python文件夹中文件读取踩坑

Posted 2021-12-06 jqpy1994

Q:

进行数据集图片预处理时，初始命名如下图（Fig1左），发现读取文件时，读取的结构并非如所设想的那样顺序读取

Fig 1

A：

pyhton读取文件的时候，按照文件名的ascii码中的顺序进行逐位排序，于是编写以下代码，更改命名规则，程序如下

 1 # -*- coding:utf-8 -*-
 2 # ------------------------------
 3 # @Time     :2019/5/26 10:55
 4 # @Author   :jonie
 5 # @Email    :[email protected]
 6 # @File     :重命名.py
 7 # Description:实现形如"类名_000x.jpg"的命名方式
 8 # ------------------------------
 9 import os
10 def re_name(path_name):
11     num = 0
12     for dir_image in os.listdir(path_name):  # os.listdir() 方法用于返回指定的文件夹包含的文件或文件夹的名字的列表
13     # print(dir_image)
14         full_path = os.path.abspath(os.path.join(path_name, dir_image))
15         # print(full_path)
16 
17         if os.path.isdir(full_path):  # 如果是文件夹，继续递归调用
18             if not os.path.exists(os.path.join(r"D:\\Jonie_Project_scc15\\dataset\\scene_categories",dir_image)):
19                 os.makedirs(os.path.join(r"D:\\Jonie_Project_scc15\\dataset\\scene_categories",dir_image))
20             # print(os.path.join(r"D:\\Jonie_Project_ot8\\data1\\scene_categories",dir_image))
21             print("isdir")
22             re_name(full_path)
23             # print(full_path)
24         else:  # 如果是文件了
25             # pass
26             # print(dir_image)
27             # print(full_path)
28             if dir_image.endswith(‘.jpg‘):
29                 cla=full_path.split(‘\\\\‘)[-2]
30                 image_name = ‘%s_%04d.jpg‘ % (cla, num)  # 注意这里图片名一定要加上扩展名，否则后面imwrite的时候会报错
31                 image_name =os.path.join("D:\\Jonie_Project_scc15","dataset\\scene_categories",cla,image_name)
32                 oldname=os.path.join("D:\\Jonie_Project_scc15","dataset\\scene_categories",cla,image_name)
33                 # print(image_name)
34                 print(full_path)
35                 os.rename(full_path, image_name)
36                 num = num + 1
37 if __name__ == ‘__main__‘:
38     re_name(r"D:\\Jonie_Project_scc15\\dataset\\scene_categories")

rename.py

再次进行读取，发现达到预期结果

Fig 2