问题:Python 3中的相对导入

我想从同一目录中的另一个文件导入函数。

有时它对我有用,from .mymodule import myfunction但有时我得到:

SystemError: Parent module '' not loaded, cannot perform relative import

有时它可与一起使用from mymodule import myfunction,但有时我也会得到:

SystemError: Parent module '' not loaded, cannot perform relative import

我不了解这里的逻辑,也找不到任何解释。这看起来完全是随机的。

有人可以向我解释所有这些背后的逻辑是什么?

I want to import a function from another file in the same directory.

Sometimes it works for me with from .mymodule import myfunction but sometimes I get a:

SystemError: Parent module '' not loaded, cannot perform relative import

Sometimes it works with from mymodule import myfunction, but sometimes I also get a:

SystemError: Parent module '' not loaded, cannot perform relative import

I don’t understand the logic here, and I couldn’t find any explanation. This looks completely random.

Could someone explain to me what’s the logic behind all this?


回答 0

不幸的是,该模块需要位于程序包内部,有时还需要作为脚本运行。知道如何实现吗?

像这样的布局很普遍…

main.py
mypackage/
    __init__.py
    mymodule.py
    myothermodule.py

mymodule.py像这样…

#!/usr/bin/env python3

# Exported function
def as_int(a):
    return int(a)

# Test function for module  
def _test():
    assert as_int('1') == 1

if __name__ == '__main__':
    _test()

……一个myothermodule.py像这样…

#!/usr/bin/env python3

from .mymodule import as_int

# Exported function
def add(a, b):
    return as_int(a) + as_int(b)

# Test function for module  
def _test():
    assert add('1', '1') == 2

if __name__ == '__main__':
    _test()

main.py这样的…

#!/usr/bin/env python3

from mypackage.myothermodule import add

def main():
    print(add('1', '1'))

if __name__ == '__main__':
    main()

…在您运行main.py或时工作正常mypackage/mymodule.py,但mypackage/myothermodule.py由于相对导入而失败,…

from .mymodule import as_int

您应该运行它的方式是…

python3 -m mypackage.myothermodule

…但是有些冗长,并且与像这样的shebang行不能很好地融合在一起#!/usr/bin/env python3

假设名称mymodule在全球范围内是唯一的,这种情况下最简单的解决方法是避免使用相对导入,而只需使用…

from mymodule import as_int

…尽管它不是唯一的,或者您的包结构更复杂,您仍需要在中包含包含包目录的目录PYTHONPATH,并按以下步骤进行操作…

from mypackage.mymodule import as_int

…或者如果您希望它“开箱即用”运行,则可以PYTHONPATH使用此方法首先获取输入代码…

import sys
import os

PACKAGE_PARENT = '..'
SCRIPT_DIR = os.path.dirname(os.path.realpath(os.path.join(os.getcwd(), os.path.expanduser(__file__))))
sys.path.append(os.path.normpath(os.path.join(SCRIPT_DIR, PACKAGE_PARENT)))

from mypackage.mymodule import as_int

这有点痛苦,但是有一个线索可以说明为什么某位Guido van Rossum写的电子邮件中

我对此表示怀疑,也对任何其他提议的__main__ 机械装置都为-1 。唯一的用例似乎是正在运行的脚本,它们恰好位于模块目录中,我一直将其视为反模式。为了让我改变主意,您必须说服我不要。

在程序包中运行脚本是否是反模式是主观的,但是就我个人而言,我发现它在包含一些自定义wxPython小部件的程序包中非常有用,因此我可以为任何源文件运行脚本以wx.Frame仅显示包含该小部件用于测试目的。

unfortunately, this module needs to be inside the package, and it also needs to be runnable as a script, sometimes. Any idea how I could achieve that?

It’s quite common to have a layout like this…

main.py
mypackage/
    __init__.py
    mymodule.py
    myothermodule.py

…with a mymodule.py like this…

#!/usr/bin/env python3

# Exported function
def as_int(a):
    return int(a)

# Test function for module  
def _test():
    assert as_int('1') == 1

if __name__ == '__main__':
    _test()

…a myothermodule.py like this…

#!/usr/bin/env python3

from .mymodule import as_int

# Exported function
def add(a, b):
    return as_int(a) + as_int(b)

# Test function for module  
def _test():
    assert add('1', '1') == 2

if __name__ == '__main__':
    _test()

…and a main.py like this…

#!/usr/bin/env python3

from mypackage.myothermodule import add

def main():
    print(add('1', '1'))

if __name__ == '__main__':
    main()

…which works fine when you run main.py or mypackage/mymodule.py, but fails with mypackage/myothermodule.py, due to the relative import…

from .mymodule import as_int

The way you’re supposed to run it is…

python3 -m mypackage.myothermodule

…but it’s somewhat verbose, and doesn’t mix well with a shebang line like #!/usr/bin/env python3.

The simplest fix for this case, assuming the name mymodule is globally unique, would be to avoid using relative imports, and just use…

from mymodule import as_int

…although, if it’s not unique, or your package structure is more complex, you’ll need to include the directory containing your package directory in PYTHONPATH, and do it like this…

from mypackage.mymodule import as_int

…or if you want it to work “out of the box”, you can frob the PYTHONPATH in code first with this…

import sys
import os

PACKAGE_PARENT = '..'
SCRIPT_DIR = os.path.dirname(os.path.realpath(os.path.join(os.getcwd(), os.path.expanduser(__file__))))
sys.path.append(os.path.normpath(os.path.join(SCRIPT_DIR, PACKAGE_PARENT)))

from mypackage.mymodule import as_int

It’s kind of a pain, but there’s a clue as to why in an email written by a certain Guido van Rossum…

I’m -1 on this and on any other proposed twiddlings of the __main__ machinery. The only use case seems to be running scripts that happen to be living inside a module’s directory, which I’ve always seen as an antipattern. To make me change my mind you’d have to convince me that it isn’t.

Whether running scripts inside a package is an antipattern or not is subjective, but personally I find it really useful in a package I have which contains some custom wxPython widgets, so I can run the script for any of the source files to display a wx.Frame containing only that widget for testing purposes.


回答 1

说明

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__’), 则相对导入的解析就好像该模块是顶级模块一样,无论该模块实际位于文件系统上的哪个位置。

在某些时候,PEP 338PEP 328冲突:

…相对导入依赖于__name__来确定当前模块在包层次结构中的位置。在主模块中,__name__的值始终为‘__main__’,因此显式相对导入将始终失败(因为它们仅适用于包中的模块)

为了解决这个问题,PEP 366引入了顶层变量__package__

通过添加新的模块级别属性,如果使用-m 开关执行模块,则该PEP允许相对导入自动进行。当按名称执行文件时,模块本身中的少量样板文件将允许相对导入工作。[…]如果存在[属性],则相对导入将基于此属性而不是模块__name__属性。[…]当通过其文件名指定主模块时,__package__属性将设置为None。[…] 当导入系统在未设置__package__的模块(或将其设置为None)的模块中遇到显式相对导入时,它将计算并存储正确的值__name __。rpartition(’。’)[0]用于常规模块__ name__用于程序包初始化模块)

(强调我的)

如果__name__'__main__',则__name__.rpartition('.')[0]返回空字符串。这就是为什么错误描述中有空字符串文字的原因:

SystemError: Parent module '' not loaded, cannot perform relative import

CPython PyImport_ImportModuleLevelObject函数的相关部分:

if (PyDict_GetItem(interp->modules, package) == NULL) {
    PyErr_Format(PyExc_SystemError,
            "Parent module %R not loaded, cannot perform relative "
            "import", package);
    goto error;
}

如果CPython packageinterp->modules(可通过访问)中找不到(包的名称),则会引发此异常sys.modules。由于sys.modules“将模块名称映射到已经加载的模块的字典”,因此现在很清楚,必须在执行相对导入之前显式绝对导入父模块

注意:问题18018中 的补丁添加了另一个ifblock,它将在以上代码之前执行:

if (PyUnicode_CompareWithASCIIString(package, "") == 0) {
    PyErr_SetString(PyExc_ImportError,
            "attempted relative import with no known parent package");
    goto error;
} /* else if (PyDict_GetItem(interp->modules, package) == NULL) {
    ...
*/

如果package(与上面相同)为空字符串,则错误消息将为

ImportError: attempted relative import with no known parent package

但是,您只会在Python 3.6或更高版本中看到它。

解决方案1:使用-m运行脚本

考虑一个目录(这是一个Python ):

.
├── package
│   ├── __init__.py
│   ├── module.py
│   └── standalone.py

软件包中的所有文件均以相同的两行代码开头:

from pathlib import Path
print('Running' if __name__ == '__main__' else 'Importing', Path(__file__).resolve())

我加入这两行只是为了使操作顺序显而易见。我们可以完全忽略它们,因为它们不会影响执行。

__init__.pymodule.py仅包含这两行(即,它们实际上是空的)。

standalone.py另外尝试通过相对导入来导入module.py

from . import module  # explicit relative import

我们深知这/path/to/python/interpreter package/standalone.py将失败。但是,我们可以使用-m命令行选项运行该模块,该选项“搜索sys.path命名的模块并将其内容作为__main__模块执行”

vaultah@base:~$ python3 -i -m package.standalone
Importing /home/vaultah/package/__init__.py
Running /home/vaultah/package/standalone.py
Importing /home/vaultah/package/module.py
>>> __file__
'/home/vaultah/package/standalone.py'
>>> __package__
'package'
>>> # The __package__ has been correctly set and module.py has been imported.
... # What's inside sys.modules?
... import sys
>>> sys.modules['__main__']
<module 'package.standalone' from '/home/vaultah/package/standalone.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/package/module.py'>
>>> sys.modules['package']
<module 'package' from '/home/vaultah/package/__init__.py'>

-m为您完成所有导入工作并自动设置__package__,但是您可以在

解决方案2:手动设置__package__

请把它当作概念证明而不是实际解决方案。它不适合在实际代码中使用。

PEP 366可以解决此问题,但是它不完整,因为__package__仅设置是不够的。您将需要至少在模块层次结构中导入N个先前的软件包,其中N是要搜索要导入的模块的父目录(相对于脚本目录)的数量。

从而,

  1. 将当前模块的第N个前辈的父目录添加到sys.path

  2. 从中删除当前文件的目录 sys.path

  3. 使用标准名称导入当前模块的父模块

  4. 设置__package__2的标准名称

  5. 执行相对导入

我将从解决方案1中借用文件,并添加更多子包:

package
├── __init__.py
├── module.py
└── subpackage
    ├── __init__.py
    └── subsubpackage
        ├── __init__.py
        └── standalone.py

这次standalone.py将使用以下相对导入方式从软件包中导入module.py

from ... import module  # N = 3

我们需要在该行之前加上样板代码,以使其正常工作。

import sys
from pathlib import Path

if __name__ == '__main__' and __package__ is None:
    file = Path(__file__).resolve()
    parent, top = file.parent, file.parents[3]

    sys.path.append(str(top))
    try:
        sys.path.remove(str(parent))
    except ValueError: # Already removed
        pass

    import package.subpackage.subsubpackage
    __package__ = 'package.subpackage.subsubpackage'

from ... import module # N = 3

它允许我们按文件名执行standalone.py

vaultah@base:~$ python3 package/subpackage/subsubpackage/standalone.py
Running /home/vaultah/package/subpackage/subsubpackage/standalone.py
Importing /home/vaultah/package/__init__.py
Importing /home/vaultah/package/subpackage/__init__.py
Importing /home/vaultah/package/subpackage/subsubpackage/__init__.py
Importing /home/vaultah/package/module.py

包裹在一个功能更通用的解决方案,可以发现在这里。用法示例:

if __name__ == '__main__' and __package__ is None:
    import_parents(level=3) # N = 3

from ... import module
from ...module.submodule import thing

解决方案3:使用绝对导入和设置工具

步骤是-

  1. 将显式相对导入替换为等效的绝对导入

  2. 安装package以使其可导入

例如,目录结构可以如下

.
├── project
│   ├── package
│   │   ├── __init__.py
│   │   ├── module.py
│   │   └── standalone.py
│   └── setup.py

其中setup.py

from setuptools import setup, find_packages
setup(
    name = 'your_package_name',
    packages = find_packages(),
)

其余文件是从解决方案#1借用的。

安装后,无论您的工作目录如何,都可以导入软件包(假设没有命名问题)。

我们可以修改standalone.py以利用这一优势(步骤1):

from package import module  # absolute import

将工作目录更改为project并运行/path/to/python/interpreter setup.py install --user--user将软件包安装在site-packages目录中)(步骤2):

vaultah@base:~$ cd project
vaultah@base:~/project$ python3 setup.py install --user

让我们验证一下现在可以将standalone.py作为脚本运行:

vaultah@base:~/project$ python3 -i package/standalone.py
Running /home/vaultah/project/package/standalone.py
Importing /home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/__init__.py
Importing /home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py
>>> module
<module 'package.module' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py'>
>>> import sys
>>> sys.modules['package']
<module 'package' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/__init__.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py'>

注意:如果您决定走这条路,最好使用虚拟环境来隔离安装软件包。

解决方案4:使用绝对导入和一些样板代码

坦白地说,不需要安装-您可以在脚本中添加一些样板代码以使绝对导入工作。

我将从解决方案1借用文件并更改standalone.py

  1. 尝试使用绝对导入从包中导入任何内容之前,将的父目录添加到:sys.path

    import sys
    from pathlib import Path # if you haven't already done so
    file = Path(__file__).resolve()
    parent, root = file.parent, file.parents[1]
    sys.path.append(str(root))
    
    # Additionally remove the current file's directory from sys.path
    try:
        sys.path.remove(str(parent))
    except ValueError: # Already removed
        pass
  2. 用绝对导入替换相对导入:

    from package import module  # absolute import

standalone.py运行没有问题:

vaultah@base:~$ python3 -i package/standalone.py
Running /home/vaultah/package/standalone.py
Importing /home/vaultah/package/__init__.py
Importing /home/vaultah/package/module.py
>>> module
<module 'package.module' from '/home/vaultah/package/module.py'>
>>> import sys
>>> sys.modules['package']
<module 'package' from '/home/vaultah/package/__init__.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/package/module.py'>

我认为我应该警告您:请不要这样做,尤其是在您的项目结构复杂的情况下。


作为附带说明,PEP 8建议使用绝对导入,但指出在某些情况下,显式相对导入是可以接受的:

建议使用绝对导入,因为它们通常更具可读性,并且往往表现得更好(或至少会提供更好的错误消息)。[…]但是,显式相对导入是绝对导入的一种可接受的替代方法,尤其是在处理复杂的包装布局时,使用绝对导入会不必要地冗长。

Explanation

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.

At some point PEP 338 conflicted with PEP 328:

… relative imports rely on __name__ to determine the current module’s position in the package hierarchy. In a main module, the value of __name__ is always ‘__main__’, so explicit relative imports will always fail (as they only work for a module inside a package)

and to address the issue, PEP 366 introduced the top level variable __package__:

By adding a new module level attribute, this PEP allows relative imports to work automatically if the module is executed using the -m switch. A small amount of boilerplate in the module itself will allow the relative imports to work when the file is executed by name. […] When it [the attribute] is present, relative imports will be based on this attribute rather than the module __name__ attribute. […] When the main module is specified by its filename, then the __package__ attribute will be set to None. […] When the import system encounters an explicit relative import in a module without __package__ set (or with it set to None), it will calculate and store the correct value (__name__.rpartition(‘.’)[0] for normal modules and __name__ for package initialisation modules)

(emphasis mine)

If the __name__ is '__main__', __name__.rpartition('.')[0] returns empty string. This is why there’s empty string literal in the error description:

SystemError: Parent module '' not loaded, cannot perform relative import

The relevant part of the CPython’s PyImport_ImportModuleLevelObject function:

if (PyDict_GetItem(interp->modules, package) == NULL) {
    PyErr_Format(PyExc_SystemError,
            "Parent module %R not loaded, cannot perform relative "
            "import", package);
    goto error;
}

CPython raises this exception if it was unable to find package (the name of the package) in interp->modules (accessible as sys.modules). Since sys.modules is “a dictionary that maps module names to modules which have already been loaded”, it’s now clear that the parent module must be explicitly absolute-imported before performing relative import.

Note: The patch from the issue 18018 has added another if block, which will be executed before the code above:

if (PyUnicode_CompareWithASCIIString(package, "") == 0) {
    PyErr_SetString(PyExc_ImportError,
            "attempted relative import with no known parent package");
    goto error;
} /* else if (PyDict_GetItem(interp->modules, package) == NULL) {
    ...
*/

If package (same as above) is empty string, the error message will be

ImportError: attempted relative import with no known parent package

However, you will only see this in Python 3.6 or newer.

Solution #1: Run your script using -m

Consider a directory (which is a Python package):

.
├── package
│   ├── __init__.py
│   ├── module.py
│   └── standalone.py

All of the files in package begin with the same 2 lines of code:

from pathlib import Path
print('Running' if __name__ == '__main__' else 'Importing', Path(__file__).resolve())

I’m including these two lines only to make the order of operations obvious. We can ignore them completely, since they don’t affect the execution.

__init__.py and module.py contain only those two lines (i.e., they are effectively empty).

standalone.py additionally attempts to import module.py via relative import:

from . import module  # explicit relative import

We’re well aware that /path/to/python/interpreter package/standalone.py will fail. However, we can run the module with the -m command line option that will “search sys.path for the named module and execute its contents as the __main__ module”:

vaultah@base:~$ python3 -i -m package.standalone
Importing /home/vaultah/package/__init__.py
Running /home/vaultah/package/standalone.py
Importing /home/vaultah/package/module.py
>>> __file__
'/home/vaultah/package/standalone.py'
>>> __package__
'package'
>>> # The __package__ has been correctly set and module.py has been imported.
... # What's inside sys.modules?
... import sys
>>> sys.modules['__main__']
<module 'package.standalone' from '/home/vaultah/package/standalone.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/package/module.py'>
>>> sys.modules['package']
<module 'package' from '/home/vaultah/package/__init__.py'>

-m does all the importing stuff for you and automatically sets __package__, but you can do that yourself in the

Solution #2: Set __package__ manually

Please treat it as a proof of concept rather than an actual solution. It isn’t well-suited for use in real-world code.

PEP 366 has a workaround to this problem, however, it’s incomplete, because setting __package__ alone is not enough. You’re going to need to import at least N preceding packages in the module hierarchy, where N is the number of parent directories (relative to the directory of the script) that will be searched for the module being imported.

Thus,

  1. Add the parent directory of the Nth predecessor of the current module to sys.path

  2. Remove the current file’s directory from sys.path

  3. Import the parent module of the current module using its fully-qualified name

  4. Set __package__ to the fully-qualified name from 2

  5. Perform the relative import

I’ll borrow files from the Solution #1 and add some more subpackages:

package
├── __init__.py
├── module.py
└── subpackage
    ├── __init__.py
    └── subsubpackage
        ├── __init__.py
        └── standalone.py

This time standalone.py will import module.py from the package package using the following relative import

from ... import module  # N = 3

We’ll need to precede that line with the boilerplate code, to make it work.

import sys
from pathlib import Path

if __name__ == '__main__' and __package__ is None:
    file = Path(__file__).resolve()
    parent, top = file.parent, file.parents[3]

    sys.path.append(str(top))
    try:
        sys.path.remove(str(parent))
    except ValueError: # Already removed
        pass

    import package.subpackage.subsubpackage
    __package__ = 'package.subpackage.subsubpackage'

from ... import module # N = 3

It allows us to execute standalone.py by filename:

vaultah@base:~$ python3 package/subpackage/subsubpackage/standalone.py
Running /home/vaultah/package/subpackage/subsubpackage/standalone.py
Importing /home/vaultah/package/__init__.py
Importing /home/vaultah/package/subpackage/__init__.py
Importing /home/vaultah/package/subpackage/subsubpackage/__init__.py
Importing /home/vaultah/package/module.py

A more general solution wrapped in a function can be found here. Example usage:

if __name__ == '__main__' and __package__ is None:
    import_parents(level=3) # N = 3

from ... import module
from ...module.submodule import thing

Solution #3: Use absolute imports and setuptools

The steps are –

  1. Replace explicit relative imports with equivalent absolute imports

  2. Install package to make it importable

For instance, the directory structure may be as follows

.
├── project
│   ├── package
│   │   ├── __init__.py
│   │   ├── module.py
│   │   └── standalone.py
│   └── setup.py

where setup.py is

from setuptools import setup, find_packages
setup(
    name = 'your_package_name',
    packages = find_packages(),
)

The rest of the files were borrowed from the Solution #1.

Installation will allow you to import the package regardless of your working directory (assuming there’ll be no naming issues).

We can modify standalone.py to use this advantage (step 1):

from package import module  # absolute import

Change your working directory to project and run /path/to/python/interpreter setup.py install --user (--user installs the package in your site-packages directory) (step 2):

vaultah@base:~$ cd project
vaultah@base:~/project$ python3 setup.py install --user

Let’s verify that it’s now possible to run standalone.py as a script:

vaultah@base:~/project$ python3 -i package/standalone.py
Running /home/vaultah/project/package/standalone.py
Importing /home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/__init__.py
Importing /home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py
>>> module
<module 'package.module' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py'>
>>> import sys
>>> sys.modules['package']
<module 'package' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/__init__.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/.local/lib/python3.6/site-packages/your_package_name-0.0.0-py3.6.egg/package/module.py'>

Note: If you decide to go down this route, you’d be better off using virtual environments to install packages in isolation.

Solution #4: Use absolute imports and some boilerplate code

Frankly, the installation is not necessary – you could add some boilerplate code to your script to make absolute imports work.

I’m going to borrow files from Solution #1 and change standalone.py:

  1. Add the parent directory of package to sys.path before attempting to import anything from package using absolute imports:

    import sys
    from pathlib import Path # if you haven't already done so
    file = Path(__file__).resolve()
    parent, root = file.parent, file.parents[1]
    sys.path.append(str(root))
    
    # Additionally remove the current file's directory from sys.path
    try:
        sys.path.remove(str(parent))
    except ValueError: # Already removed
        pass
    
  2. Replace the relative import by the absolute import:

    from package import module  # absolute import
    

standalone.py runs without problems:

vaultah@base:~$ python3 -i package/standalone.py
Running /home/vaultah/package/standalone.py
Importing /home/vaultah/package/__init__.py
Importing /home/vaultah/package/module.py
>>> module
<module 'package.module' from '/home/vaultah/package/module.py'>
>>> import sys
>>> sys.modules['package']
<module 'package' from '/home/vaultah/package/__init__.py'>
>>> sys.modules['package.module']
<module 'package.module' from '/home/vaultah/package/module.py'>

I feel that I should warn you: try not to do this, especially if your project has a complex structure.


As a side note, PEP 8 recommends the use of absolute imports, but states that in some scenarios explicit relative imports are acceptable:

Absolute imports are recommended, as they are usually more readable and tend to be better behaved (or at least give better error messages). […] However, explicit relative imports are an acceptable alternative to absolute imports, especially when dealing with complex package layouts where using absolute imports would be unnecessarily verbose.


回答 2

将其放入包的__init__.py文件中

# For relative imports to work in Python 3.6
import os, sys; sys.path.append(os.path.dirname(os.path.realpath(__file__)))

假设您的包裹是这样的:

├── project
   ├── package
      ├── __init__.py
      ├── module1.py
      └── module2.py
   └── setup.py

现在在包中使用常规导入,例如:

# in module2.py
from module1 import class1

这适用于python 2和3。

Put this inside your package’s __init__.py file:

# For relative imports to work in Python 3.6
import os, sys; sys.path.append(os.path.dirname(os.path.realpath(__file__)))

Assuming your package is like this:

├── project
│   ├── package
│   │   ├── __init__.py
│   │   ├── module1.py
│   │   └── module2.py
│   └── setup.py

Now use regular imports in you package, like:

# in module2.py
from module1 import class1

This works in both python 2 and 3.


回答 3

我遇到了这个问题。黑客的解决方法是通过if / else块导入,如下所示:

#!/usr/bin/env python3
#myothermodule

if __name__ == '__main__':
    from mymodule import as_int
else:
    from .mymodule import as_int


# Exported function
def add(a, b):
    return as_int(a) + as_int(b)

# Test function for module  
def _test():
    assert add('1', '1') == 2

if __name__ == '__main__':
    _test()

I ran into this issue. A hack workaround is importing via an if/else block like follows:

#!/usr/bin/env python3
#myothermodule

if __name__ == '__main__':
    from mymodule import as_int
else:
    from .mymodule import as_int


# Exported function
def add(a, b):
    return as_int(a) + as_int(b)

# Test function for module  
def _test():
    assert add('1', '1') == 2

if __name__ == '__main__':
    _test()

回答 4

希望这对那里的某人有价值-我浏览了六堆stackoverflow帖子,试图找出与上面上面发布的内容类似的相对进口量。我按照建议设置了所有内容,但仍在执行ModuleNotFoundError: No module named 'my_module_name'

由于我只是在本地开发和玩耍,所以我没有创建/运行setup.py文件。我也没有明显地设置了我PYTHONPATH

我意识到,当我像在模块位于同一目录中那样运行代码时,找不到模块:

$ python3 test/my_module/module_test.py                                                                                                               2.4.0
Traceback (most recent call last):
  File "test/my_module/module_test.py", line 6, in <module>
    from my_module.module import *
ModuleNotFoundError: No module named 'my_module'

但是,当我明确指定路径时,事情开始起作用:

$ PYTHONPATH=. python3 test/my_module/module_test.py                                                                                                  2.4.0
...........
----------------------------------------------------------------------
Ran 11 tests in 0.001s

OK

因此,如果有人尝试了一些建议,则认为他们的代码结构正确,并且如果您不将当前目录导出到PYTHONPATH,则仍然遇到与我类似的情况:

  1. 运行您的代码,并明确包含如下路径: $ PYTHONPATH=. python3 test/my_module/module_test.py
  2. 为避免调用PYTHONPATH=.,请创建一个setup.py内容如下的文件,然后运行python setup.py development以将软件包添加到路径中:
# setup.py
from setuptools import setup, find_packages

setup(
    name='sample',
    packages=find_packages()
)

Hopefully, this will be of value to someone out there – I went through half a dozen stackoverflow posts trying to figure out relative imports similar to whats posted above here. I set up everything as suggested but I was still hitting ModuleNotFoundError: No module named 'my_module_name'

Since I was just developing locally and playing around, I hadn’t created/run a setup.py file. I also hadn’t apparently set my PYTHONPATH.

I realized that when I ran my code as I had been when the tests were in the same directory as the module, I couldn’t find my module:

$ python3 test/my_module/module_test.py                                                                                                               2.4.0
Traceback (most recent call last):
  File "test/my_module/module_test.py", line 6, in <module>
    from my_module.module import *
ModuleNotFoundError: No module named 'my_module'

However, when I explicitly specified the path things started to work:

$ PYTHONPATH=. python3 test/my_module/module_test.py                                                                                                  2.4.0
...........
----------------------------------------------------------------------
Ran 11 tests in 0.001s

OK

So, in the event that anyone has tried a few suggestions, believes their code is structured correctly and still finds themselves in a similar situation as myself try either of the following if you don’t export the current directory to your PYTHONPATH:

  1. Run your code and explicitly include the path like so: $ PYTHONPATH=. python3 test/my_module/module_test.py
  2. To avoid calling PYTHONPATH=., create a setup.py file with contents like the following and run python setup.py development to add packages to the path:
# setup.py
from setuptools import setup, find_packages

setup(
    name='sample',
    packages=find_packages()
)

回答 5

我需要从主项目目录运行python3才能使其工作。

例如,如果项目具有以下结构:

project_demo/
├── main.py
├── some_package/
   ├── __init__.py
   └── project_configs.py
└── test/
    └── test_project_configs.py

我将在文件夹project_demo /中运行python3 ,然后执行

from some_package import project_configs

I needed to run python3 from the main project directory to make it work.

For example, if the project has the following structure:

project_demo/
├── main.py
├── some_package/
│   ├── __init__.py
│   └── project_configs.py
└── test/
    └── test_project_configs.py

Solution

I would run python3 inside folder project_demo/ and then perform a

from some_package import project_configs

回答 6

为了解决这个问题,我设计了带有重新包装软件包的解决方案,该解决方案已经为我服务了一段时间。它将上层目录添加到lib路径:

import repackage
repackage.up()
from mypackage.mymodule import myfunction

重新打包可以使用智能策略(检查调用堆栈)进行相对导入,从而在各种情况下都起作用。

To obviate this problem, I devised a solution with the repackage package, which has worked for me for some time. It adds the upper directory to the lib path:

import repackage
repackage.up()
from mypackage.mymodule import myfunction

Repackage can make relative imports that work in a wide range of cases, using an intelligent strategy (inspecting the call stack).


回答 7

如果两个软件包都在您的导入路径(sys.path)中,并且您想要的模块/类在example / example.py中,则在没有相对导入的情况下访问该类,请尝试:

from example.example import fkt

if both packages are in your import path (sys.path), and the module/class you want is in example/example.py, then to access the class without relative import try:

from example.example import fkt

回答 8

我认为最好的解决方案是为您的模块创建一个软件包: 是有关如何执行操作的更多信息。

有了软件包后,您就不必担心相对导入了,就可以进行绝对导入。

I think the best solution is to create a package for your module: Here is more info on how to do it.

Once you have a package you don’t need to worry about relative import, you can just do absolute imports.


回答 9

我有一个类似的问题:我需要一个Linux服务和cgi插件,它们使用公共常量进行协作。做到这一点的“自然”方法是将它们放在程序包的init .py中,但是我无法使用-m参数启动cgi插件。

我的最终解决方案与上述解决方案2类似:

import sys
import pathlib as p
import importlib

pp = p.Path(sys.argv[0])
pack = pp.resolve().parent

pkg = importlib.import_module('__init__', package=str(pack))

缺点是必须在常量(或通用函数)前加上pkg:

print(pkg.Glob)

I had a similar problem: I needed a Linux service and cgi plugin which use common constants to cooperate. The ‘natural’ way to do this is to place them in the init.py of the package, but I cannot start the cgi plugin with the -m parameter.

My final solution was similar to Solution #2 above:

import sys
import pathlib as p
import importlib

pp = p.Path(sys.argv[0])
pack = pp.resolve().parent

pkg = importlib.import_module('__init__', package=str(pack))

The disadvantage is that you must prefix the constants (or common functions) with pkg:

print(pkg.Glob)

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