Pandas 将文件名存储在列表中，并将所有文件读入数据框

Question

您好，我正在尝试使用 glob 和 re 搜索指定的文件名。找到匹配项后，我会将匹配文件列表存储到文本文件中。我需要的是我想将所有这些匹配的文件名提取到一个列表中，以便我可以将它们存储到一个数据框中。

import re
import sys
import os
import glob
import pandas as pd
import logging

with open('files_matched.txt', 'w') as f_matched, open('files_not_matched.txt','w') as f_notmatched:
    try:
        for file in glob.glob('*.csv'):
                r = re.search(r'abc_sales(20[0-9][0-9])-([1-9]|1[0-2]|0[0-9])-([1-9]|1[0-9]|2[0-9]|3[0-1]|0[0-9])-[0-9]{2}_[a-z0-9]{3,5}.csv', file)
                if r:
                    filename=[file] # unable to store list of files 
                    match=f'File matched:{file}'
                    f_matched.write(match+'\n')
                else:
                    not_match=f'File not matched:{file}'
                    f_notmatched.write(not_match + '\n')
    except Exception as e:
        pass

df=[pd.read_csv(f,encoding='ISO-8859-1',error_bad_lines=False,engine='python') for f in filename]

for df_new, f in zip(df,filename):
      df_new['f'] = f
combined_df = pd.concat(df, ignore_index=False)
combined_df.head()

Answer 1

通过append 创建新的文件列表，然后通过DataFrame.assign 创建新列：

filenames = []
with open('files_matched.txt', 'w') as f_matched, open('files_not_matched.txt','w') as f_notmatched:
    try:
        for file in glob.glob('*.csv'):
                r = re.search(r'abc_sales(20[0-9][0-9])-([1-9]|1[0-2]|0[0-9])-([1-9]|1[0-9]|2[0-9]|3[0-1]|0[0-9])-[0-9]{2}_[a-z0-9]{3,5}.csv', file)
                if r:
                    filenames.append(file) # add filename to list 
                    match=f'File matched:{file}'
                    f_matched.write(match+'\n')
                else:
                    not_match=f'File not matched:{file}'
                    f_notmatched.write(not_match + '\n')
    except Exception as e:
        pass

#add new column by assign
df=[pd.read_csv(f,
                encoding='ISO-8859-1',
                error_bad_lines=False,
                engine='python').assign(f = f) 
    for f in filenames]

combined_df = pd.concat(df, ignore_index=False)