os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir)) 是什么意思？Python
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What does os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir)) mean? python
提问by alvas
In several SO's question there is these lines to access the parent directory of the code, e.g. os.path.join(os.path.dirname(__file__)) returns nothingand os.path.join(os.path.dirname(__file__)) returns nothing
在几个 SO 的问题中有这些行来访问代码的父目录，例如os.path.join(os.path.dirname(__file__)) 不返回任何内容而os.path.join(os.path.dirname(__file__) ) 不返回任何内容
import os, sys
parentddir = os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))
sys.path.append(parentddir)
I understand that os.path.abspath()returns absolute path of something and sys.path.append()adds the path for the code to access. but what is this cryptic line below, what does it really mean?
我知道os.path.abspath()返回某些东西的绝对路径并sys.path.append()添加要访问的代码的路径。但是下面这条神秘的线是什么，它到底是什么意思？
os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))
Is there another way to achieve the same purpose of appending the parent directory of the where the code?
是否有另一种方法可以达到附加代码所在父目录的相同目的？
This problem happens because I am calling functions across directories and sometimes they share the same file name, e.g. script1/utils.pyand script2/utils.py. I am calling a function from script1/test.pywhich calls script2/something.pycontains a function that calls script2/utils.pyand the following code
发生此问题是因为我跨目录调用函数，有时它们共享相同的文件名，例如script1/utils.py和script2/utils.py. 我正在调用一个函数，script1/test.py其中的调用script2/something.py包含一个调用函数script2/utils.py和以下代码
script1/
utils.py
src/
test.py
script2/
utils.py
code/
something.py
test.py
测试文件
from script2.code import something
import sys
sys.path.append('../')
import utils
something.foobar()
something.py
东西.py
import os, sys
parentddir = os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))
sys.path.append(parentddir)
import utils
def foobar():
utils.somefunc()
采纳答案by Paulo Bu
That is a clever way to refer to paths regardless of the script location. The crypticline you're referring is:
无论脚本位置如何，这是一种引用路径的巧妙方法。您所指的神秘路线是：
os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))
There are 3 methods and a 2 constants present:
有 3 种方法和 2 个常量：
abspathreturns absolute path of a pathjoinjoin to path stringsdirnamereturns the directory of a file__file__refers to thescript's file namepardirreturns the representation of a parent directory in the OS (usually..)
abspath返回路径的绝对路径join加入路径字符串dirname返回文件的目录__file__指的是script的文件名pardir返回操作系统中父目录的表示（通常为..）
Thus, the expression returns the full path name of the executing scriptin a multiplatform-safeway. No need to hardwireany directions, that's why it is so useful.
因此，该表达式以多平台安全的方式返回执行脚本的完整路径名。无需硬连线任何方向，这就是它如此有用的原因。
There might be other approaches to get a parent directory of where a file is located, for example, programs have the concept of current working directory, os.getcwd(). So doing os.getcwd()+'/..'might work. But this is very dangerous, because working directories can be changed.
可能还有其他方法来获取文件所在位置的父目录，例如，程序具有当前工作目录的概念，os.getcwd(). 所以这样做os.getcwd()+'/..'可能会奏效。但这非常危险，因为工作目录可以更改。
Also, if the file is intended to be imported, the working directory will point to the importing file, not the importee, but __file__always points to the actual module's file so it is safer.
此外，如果要导入文件，工作目录将指向导入文件，而不是导入者，但__file__始终指向实际模块的文件，因此更安全。
Hope this helps!
希望这可以帮助！
Edit: P.S. - Python 3 greatly simplifies this situation by letting us treat paths in an object-oriented manner, so the above line becomes:
编辑：PS - Python 3 通过让我们以面向对象的方式处理路径，大大简化了这种情况，因此上面的行变为：
from pathlib import Path
Path(__file__).resolve().parent.parent
回答by praveen
__file__represents the file the code is executing from
__file__表示代码正在执行的文件
os.path.dirname(__file__)gives you the directory the file is in
os.path.dirname(__file__)给你文件所在的目录
os.path.pardirstands for ".." which means one directory above the current one
os.path.pardir代表“..”，表示在当前目录之上一个目录
os.path.join(os.path.dirname(__file__), os.path.pardir)joins the directory name and ".."
os.path.join(os.path.dirname(__file__), os.path.pardir)加入目录名和“..”
os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))resolves the above path and gives you an absolute path for the parent directory of the directory your file is in
os.path.abspath(os.path.join(os.path.dirname(__file__), os.path.pardir))解析上述路径并为您提供文件所在目录的父目录的绝对路径
