在文件夾中讀取同一個文件的文件

Question

我需要一點幫助來完成我的程序。 我在一個文件夾中有20個相同類型的文件，字符串與相應的值。 有沒有辦法創建一個可以這種方式打開所有文件的函數
file1 = [line.strip() for line in open("/Python34/elez/file1.txt", "r")]？

我希望我解釋得很好。 謝謝！

Answer 1

from os import listdir 
from os.path import join, isfile 

def contents(filepath): 
    with open(filepath) as f: 
     return f.read() 

directory = '/Python34/elez' 

all_file_contents = [contents(join(directory, filename)) 
        for filename in listdir(directory) 
        if isfile(join(directory, filename)] 

Answer 2

喜格列佛，這是我怎麼會做它：

import os 

all_files = [] ## create a list to keep all the lines for all files 


for file in os.listdir('./'): ## use list dir to list all files in the dir 
    with open(file, 'r') as f: ## use with to open file 
     fields = [line.strip() for line in f] ## list comprehension to finish reading the field 
     all_fields.extend(fields) ## store in big list 

更多有關使用with語句來打開和讀取文件，請參考此答案Correct way to write to files?