Python open() 需要完整路径

Question

这个问题已经有答案了:

open() gives FileNotFoundError / IOError: '[Errno 2] No such file or directory' (10 个回答)

已关闭 5 个月前。

我正在编写一个脚本来读取 csv 文件。 csv 文件和脚本位于同一目录中。但是当我尝试打开该文件时，它给了我 FileNotFoundError: [Errno 2] No such file or directory: 'zipcodes.csv' 。我用来读取文件的代码是

with open('zipcodes.csv', 'r') as zipcode_file:
    reader = csv.DictReader(zipcode_file)

如果我给出文件的完整路径，它就会起作用。为什么open()需要文件的完整路径？

Answer 1

来自documentation :

open(file, mode='r', buffering=-1, encoding=None, errors=None,newline=None, closefd=True, opener=None)

file is a path-like object giving the pathname (absolute or relativeto the current working directory) of the file to be opened or aninteger file descriptor of the file to be wrapped.

因此，如果您要打开的文件不在运行脚本的当前文件夹中，您可以使用绝对路径，或者通过使用以下方式获取工作目录或/和绝对路径:

import os
# Look to the path of your current working directory
working_directory = os.getcwd()
# Or: file_path = os.path.join(working_directory, 'my_file.py')
file_path = working_directory + 'my_file.py'

或者，您可以在运行脚本时检索绝对路径，使用:

import os
# Look for your absolute directory path
absolute_path = os.path.dirname(os.path.abspath(__file__))
# Or: file_path = os.path.join(absolute_path, 'folder', 'my_file.py')
file_path = absolute_path + '/folder/my_file.py'

如果您想与操作系统无关，那么您可以使用:

file_path = os.path.join(absolute_path, folder, my_file.py)