python批量修改文件名
代码： 代码示例:
#coding:utf-8
#批量修改文件名
import os import re import datetime
re_st = r'(\d+)\+\s?\((\d+)\)'
#用于匹配旧的文件名,需含分组 re_match_old_file_name = re.compile(re_st)
#要修改的目录 WORKING_PATH = r'F:\Gallery'
#------------------------------------------
def rename_fomat(name):
"""
文件重命名格式组织模块(一般修改这里就可以了)
NOTE:返回类型必须是unicode
"""
if name:
re_rn = re_match_old_file_name.findall(name)
if re_rn and re_rn != []:
re_rn = re_rn[0]
num = int(re_rn)
new_nm = u'NO.%04d' % ( num)
return new_nm
#------------------------------------------
def logs(error):
"""
错误记录
"""
log = ''
LOG_FILE = open(r'./log.txt', 'a')
live_info ="""
==========
Time : %s
title : %s
Path :
%s
==========
""" % (
datetime.datetime.now(),
str(error['title']),
str(error['index']),
)
log += live_info
errors = error['error_paths']
for item in errors:
item = '%s\n' % item
log += item
log = log.encode('utf-8')
try:
LOG_FILE.write(log)
except IOError:
print u'写入日志失败'
finally:
LOG_FILE.close()
#---------------------------------------
def rename(old, new):
"""
文件重命名模块
return:
0:rename success
1:the new path is exists
-1:rename failed
"""
if not os.path.exists(new):
try:
os.renames(old, new)
return 0
except IOError:
print 'path error:', new
return -1
else:
return 1
#---------------------------------------
def get_dirs(path):
"""
获取目录列表
"""
if os.path.exists(path):
return os.listdir(path)
else:
return -1
#------------------------------------------
def get_input_result(word, choice):
"""
获取正确的输入结果
"""
correct_result = set(choice)
word = '===%s?\n[in]:' % (word)
while True:
in_choice = raw_input(word)
if in_choice in correct_result: return in_choice
#---------------------------------------------
def batch_rename(index, dirs = []):
"""
批量修改文件
"""
index = unicode(index)
errors = []
if dirs == []:
dirs = get_dirs(path = index)
if dirs and dirs != []:
for item in dirs:
item = unicode(item)
new_name = rename_fomat(item)
if new_name :
old_pt = u'%s\\%s'% (index, item)
new_pt = u'%s\\%s'% (index, new_name)
res_rn = rename(old_pt, new_pt)
if res_rn != 0:
errors.append(item)
else:
errors.append(item)
if errors and errors != []:
print 'Rename Failed:'
logs({
'index': index,
'title': 'Rename Failed' ,
'error_paths': errors,
})
for i, item in enumerate(errors):
print item, '|',
if i % 5 == 4:
print ''
print ''
else:
return -1
#----------------------------------------------
def batch_rename_test(index):
"""
测试
返回过滤结果
"""
index = unicode(index)
errors = []
correct = []
dirs = get_dirs(path = index)
if dirs and dirs != []:
for x, item in enumerate(dirs):
item = unicode(item)
new_name = rename_fomat(item)
if new_name :
correct.append(item)
old_pt = u'%s\\%s'% (index, item)
new_pt = u'%s\\%s'% (index, new_name)
print '[%d]O: %s' % ( x + 1, old_pt)
print '[%d]N: %s' % ( x + 1, new_pt)
else:
errors.append(item)
if errors and errors != []:
print 'Not Match:'
logs({
'index': index,
'title': 'Not Match',
'error_paths': errors,
})
for i, item in enumerate(errors):
print item, '|',
if i % 5 == 4:
print ''
print ''
return correct
#--------------------------------
def manage(index):
"""
程序组织块
"""
file_filter = batch_
rename_test(index)
do_choice = get_input_result(
word = 'Do with this(y / n)',
choice = ['y', 'n']
)
if do_choice == 'y':
batch_rename(index, dirs= file_filter)
print 'Finished !'
if __name__ == '__main__':
path = WORKING_PATH
manage(index = path)

python批量修改文件名
要点：

文件名获取
文件名后缀按照规定字符'.'分割
文件重命名: 字符串与数字的连接

代码如下
#encoding=gbk
import os

def main (path):
    filename_list = os.listdir(path) 
    """os.listdir(path) 扫描路径的文件，将文件名存入存入列表"""

    a = 0
    for i in filename_list:
        used_name = path + filename_list[a]
        new_name = path + str(a) + used_name[used_name.index('.'):] # 保留原后缀
        os.rename(used_name, new_name)
        print("文件%s重命名成功，新的文件名为%s" %(used_name,new_name))
        a += 1

if __name__=='__main__':
    path="C:/Users/HeBin/Pictures/Screenshots/" # 目标路径
    main(path)

利用Python批量修改文件名

import os
import glob


def main():
    folder = '../../datasets/DIV2K/DIV2K_train_LR_mild' # 选择文件夹
    DIV2K(folder)
    print('Finished.')


def DIV2K(path):
    img_path_l = glob.glob(os.path.join(path, '*')) #读取文件名
    for img_path in img_path_l:
        new_path = img_path.replace('x4m', '') # 去除文件名中的'x4m'
        os.rename(img_path, new_path)


if __name__ == "__main__":
    main()

 

python3实现批量修改文件名
以批量修改某文件夹下所有图片名称为例，注释超详细，万能模板，读者可举一反三，适当修改模板，效果显著！
#批量修改文件名
#批量修改图片文件名
import os
import re
import sys
def renameall():
	fileList = os.listdir(r"C:\Users\Administrator\Desktop\stars")		#待修改文件夹
	print("修改前："+str(fileList))		#输出文件夹中包含的文件
	currentpath = os.getcwd()		#得到进程当前工作目录
	os.chdir(r"C:\Users\Administrator\Desktop\stars")		#将当前工作目录修改为待修改文件夹的位置
	num=1		#名称变量
	for fileName in fileList:		#遍历文件夹中所有文件
		pat=".+\.(jpg|png|gif)"		#匹配文件名正则表达式
		pattern = re.findall(pat,fileName)		#进行匹配
		os.rename(fileName,(str(num)+'.'+pattern[0]))		#文件重新命名
		num = num+1		#改变编号，继续下一项
	print("---------------------------------------------------")
	os.chdir(currentpath)		#改回程序运行前的工作目录
	sys.stdin.flush()		#刷新
	print("修改后："+str(os.listdir(r"C:\Users\Administrator\Desktop\stars")))		#输出修改后文件夹中包含的文件
renameall()
运行程序：
修改过后文件夹中显示情况：
修改成功，我们非常满意！

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