Question

我编写了以下代码来迭代文件夹及其文件，并将每个文件夹中的每个文件重命名为文件夹中的文件索引。 E.g每个文件夹中的第一个文件将命名为1.JPG，第二个文件将命名为2.JPG，依此类推。文件夹名称是从1到82的整数。我需要文件夹名称来指定os.rename()中的路径，并且计划从dirs列表中获取它，因为os.walk(path)不会遍历目录按顺序。

代码：

import os
import sys

path='/home/srilatha/Desktop/Research_intern/Data_sets/Final'
i=0

for root, dirs, files in os.walk(path):
    print(dirs)
    print(dirs[i])
    #folder_name=dirs[0]
    #print(folder_name)
    j=0
    for name in sorted(files):
        j+=1
        #print('j=')
        #print(j)
        print(name)
        new=str(j)
        new_name=new+'.JPG'
        print(new_name)
        #os.rename(name,new_name)


    i+=1

错误讯息：

/usr/bin/python3.4 /home/srilatha/PycharmProjects/Research_Intern/Sort_images_into_folders.py
['9', '43', '78', '7', '51', '15', '4', '68', '48', '67', '27', '16', '55', '20', '57', '38', '47', '18', '77', '82', '12', '65', '25', '59', '49', '30', '36', '79', '71', '17', '22', '42', '40', '73', '19', '24', '10', '37', '32', '3', '64', '62', '58', '13', '72', '2', '14', '70', '11', '66', '69', '50', '54', '34', '5', '52', '81', '26', '39', '60', '1', '56', '33', '80', '23', '53', '44', '45', '29', '41', '28', '35', '6', '46', '31', '8', '63', '75', '61', '76', '74', '21']
9
[]
Traceback (most recent call last):
  File "/home/srilatha/PycharmProjects/Research_Intern/Sort_images_into_folders.py", line 9, in <module>
    print(dirs[i])
IndexError: list index out of range

Answer 1

我想你想要这样的东西？

# Import the os module, for the os.walk function
import os

# Set the directory you want to start from
rootDir = '/Users/heinst'
for dirName, subdirList, fileList in os.walk(rootDir):
    print('Found directory: %s' % dirName)
    i = 0
    for fname in fileList:
        print '\t{0} -> {1}'.format(fname, str(i) + os.path.splitext(fname)[1])
        i += 1

python中列表的索引错误

1 个答案: