Answers:
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)')
上面的示例会将dir中的所有*.doc
文件都转换c:\temp\xx
为new(%s).doc
,其中%s
是文件的先前基本名称(不带扩展名)。
我更喜欢为每次替换编写一个小的内衬,而不是编写更加通用和复杂的代码。例如:
这会将当前目录中任何非隐藏文件中的所有下划线都用连字符替换
import os
[os.rename(f, f.replace('_', '-')) for f in os.listdir('.') if not f.startswith('.')]
rename
:(
no such file error
只记得os.rename
需要完整的路
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(假设这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")
和更多。
我用它来简单地重命名文件夹的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)))
我用new_str替换所有出现的old_str情况。
尝试:http : //www.mattweber.org/2007/03/04/python-script-renamepy/
我喜欢用某种方式命名我的音乐,电影和图片文件。当我从Internet下载文件时,它们通常不遵循我的命名约定。我发现自己手动重命名每个文件以适合我的风格。这真的很快就过去了,所以我决定编写一个程序为我做。
该程序可以将文件名转换为所有小写字母,将文件名中的字符串替换为所需内容,并从文件名的开头或后面修剪任意数量的字符。
该程序的源代码也可用。
我自己编写了一个python脚本。它以存在文件的目录路径和要使用的命名模式作为参数。但是,它通过给您提供的命名模式附加一个增量数字(1、2、3等)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
希望这对您有用。
在您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
os.chdir(path_of_directory)
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,但是我想在目录中所有文件的文件名的开头添加文本,并使用类似的方法。请参见下面的示例:
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]))
至于我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
# 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
如果要在编辑器(例如vim)中修改文件名,单击库将随命令一起提供,该命令click.edit()
可用于接收来自编辑器的用户输入。这是如何使用它来重构目录中文件的示例。
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,但它减少了此脚本的易变性,并提供了更多选项,例如递归重构。这是github页面的链接。如果您喜欢命令行应用程序,并且对文件名进行一些快速编辑,这很有用。(我的应用程序是类似于中发现的“bulkrename”命令游侠)。
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
%
命令中如何使用符号os.path.join(dir, titlePattern % title + ext)
?我知道%
用于模运算,也用作格式运算符。但通常后跟s
或f
来指定格式。为什么%
在上述命令后立即没有任何内容(空格)?