如何遍歷固定數量的文件，直到文件夾的最后一個文件？

Question

我需要迭代的文件夾中有大量文本文件（超過2000個）。 這是我目前設法執行的操作：

import os
filepath='E:\Data'
save_path='E:\Results'

for file in os.listdir(filepath):
   if file.endswith('.txt'):
     with open(os.path.join(filepath,file),'r') as myfile:
          for eachline in myfile:
              MainID=eachline[:6]
                  if MainID=='AKJ':
                     for field in eachline.split():
                          MainID=field.split(',')[1]
                          Origin=field.split(',')[9]
                          Price=field.split(',')[13]
                          fo1=open(os.path.join(save_path,file),'a')
                          fo1.write('%s,%s,%s\n' %(MainID,Origin,Price))
                          fo1.close()

但是，我需要對前100個文件，然后對隨后的100個文件等執行我的過程，直到文件夾結束，而不是像上面的代碼一樣一次遍歷所有文件直到結束。 任何幫助，將不勝感激。

Answer 1

files = [file for file in os.listdir(filepath) if file.endswith('.txt')]
batchsize = 100
index = 0
remaining = len(files)
while remaining > 0:
    batch = min(remaining, batchsize)
    print('NEW BATCH')
    for file in files[index:index+batch]:
        with open(os.path.join(filepath, file), 'r') as myfile:
            print(' ', file)

    index += batch
    remaining -= batch

Answer 2

complete_file_paths = [os.path.join(filepath,file) for file in os.listdir(filepath)]
chunks_of_100 = (complete_file_paths[i:i+100] for i in range(0,len(complete_file_paths),100))

for chunk in chunks_of_100:
      print chunk