问题:如何在Python中进行相对导入?
想象一下这个目录结构:
app/
__init__.py
sub1/
__init__.py
mod1.py
sub2/
__init__.py
mod2.py
我正在编码mod1
,我需要从中导入一些东西mod2
。我该怎么办?
我尝试过,from ..sub2 import mod2
但是得到了“未打包的相对导入尝试”。
我四处搜寻,但只发现“ sys.path
操纵”骇客。有没有一种干净的方法?
编辑:我所有__init__.py
的当前为空
EDIT2:我想这样做,因为SUB2包含了为子包(共享类sub1
,subX
等等)。
Edit3:我要寻找的行为与PEP 366中描述的相同(感谢John B)
Imagine this directory structure:
app/
__init__.py
sub1/
__init__.py
mod1.py
sub2/
__init__.py
mod2.py
I’m coding mod1
, and I need to import something from mod2
. How should I do it?
I tried from ..sub2 import mod2
but I’m getting an “Attempted relative import in non-package”.
I googled around but found only “sys.path
manipulation” hacks. Isn’t there a clean way?
Edit: all my __init__.py
‘s are currently empty
Edit2: I’m trying to do this because sub2 contains classes that are shared across sub packages (sub1
, subX
, etc.).
Edit3: The behaviour I’m looking for is the same as described in PEP 366 (thanks John B)
回答 0
每个人似乎都想告诉您应该做什么,而不仅仅是回答问题。
问题是您通过将mod1.py作为参数传递给解释器,从而以“ __main__”的身份运行模块。
从PEP 328:
相对导入使用模块的__name__属性来确定该模块在包层次结构中的位置。如果模块的名称不包含任何包信息(例如,将其设置为“ __main__”),则相对导入将被视为模块是顶级模块,而不论该模块实际位于文件系统上的哪个位置。
在Python 2.6中,他们添加了相对于主模块引用模块的功能。 PEP 366说明了更改。
更新:根据Nick Coghlan的建议,推荐的替代方法是使用-m开关运行软件包中的模块。
Everyone seems to want to tell you what you should be doing rather than just answering the question.
The problem is that you’re running the module as ‘__main__’ by passing the mod1.py as an argument to the interpreter.
From PEP 328:
Relative imports use a module’s __name__ attribute to determine that module’s position in the package hierarchy. If the module’s name does not contain any package information (e.g. it is set to ‘__main__’) then relative imports are resolved as if the module were a top level module, regardless of where the module is actually located on the file system.
In Python 2.6, they’re adding the ability to reference modules relative to the main module. PEP 366 describes the change.
Update: According to Nick Coghlan, the recommended alternative is to run the module inside the package using the -m switch.
回答 1
这是对我有效的解决方案:
我执行as的相对导入 from ..sub2 import mod2
,然后,如果要运行,mod1.py
则转到的父目录,app
并使用python -m开关as运行模块 python -m app.sub1.mod1
。
相对导入发生此问题的真正原因是,相对导入通过获取__name__
模块的属性起作用。如果模块直接运行,则__name__
设置为__main__
,并且不包含有关包结构的任何信息。并且,这就是为什么python抱怨该relative import in non-package
错误的原因。
因此,通过使用-m开关,您可以将包结构信息提供给python,通过它可以成功解析相对导入。
在进行相对导入时,我多次遇到此问题。而且,在阅读了所有先前的答案之后,我仍然无法弄清楚如何解决此问题,而无需在所有文件中添加样板代码。(尽管有些评论确实很有帮助,这要感谢@ncoghlan和@XiongChiamiov)
希望这对正在解决相对进口问题的人有所帮助,因为通过PEP确实很有趣。
Here is the solution which works for me:
I do the relative imports as from ..sub2 import mod2
and then, if I want to run mod1.py
then I go to the parent directory of app
and run the module using the python -m switch as python -m app.sub1.mod1
.
The real reason why this problem occurs with relative imports, is that relative imports works by taking the __name__
property of the module. If the module is being directly run, then __name__
is set to __main__
and it doesn’t contain any information about package structure. And, thats why python complains about the relative import in non-package
error.
So, by using the -m switch you provide the package structure information to python, through which it can resolve the relative imports successfully.
I have encountered this problem many times while doing relative imports. And, after reading all the previous answers, I was still not able to figure out how to solve it, in a clean way, without needing to put boilerplate code in all files. (Though some of the comments were really helpful, thanks to @ncoghlan and @XiongChiamiov)
Hope this helps someone who is fighting with relative imports problem, because going through PEP is really not fun.
回答 2
main.py
setup.py
app/ ->
__init__.py
package_a/ ->
__init__.py
module_a.py
package_b/ ->
__init__.py
module_b.py
- 你跑
python main.py
。
main.py
确实: import app.package_a.module_a
module_a.py
确实 import app.package_b.module_b
另外2或3可以使用: from app.package_a import module_a
只要您app
在PYTHONPATH中具有此功能,便可以使用。main.py
可能在那时的任何地方。
因此,您编写了一个setup.py
将整个应用程序包和子包复制(安装)到目标系统的python文件夹以及main.py
目标系统的脚本文件夹的代码。
main.py
setup.py
app/ ->
__init__.py
package_a/ ->
__init__.py
module_a.py
package_b/ ->
__init__.py
module_b.py
- You run
python main.py
.
main.py
does: import app.package_a.module_a
module_a.py
does import app.package_b.module_b
Alternatively 2 or 3 could use: from app.package_a import module_a
That will work as long as you have app
in your PYTHONPATH. main.py
could be anywhere then.
So you write a setup.py
to copy (install) the whole app package and subpackages to the target system’s python folders, and main.py
to target system’s script folders.
回答 3
“ Guido将程序包中正在运行的脚本视为反模式”(被拒绝的
PEP-3122)
我花了很多时间试图找到解决方案,在Stack Overflow上阅读了相关文章,并对自己说:“一定有更好的方法!”。好像没有。
“Guido views running scripts within a package as an anti-pattern” (rejected
PEP-3122)
I have spent so much time trying to find a solution, reading related posts here on Stack Overflow and saying to myself “there must be a better way!”. Looks like there is not.
回答 4
这可以100%解决:
在app / main.py中导入settings / local_setting.py:
main.py:
import sys
sys.path.insert(0, "../settings")
try:
from local_settings import *
except ImportError:
print('No Import')
This is solved 100%:
Import settings/local_setting.py in app/main.py:
main.py:
import sys
sys.path.insert(0, "../settings")
try:
from local_settings import *
except ImportError:
print('No Import')
回答 5
def import_path(fullpath):
"""
Import a file with full path specification. Allows one to
import from anywhere, something __import__ does not do.
"""
path, filename = os.path.split(fullpath)
filename, ext = os.path.splitext(filename)
sys.path.append(path)
module = __import__(filename)
reload(module) # Might be out of date
del sys.path[-1]
return module
我正在使用此代码片段从路径导入模块,希望对您有所帮助
def import_path(fullpath):
"""
Import a file with full path specification. Allows one to
import from anywhere, something __import__ does not do.
"""
path, filename = os.path.split(fullpath)
filename, ext = os.path.splitext(filename)
sys.path.append(path)
module = __import__(filename)
reload(module) # Might be out of date
del sys.path[-1]
return module
I’m using this snippet to import modules from paths, hope that helps
回答 6
nosklo's
举例说明答案
注意:所有__init__.py
文件均为空。
main.py
app/ ->
__init__.py
package_a/ ->
__init__.py
fun_a.py
package_b/ ->
__init__.py
fun_b.py
app / package_a / fun_a.py
def print_a():
print 'This is a function in dir package_a'
app / package_b / fun_b.py
from app.package_a.fun_a import print_a
def print_b():
print 'This is a function in dir package_b'
print 'going to call a function in dir package_a'
print '-'*30
print_a()
main.py
from app.package_b import fun_b
fun_b.print_b()
如果运行,$ python main.py
它将返回:
This is a function in dir package_b
going to call a function in dir package_a
------------------------------
This is a function in dir package_a
- main.py可以:
from app.package_b import fun_b
- fun_b.py确实
from app.package_a.fun_a import print_a
所以文件夹中的文件package_b
使用了文件夹中的文件package_a
,这就是您想要的。对??
explanation of nosklo's
answer with examples
note: all __init__.py
files are empty.
main.py
app/ ->
__init__.py
package_a/ ->
__init__.py
fun_a.py
package_b/ ->
__init__.py
fun_b.py
app/package_a/fun_a.py
def print_a():
print 'This is a function in dir package_a'
app/package_b/fun_b.py
from app.package_a.fun_a import print_a
def print_b():
print 'This is a function in dir package_b'
print 'going to call a function in dir package_a'
print '-'*30
print_a()
main.py
from app.package_b import fun_b
fun_b.print_b()
if you run $ python main.py
it returns:
This is a function in dir package_b
going to call a function in dir package_a
------------------------------
This is a function in dir package_a
- main.py does:
from app.package_b import fun_b
- fun_b.py does
from app.package_a.fun_a import print_a
so file in folder package_b
used file in folder package_a
, which is what you want. Right??
回答 7
不幸的是,这是一个sys.path hack,但是效果很好。
我在另一层遇到了这个问题:我已经有一个具有指定名称的模块,但这是错误的模块。
我想要做的是以下操作(我正在使用的模块是module3):
mymodule\
__init__.py
mymodule1\
__init__.py
mymodule1_1
mymodule2\
__init__.py
mymodule2_1
import mymodule.mymodule1.mymodule1_1
请注意,我已经安装了mymodule,但是在我的安装中我没有“ mymodule1”
而且我会收到一个ImportError,因为它试图从安装的模块中导入。
我试图做一个sys.path.append,但是没有用。起作用的是sys.path.insert
if __name__ == '__main__':
sys.path.insert(0, '../..')
有点骇客,但一切正常!因此请记住,如果要决定覆盖其他路径,则需要使用sys.path.insert(0,pathname)使其起作用!这对我来说是一个非常令人沮丧的症结所在,有人说要在sys.path中使用“ append”功能,但是如果您已经定义了一个模块,那是行不通的(我发现这是非常奇怪的行为)
This is unfortunately a sys.path hack, but it works quite well.
I encountered this problem with another layer: I already had a module of the specified name, but it was the wrong module.
what I wanted to do was the following (the module I was working from was module3):
mymodule\
__init__.py
mymodule1\
__init__.py
mymodule1_1
mymodule2\
__init__.py
mymodule2_1
import mymodule.mymodule1.mymodule1_1
Note that I have already installed mymodule, but in my installation I do not have “mymodule1”
and I would get an ImportError because it was trying to import from my installed modules.
I tried to do a sys.path.append, and that didn’t work. What did work was a sys.path.insert
if __name__ == '__main__':
sys.path.insert(0, '../..')
So kind of a hack, but got it all to work!
So keep in mind, if you want your decision to override other paths then you need to use sys.path.insert(0, pathname) to get it to work! This was a very frustrating sticking point for me, allot of people say to use the “append” function to sys.path, but that doesn’t work if you already have a module defined (I find it very strange behavior)
回答 8
让我将其放在此处以供我自己参考。我知道这不是很好的Python代码,但是我需要一个脚本来处理我正在处理的项目,所以我想将脚本放在scripts
目录中。
import os.path
import sys
sys.path.append(os.path.abspath(os.path.join(os.path.dirname(__file__), "..")))
Let me just put this here for my own reference. I know that it is not good Python code, but I needed a script for a project I was working on and I wanted to put the script in a scripts
directory.
import os.path
import sys
sys.path.append(os.path.abspath(os.path.join(os.path.dirname(__file__), "..")))
回答 9
正如@EvgeniSergeev在对OP的注释中所说,您可以使用以下.py
命令从任意位置的文件导入代码:
import imp
foo = imp.load_source('module.name', '/path/to/file.py')
foo.MyClass()
这是从此SO答案中获取的。
As @EvgeniSergeev says in the comments to the OP, you can import code from a .py
file at an arbitrary location with:
import imp
foo = imp.load_source('module.name', '/path/to/file.py')
foo.MyClass()
This is taken from this SO answer.
回答 10
回答 11
从Python文档中,
在Python 2.5中,您可以使用from __future__ import absolute_import
指令将导入的行为切换为绝对导入。在将来的版本(可能是Python 2.7)中,此绝对导入行为将成为默认设置。一旦将绝对导入设置为默认设置,import string
就将始终找到标准库的版本。建议用户应尽可能多地开始使用绝对导入,因此最好开始编写from pkg import string
代码
From Python doc,
In Python 2.5, you can switch import‘s behaviour to absolute imports using a from __future__ import absolute_import
directive. This absolute- import behaviour will become the default in a future version (probably Python 2.7). Once absolute imports are the default, import string
will always find the standard library’s version. It’s suggested that users should begin using absolute imports as much as possible, so it’s preferable to begin writing from pkg import string
in your code
回答 12
我发现将“ PYTHONPATH”环境变量设置为顶层文件夹更容易:
bash$ export PYTHONPATH=/PATH/TO/APP
然后:
import sub1.func1
#...more import
当然,PYTHONPATH是“全局”的,但它并没有给我带来麻烦。
I found it’s more easy to set “PYTHONPATH” enviroment variable to the top folder:
bash$ export PYTHONPATH=/PATH/TO/APP
then:
import sub1.func1
#...more import
of course, PYTHONPATH is “global”, but it didn’t raise trouble for me yet.
回答 13
除了John B所说的,似乎设置__package__
变量应该有帮助,而不是更改变量__main__
可能会搞砸其他事情。但是据我测试,它并不能完全正常工作。
我遇到了同样的问题sys.path
,而据我所知,PEP 328或366都无法完全解决问题,因为两者在一天结束时都需要将包装的标头包括在其中。
我还要提及的是,我没有找到如何格式化应该放入这些变量的字符串的格式。是"package_head.subfolder.module_name"
还是什么?
On top of what John B said, it seems like setting the __package__
variable should help, instead of changing __main__
which could screw up other things. But as far as I could test, it doesn’t completely work as it should.
I have the same problem and neither PEP 328 or 366 solve the problem completely, as both, by the end of the day, need the head of the package to be included in sys.path
, as far as I could understand.
I should also mention that I did not find how to format the string that should go into those variables. Is it "package_head.subfolder.module_name"
or what?
回答 14
您必须将模块的路径附加到PYTHONPATH
:
export PYTHONPATH="${PYTHONPATH}:/path/to/your/module/"
You have to append the module’s path to PYTHONPATH
:
export PYTHONPATH="${PYTHONPATH}:/path/to/your/module/"