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批量重命名目录中的文件

发布于 2024-07-06 23:18:56 字数 211 浏览 11 评论 0原文

有没有一种简单的方法可以使用 Python 重命名目录中已包含的一组文件？

示例：我有一个充满 *.doc 文件的目录，我想以一致的方式重命名它们。

X.doc-> “新（X）.doc”

Y.doc-> “新（Y）.doc”

原文

Is there an easy way to rename a group of files already contained in a directory, using Python?

Example: I have a directory full of *.doc files and I want to rename them in a consistent way.

X.doc -> "new(X).doc"

Y.doc -> "new(Y).doc"

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评论（14）

独木成林 2024-07-13 23:18:57

如果您想在编辑器（例如 vim）中修改文件名，请单击 库附带命令 click.edit()，可用于从编辑器接收用户输入。 以下是如何使用它来重构目录中的文件的示例。

import click
from pathlib import Path

# current directory
direc_to_refactor = Path(".")

# list of old file paths
old_paths = list(direc_to_refactor.iterdir())

# list of old file names
old_names = [str(p.name) for p in old_paths]

# modify old file names in an editor,
# and store them in a list of new file names
new_names = click.edit("\n".join(old_names)).split("\n")

# refactor the old file names
for i in range(len(old_paths)):
    old_paths[i].replace(direc_to_refactor / new_names[i])

我编写了一个使用相同技术的命令行应用程序，但这减少了该脚本的波动性，并提供了更多选项，例如递归重构。 以下是 github 页面 的链接。 如果您喜欢命令行应用程序，并且有兴趣对文件名进行一些快速编辑，这非常有用。 （我的应用程序类似于 ranger 中的“bulkrename”命令）。

If you would like to modify file names in an editor (such as vim), the click library comes with the command click.edit(), which can be used to receive user input from an editor. Here is an example of how it can be used to refactor files in a directory.

import click
from pathlib import Path

# current directory
direc_to_refactor = Path(".")

# list of old file paths
old_paths = list(direc_to_refactor.iterdir())

# list of old file names
old_names = [str(p.name) for p in old_paths]

# modify old file names in an editor,
# and store them in a list of new file names
new_names = click.edit("\n".join(old_names)).split("\n")

# refactor the old file names
for i in range(len(old_paths)):
    old_paths[i].replace(direc_to_refactor / new_names[i])

I wrote a command line application that uses the same technique, but that reduces the volatility of this script, and comes with more options, such as recursive refactoring. Here is the link to the github page. This is useful if you like command line applications, and are interested in making some quick edits to file names. (My application is similar to the "bulkrename" command found in ranger).

回复 原文
聊慰 2024-07-13 23:18:57

该代码将起作用

该函数精确地采用两个参数 f_path 作为重命名文件的路径，并将 new_name 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

This code will work

The function exactly takes two arguments f_patth as your path to rename file and new_name as your new name to the file.

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
回复 原文
最佳男配角 2024-07-13 23:18:57

基于 Cesar Canassa 上面评论构建。

import os
[os.rename(f, f.replace(f[f.find('___'):], '')) for f in os.listdir('.') if not f.startswith('.')]

这将找到三个下划线 (_) 并将它们及其后面的所有内容替换为空 ('')。

Building off of Cesar Canassa comment above.

import os
[os.rename(f, f.replace(f[f.find('___'):], '')) for f in os.listdir('.') if not f.startswith('.')]

This will find three underscores (_) and replace them and everything after them with nothing ('').

回复 原文
千秋岁 2024-07-13 23:18:56

我更喜欢为我必须做的每次替换编写一小段代码，而不是编写更通用和复杂的代码。 例如：

这会将当前目录中任何非隐藏文件中的所有下划线替换为连字符

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

I prefer writing small one liners for each replace I have to do instead of making a more generic and complex code. E.g.:

This replaces all underscores with hyphens in any non-hidden file in the current directory

import os
[os.rename(f, f.replace('_', '-')) for f in os.listdir('.') if not f.startswith('.')]
回复 原文
花期渐远 2024-07-13 23:18:56

这种重命名非常简单，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 c:\temp\xx 目录中的 *.doc 文件复制到 new(%s).doc，其中 %s 是文件以前的基本名称（不带扩展名）。

Such renaming is quite easy, for example with os and glob modules:

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))

You could then use it in your example like this:

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

The above example will convert all *.doc files in c:\temp\xx dir to new(%s).doc, where %s is the previous base name of the file (without extension).

回复 原文
ゝ偶尔ゞ 2024-07-13 23:18:56

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（假设它是文件所在的当前目录）：

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

但您也可以回滚到初始文件名：

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

等等。

If you don't mind using regular expressions, then this function would give you much power in renaming files:

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))

So in your example, you could do (assuming it's the current directory where the files are):

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

but you could also roll back to the initial filenames:

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

and more.

回复 原文
总以为 2024-07-13 23:18:56

我用这个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)))

我将用 new_str 替换所有出现的 old_str 。

I have this to simply rename all files in subfolders of folder

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)))

I am replacing all occurences of old_str with any case by new_str.

回复 原文
把人绕傻吧 2024-07-13 23:18:56

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。
当我从以下位置下载文件时
互联网，他们通常不关注我的
命名约定。 我寻找到了自我
手动重命名每个文件以适合我的
风格。 这很快就过时了，所以我
决定写一个程序来做到这一点
对我来说。

该程序可以转换文件名
全部小写，替换字符串
文件名可以是你想要的任何内容，
并从中修剪任意数量的字符
文件名的前面或后面。

该程序的源代码也可用。

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

I like to have my music, movie, and
picture files named a certain way.
When I download files from the
internet, they usually don’t follow my
naming convention. I found myself
manually renaming each file to fit my
style. This got old realy fast, so I
decided to write a program to do it
for me.

This program can convert the filename
to all lowercase, replace strings in
the filename with whatever you want,
and trim any number of characters from
the front or back of the filename.

The program's source code is also available.

回复 原文
寂寞陪衬 2024-07-13 23:18:56

我自己写了一个python脚本。 它将文件所在目录的路径以及要使用的命名模式作为参数。 但是，它会通过将增量数字（1、2、3 等）附加到您提供的命名模式来进行重命名。

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。

I've written a python script on my own. It takes as arguments the path of the directory in which the files are present and the naming pattern that you want to use. However, it renames by attaching an incremental number (1, 2, 3 and so on) to the naming pattern you give.

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

Hope this works for you.

回复 原文
美男兮 2024-07-13 23:18:56

位于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

Be in the directory where you need to perform the renaming.

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
回复 原文
不必你懂 2024-07-13 23:18:56
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 

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
回复 原文
给不了的爱 2024-07-13 23:18:56

我遇到了类似的问题，但我想将文本附加到目录中所有文件的文件名开头并使用类似的方法。 请参阅下面的示例：

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]))

I had a similar problem, but I wanted to append text to the beginning of the file name of all files in a directory and used a similar method. See example below:

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]))
回复 原文
谜兔 2024-07-13 23:18:56

对于我来说，在我的目录中我有多个子目录，每个子目录有很多图像我想将所有子目录图像更改为 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

as to me in my directory I have multiple subdir, each subdir has lots of images I want to change all the subdir images to 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

(mys own answer)https://stackoverflow.com/a/45734381/6329006

回复 原文
烟织青萝梦 2024-07-13 23:18:56
#  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

#  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
回复 原文
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