如何使在中间具有可变输入的路径在python的输入文件中读取 [英] How to make a path that has variable input in the middle read in a input file in python

问题描述

这应该很容易，不确定为什么我无法使它正常工作.我正在尝试导入大量的.txt文件，作为更大的过程的一部分，例如:

This should be pretty easy, not sure why I can't get it to work. I am trying to import a ton of .txt files as part of a larger process like so:

    path = "C:/Users/A/B/"

    with open(path + "*full.txt","r") as f:
        contents =f.read()
        print(contents)  

我只是尝试导入此文件夹路径中的所有.txt文件(其中有很多)，当我这样做时，我会得到:

I am just trying to import all .txt files (there are a ton of them) in this folder path, when I do this I get:

OSError: [Errno 22] Invalid argument: 

中间的字符串在每个文件之间是不同的，因此，在完整文件前的* 它在参数后列出了路径(出于隐私原因，我会省略它，但您会明白这一点)，而且我知道该路径是正确的，为什么会给我这个错误?

There are strings in the middle that are different between each file hence the * before the full it lists the path after argument (for privacy reasons I will leave it out but you get the point) and I know that the path is correct, why is it giving me this error?

推荐答案

您不能在open()中使用*. open()只能打开一个名称完全相同的文件.

You can't use * in open(). open() can open only one file with exact name.

您必须获取目录中的所有名称，并使用for -loop单独打开每个文件.

You have to get all names in directory and use for-loop to open every file separatelly.

使用glob.glob():

import glob

path = "C:/Users/A/B/"

for fullname in glob.glob( path + "*full.txt" ):
    with open(fullname, "r") as f:
        contents = f.read()
        print(contents)

使用os.listdir():

import os

path = "C:/Users/A/B/"

for name in os.listdir(path):
    if name.endswith("full.txt"):
        fullname = os.path.join(path, name):
        with open(fullname, "r") as f:
            contents = f.read()
            print(contents)

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