例如,使用 os 和 glob 模块:

import glob, os

def rename(dir, pattern, titlePattern):
    for pathAndFilename in glob.iglob(os.path.join(dir, pattern)):
        title, ext = os.path.splitext(os.path.basename(pathAndFilename))
        os.rename(pathAndFilename, 
                  os.path.join(dir, titlePattern % title + ext))

然后您可以在示例中使用它,如下所示:

rename(r'c:\temp\xx', r'*.doc', r'new(%s)')

上面的示例将转换所有 *.doc 文件在 c:\temp\xx 迪尔 new(%s).doc 在哪里 %s 是文件的前一个基名称(不带扩展名)。

我更喜欢为我必须做的每个替换编写一个小的一行程序,而不是生成一个更通用和复杂的代码。例如。:

这将用当前目录中任何非隐藏文件中的连字符替换所有下划线

import os
[os.rename(f, f.replace('_', '-')) for f in os.listdir('.') if not f.startswith('.')]

如果您不介意使用正则表达式,那么此函数将为您提供重命名文件的强大功能:

import re, glob, os

def renamer(files, pattern, replacement):
    for pathname in glob.glob(files):
        basename= os.path.basename(pathname)
        new_filename= re.sub(pattern, replacement, basename)
        if new_filename != basename:
            os.rename(
              pathname,
              os.path.join(os.path.dirname(pathname), new_filename))

因此,在您的示例中,您可以这样做(假设文件所在的目录为当前目录):

renamer("*.doc", r"^(.*)\.doc$", r"new(\1).doc")

但也可以回滚到初始文件名:

renamer("*.doc", r"^new\((.*)\)\.doc", r"\1.doc")

还有更多。

我可以简单地重命名文件夹子文件夹中的所有文件

import os

def replace(fpath, old_str, new_str):
    for path, subdirs, files in os.walk(fpath):
        for name in files:
            if(old_str.lower() in name.lower()):
                os.rename(os.path.join(path,name), os.path.join(path,
                                            name.lower().replace(old_str,new_str)))

我正在用任何一种情况来代替旧的情况。

尝试: http://www.mattweber.org/2007/03/04/python-script-renamepy/

我喜欢我的音乐、电影和 以某种方式命名的图片文件。 当我从 互联网,他们通常不遵循我的 命名约定。我发现了我自己 手动重命名每个文件以适应我的 风格。这真是太快了,所以我 决定写一个程序来做这件事 为了我。

这个程序可以转换文件名 全部小写,替换中的字符串 文件名随你的需要, 并从中删除任意数量的字符 文件名的前面或后面。

程序的源代码也可用。

我自己写了一个python脚本。它将文件所在目录的路径和要使用的命名模式作为参数。但是,它通过将一个递增的数字(1、2、3等)附加到您提供的命名模式来重命名。

import os
import sys

# checking whether path and filename are given.
if len(sys.argv) != 3:
    print "Usage : python rename.py <path> <new_name.extension>"
    sys.exit()

# splitting name and extension.
name = sys.argv[2].split('.')
if len(name) < 2:
    name.append('')
else:
    name[1] = ".%s" %name[1]

# to name starting from 1 to number_of_files.
count = 1

# creating a new folder in which the renamed files will be stored.
s = "%s/pic_folder" % sys.argv[1]
try:
    os.mkdir(s)
except OSError:
    # if pic_folder is already present, use it.
    pass

try:
    for x in os.walk(sys.argv[1]):
        for y in x[2]:
            # creating the rename pattern.
            s = "%spic_folder/%s%s%s" %(x[0], name[0], count, name[1])
            # getting the original path of the file to be renamed.
            z = os.path.join(x[0],y)
            # renaming.
            os.rename(z, s)
            # incrementing the count.
            count = count + 1
except OSError:
    pass

希望这对你有用。

directoryName = "Photographs"
filePath = os.path.abspath(directoryName)
filePathWithSlash = filePath + "\\"

for counter, filename in enumerate(os.listdir(directoryName)):

    filenameWithPath = os.path.join(filePathWithSlash, filename)

    os.rename(filenameWithPath, filenameWithPath.replace(filename,"DSC_" + \
          str(counter).zfill(4) + ".jpg" ))

# e.g. filename = "photo1.jpg", directory = "c:\users\Photographs"        
# The string.replace call swaps in the new filename into 
# the current filename within the filenameWitPath string. Which    
# is then used by os.rename to rename the file in place, using the  
# current (unmodified) filenameWithPath.

# os.listdir delivers the filename(s) from the directory
# however in attempting to "rename" the file using os 
# a specific location of the file to be renamed is required.

# this code is from Windows 

我有一个类似的问题,但我想在目录中所有文件的文件名的开头追加文本,并使用了类似的方法。见下例:

folder = r"R:\mystuff\GIS_Projects\Website\2017\PDF"

import os


for root, dirs, filenames in os.walk(folder):


for filename in filenames:  
    fullpath = os.path.join(root, filename)  
    filename_split = os.path.splitext(filename) # filename will be filename_split[0] and extension will be filename_split[1])
    print fullpath
    print filename_split[0]
    print filename_split[1]
    os.rename(os.path.join(root, filename), os.path.join(root, "NewText_2017_" + filename_split[0] + filename_split[1]))

在需要执行重命名的目录中。

import os
# get the file name list to nameList
nameList = os.listdir() 
#loop through the name and rename
for fileName in nameList:
    rename=fileName[15:28]
    os.rename(fileName,rename)
#example:
#input fileName bulk like :20180707131932_IMG_4304.JPG
#output renamed bulk like :IMG_4304.JPG

对于我的目录,我有多个子目录,每个子目录都有很多图像,我想把所有的子目录图像都改成1.jpg~n.jpg。

def batch_rename():
    base_dir = 'F:/ad_samples/test_samples/'
    sub_dir_list = glob.glob(base_dir + '*')
    # print sub_dir_list # like that ['F:/dir1', 'F:/dir2']
    for dir_item in sub_dir_list:
        files = glob.glob(dir_item + '/*.jpg')
        i = 0
        for f in files:
            os.rename(f, os.path.join(dir_item, str(i) + '.jpg'))
            i += 1

(我自己的答案) https://stackoverflow.com/a/45734381/6329006

#  another regex version
#  usage example:
#  replacing an underscore in the filename with today's date
#  rename_files('..\\output', '(.*)(_)(.*\.CSV)', '\g<1>_20180402_\g<3>')
def rename_files(path, pattern, replacement):
    for filename in os.listdir(path):
        if re.search(pattern, filename):
            new_filename = re.sub(pattern, replacement, filename)
            new_fullname = os.path.join(path, new_filename)
            old_fullname = os.path.join(path, filename)
            os.rename(old_fullname, new_fullname)
            print('Renamed: ' + old_fullname + ' to ' + new_fullname

这个代码可以用

该函数将两个参数f_patth作为重命名文件的路径,并将新名称作为文件的新名称。

import glob2
import os


def rename(f_path, new_name):
    filelist = glob2.glob(f_path + "*.ma")
    count = 0
    for file in filelist:
        print("File Count : ", count)
        filename = os.path.split(file)
        print(filename)
        new_filename = f_path + new_name + str(count + 1) + ".ma"
        os.rename(f_path+filename[1], new_filename)
        print(new_filename)
        count = count + 1