"test.py");
}

在此处获取IronPython。

If you’re willing to use IronPython, you can execute scripts directly in C#:

using IronPython.Hosting;
using Microsoft.Scripting.Hosting;

private static void doPython()
{
    ScriptEngine engine = Python.CreateEngine();
    engine.ExecuteFile(@"test.py");
}

Get IronPython here.


回答 2

从C执行Python脚本

创建一个C#项目并编写以下代码。

using System;
using System.Diagnostics;
using System.IO;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace WindowsFormsApplication1
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {
            run_cmd();
        }

        private void run_cmd()
        {

            string fileName = @"C:\sample_script.py";

            Process p = new Process();
            p.StartInfo = new ProcessStartInfo(@"C:\Python27\python.exe", fileName)
            {
                RedirectStandardOutput = true,
                UseShellExecute = false,
                CreateNoWindow = true
            };
            p.Start();

            string output = p.StandardOutput.ReadToEnd();
            p.WaitForExit();

            Console.WriteLine(output);

            Console.ReadLine();

        }
    }
}

Python sample_script

打印“ Python C#测试”

您将在C#控制台中看到“ Python C#测试”

Execute Python script from C

Create a C# project and write the following code.

using System;
using System.Diagnostics;
using System.IO;
using System.Threading.Tasks;
using System.Windows.Forms;
namespace WindowsFormsApplication1
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {
            run_cmd();
        }

        private void run_cmd()
        {

            string fileName = @"C:\sample_script.py";

            Process p = new Process();
            p.StartInfo = new ProcessStartInfo(@"C:\Python27\python.exe", fileName)
            {
                RedirectStandardOutput = true,
                UseShellExecute = false,
                CreateNoWindow = true
            };
            p.Start();

            string output = p.StandardOutput.ReadToEnd();
            p.WaitForExit();

            Console.WriteLine(output);

            Console.ReadLine();

        }
    }
}

Python sample_script

print "Python C# Test"

You will see the ‘Python C# Test’ in the console of C#.


回答 3

我遇到了同样的问题,道德大师的回答对我没有帮助。以下是基于先前答案的方法:

private void run_cmd(string cmd, string args)
{
 ProcessStartInfo start = new ProcessStartInfo();
 start.FileName = cmd;//cmd is full path to python.exe
 start.Arguments = args;//args is path to .py file and any cmd line args
 start.UseShellExecute = false;
 start.RedirectStandardOutput = true;
 using(Process process = Process.Start(start))
 {
     using(StreamReader reader = process.StandardOutput)
     {
         string result = reader.ReadToEnd();
         Console.Write(result);
     }
 }
}

例如,如果要使用cmd行参数100执行test.py ,则cmd为,@C:/Python26/python.exe而args为C://Python26//test.py 100。请注意,.py文件的路径中没有@符号。

I ran into the same problem and Master Morality’s answer didn’t do it for me. The following, which is based on the previous answer, worked:

private void run_cmd(string cmd, string args)
{
 ProcessStartInfo start = new ProcessStartInfo();
 start.FileName = cmd;//cmd is full path to python.exe
 start.Arguments = args;//args is path to .py file and any cmd line args
 start.UseShellExecute = false;
 start.RedirectStandardOutput = true;
 using(Process process = Process.Start(start))
 {
     using(StreamReader reader = process.StandardOutput)
     {
         string result = reader.ReadToEnd();
         Console.Write(result);
     }
 }
}

As an example, cmd would be @C:/Python26/python.exe and args would be C://Python26//test.py 100 if you wanted to execute test.py with cmd line argument 100. Note that the path the .py file does not have the @ symbol.


回答 4


回答 5

设置WorkingDirectory或在Argument中指定python脚本的完整路径

ProcessStartInfo start = new ProcessStartInfo();
start.FileName = "C:\\Python27\\python.exe";
//start.WorkingDirectory = @"D:\script";
start.Arguments = string.Format("D:\\script\\test.py -a {0} -b {1} ", "some param", "some other param");
start.UseShellExecute = false;
start.RedirectStandardOutput = true;
using (Process process = Process.Start(start))
{
    using (StreamReader reader = process.StandardOutput)
    {
        string result = reader.ReadToEnd();
        Console.Write(result);
    }
}

Set WorkingDirectory or specify the full path of the python script in the Argument

ProcessStartInfo start = new ProcessStartInfo();
start.FileName = "C:\\Python27\\python.exe";
//start.WorkingDirectory = @"D:\script";
start.Arguments = string.Format("D:\\script\\test.py -a {0} -b {1} ", "some param", "some other param");
start.UseShellExecute = false;
start.RedirectStandardOutput = true;
using (Process process = Process.Start(start))
{
    using (StreamReader reader = process.StandardOutput)
    {
        string result = reader.ReadToEnd();
        Console.Write(result);
    }
}

回答 6

实际上,使用IronPython在Csharp(VS)和Python之间进行集成非常容易。并没有那么复杂…正如克里斯·邓纳维(Chris Dunaway)在回答部分中所述,我开始为自己的项目建立这种整合。N非常简单。只需按照这些步骤N,您将获得结果。

步骤1:打开VS并创建新的空ConsoleApp项目。

步骤2:转到工具-> NuGet软件包管理器->软件包管理器控制台。

步骤3:在此之后,在浏览器中打开此链接并复制NuGet命令。链接:https//www.nuget.org/packages/IronPython/2.7.9

步骤4:打开以上链接后,复制PM> Install-Package IronPython -Version 2.7.9命令并将其粘贴到VS的NuGet控制台中。它将安装支持包。

步骤5:这是我用来运行存储在我的Python.exe目录中的.py文件的代码。

using IronPython.Hosting;//for DLHE
using Microsoft.Scripting.Hosting;//provides scripting abilities comparable to batch files
using System;
using System.Diagnostics;
using System.IO;
using System.Net;
using System.Net.Sockets;
class Hi
{
private static void Main(string []args)
{
Process process = new Process(); //to make a process call
ScriptEngine engine = Python.CreateEngine(); //For Engine to initiate the script
engine.ExecuteFile(@"C:\Users\daulmalik\AppData\Local\Programs\Python\Python37\p1.py");//Path of my .py file that I would like to see running in console after running my .cs file from VS.//process.StandardInput.Flush();
process.StandardInput.Close();//to close
process.WaitForExit();//to hold the process i.e. cmd screen as output
}
} 

步骤6:保存并执行代码

Actually its pretty easy to make integration between Csharp (VS) and Python with IronPython. It’s not that much complex… As Chris Dunaway already said in answer section I started to build this inegration for my own project. N its pretty simple. Just follow these steps N you will get your results.

step 1 : Open VS and create new empty ConsoleApp project.

step 2 : Go to tools –> NuGet Package Manager –> Package Manager Console.

step 3 : After this open this link in your browser and copy the NuGet Command. Link: https://www.nuget.org/packages/IronPython/2.7.9

step 4 : After opening the above link copy the PM>Install-Package IronPython -Version 2.7.9 command and paste it in NuGet Console in VS. It will install the supportive packages.

step 5 : This is my code that I have used to run a .py file stored in my Python.exe directory.

using IronPython.Hosting;//for DLHE
using Microsoft.Scripting.Hosting;//provides scripting abilities comparable to batch files
using System;
using System.Diagnostics;
using System.IO;
using System.Net;
using System.Net.Sockets;
class Hi
{
private static void Main(string []args)
{
Process process = new Process(); //to make a process call
ScriptEngine engine = Python.CreateEngine(); //For Engine to initiate the script
engine.ExecuteFile(@"C:\Users\daulmalik\AppData\Local\Programs\Python\Python37\p1.py");//Path of my .py file that I would like to see running in console after running my .cs file from VS.//process.StandardInput.Flush();
process.StandardInput.Close();//to close
process.WaitForExit();//to hold the process i.e. cmd screen as output
}
} 

step 6 : save and execute the code


回答 7

我遇到问题stdin/stout-当有效负载大小超过几千字节时,它挂起了。我不仅需要使用一些简短的参数来调用Python函数,还需要使用可能很大的自定义有效负载。

前一段时间,我编写了一个虚拟角色库,该库允许通过Redis在不同计算机上分发任务。为了调用Python代码,我添加了侦听来自Python的消息,对其进行处理并将结果返回给.NET的功能。 这是它如何工作的简要说明

它也可以在一台机器上运行,但是需要一个Redis实例。Redis增加了一些可靠性保证-有效负载被存储,直到工作确认完成为止。如果工作人员死亡,则有效负载将返回到作业队列,然后由另一个工作人员重新处理。

I am having problems with stdin/stout – when payload size exceeds several kilobytes it hangs. I need to call Python functions not only with some short arguments, but with a custom payload that could be big.

A while ago, I wrote a virtual actor library that allows to distribute task on different machines via Redis. To call Python code, I added functionality to listen for messages from Python, process them and return results back to .NET. Here is a brief description of how it works.

It works on a single machine as well, but requires a Redis instance. Redis adds some reliability guarantees – payload is stored until a worked acknowledges completion. If a worked dies, the payload is returned to a job queue and then is reprocessed by another worker.


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