如何迭代固定数量的文件，直到该文件夹​​的最后一个文件？

Question

我需要迭代的文件夹中有大量文本文件（超过2000个）。这就是我目前所做的事情：

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，我需要为前100个文件执行我的进程，然后对于以下100个文件等执行，直到文件夹结束，而不是遍历所有文件，直到结束，如同上面的代码。任何帮助将不胜感激。

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