對多個文本文件使用相同的代碼，並使用python生成多個文本文件作為輸出

Question

我有30多個文本文件。 我需要對每個文本文件進行一些處理，然后再次將它們保存在具有不同名稱的文本文件中。

示例1：precision_case_words.txt ----處理中---- precision_case_sentences.txt

示例2：random_case_words.txt ----處理---- random_case_sentences.txt

這樣，我需要對所有文本文件進行處理。

當前代碼：

new_list = []

with open('precise_case_words.txt') as inputfile:

    for line in inputfile:
        new_list.append(line)


final = open('precise_case_sentences.txt', 'w+')


for item in new_list:

    final.write("%s\n" % item)

請一直手動復制並粘貼此代碼，並每次手動更改名稱。 請為我建議一種避免使用python手動作業的解決方案。

Answer 1

假設您在當前目錄中擁有所有* _case_words.txt

import glob

in_file = glob.glob('*_case_words.txt')

prefix = [i.split('_')[0] for i in in_file]

for i, ifile in enumerate(in_file):
    data = []
    with open(ifile, 'r') as f:
        for line in f:
            data.append(line)
    with open(prefix[i] + '_case_sentence.txt' , 'w') as f:
        f.write(data)

Answer 2

這應該使您了解如何處理它：

def rename(name,suffix):
    """renames a file with one . in it by splitting and inserting suffix before the ."""
    a,b = name.split('.')             
    return ''.join([a,suffix,'.',b])  # recombine parts including suffix in it

def processFn(name):
    """Open file 'name', process it, save it under other name"""    
    # scramble data by sorting and writing anew to renamed file            
    with open(name,"r") as r,   open(rename(name,"_mang"),"w") as w:
        for line in r:
            scrambled = ''.join(sorted(line.strip("\n")))+"\n"
            w.write(scrambled)

# list of filenames, see link below for how to get them with os.listdir() 
names = ['fn1.txt','fn2.txt','fn3.txt']  

# create demo data
for name in names:
    with open(name,"w") as w:
        for i in range(12):
            w.write("someword"+str(i)+"\n")

# process files
for name in names:
    processFn(name)

有關文件列表：請參閱如何列出目錄的所有文件？

我選擇逐行讀取/寫入，您可以完全讀入一個文件，對其進行處理，然后根據自己的喜好再次輸出。

fn1.txt ：

someword0
someword1
someword2
someword3
someword4
someword5
someword6
someword7
someword8
someword9
someword10          
someword11          

進入fn1_mang.txt ：

0demoorsw
1demoorsw
2demoorsw
3demoorsw
4demoorsw
5demoorsw
6demoorsw
7demoorsw
8demoorsw
9demoorsw
01demoorsw
11demoorsw

Answer 3

就在今天，我碰巧正在編寫一些實現此目的的代碼 。