重命名文件序列答案

【问题标题】：Renaming a file sequence重命名文件序列
【发布时间】：2021-06-13 01:07:16
【问题描述】：

我有以下文件夹结构

folder1/
    file1a.txt
    file1b.txt
    file1c.txt
    file2a.txt
    file2b.txt
    file2c.txt
    file3a.txt
    file3b.txt
    file3c.txt
    file4a.txt
    file4b.txt
    file4c.txt
    file5a.txt
    file5b.txt
    file5c.txt

folder2/
    file1a.txt
    file1b.txt
    file1c.txt
    file2a.txt
    file2b.txt
    file2c.txt
    file3a.txt
    file3b.txt
    file3c.txt
    file4a.txt
    file4b.txt
    file4c.txt
    file5a.txt
    file5b.txt
    file5c.txt

我想重命名folder2 中的文件，使它们与folder1 中的文件按顺序排列——即

folder2/
    file6a.txt
    file6b.txt
    file6c.txt
    file7a.txt
    file7b.txt
    file7c.txt
    file8a.txt
    file8b.txt
    file8c.txt
    file9a.txt
    file9b.txt
    file9c.txt
    file10a.txt
    file10b.txt
    file10c.txt

我的策略是（伪代码）：

1. read in files from folder1
2. make a unique list of those files
3. count the number of files in that unique list
4. set that number to be the starting number for the files to be renamed in folder2 (e.g. 6)
5. count the number of files in folder2 and set that to be the end of the range (e.g. 10)
6. and rename the files in folder2 as shown in the example

我无法想象的步骤是重复文件夹 2 中的文件名——即file6a.txt, file6b.txt, file6c.txt。

在 BASH 中我可以运行这个：

i=1; j=6; rename file${i} file${j} folder2/file${i}*

pythonic 的做法是什么？

编辑：将问题更新为更准确，并添加伪代码来说明策略。

【问题讨论】：

将文件列表读入单个列表，对其进行排序，使用enumerate() 跟踪列表中的索引，并使用该枚举索引重命名以folder2/ 开头的文件。跨度>
你能展示一下你到目前为止所尝试的吗？
您好，我只成功读取并统计了文件夹 1 中的文件。
“告诉我如何解决这个编码问题”是off-topic for Stack Overflow。您应该发送honest attempt at the solution，然后然后询问有关它的具体问题（如有必要）。
文件目录（文件夹）中不能有两个同名的文件，因此它们的名称在它们的任何列表中都是唯一的（这意味着您的问题中显示的示例文件夹结构是不可能的） .

标签： python rename

【解决方案1】：

您可以使用我们的操作系统并执行此操作

import os
import re
files = os.listdir(PATHTOFOLDER1)
files2 = os.listdir(PATHTOFOLDER2)
lastfile = files[-1]
temp = re.findall(r'\d+', lastfile)[0]

for i in range(int(temp), len(files2)+int(temp)):
         os.rename(PATHTOFOLDER2 + files2[i], PATHTOFOLDER2 + f"file{i}.txt")

你可以这样做，那应该可以。

【讨论】：

【解决方案2】：

试试这个

import os, os.path

# simple version for working with CWD
#print (len([name for name in os.listdir('.') if os.path.isfile(name)]))

# path joining version for other paths
DIR = r"folder"
a=print (len([name for name in os.listdir(DIR) if os.path.isfile(os.path.join(DIR, name))]))
print(a)
# Python program to rename all file 
# names in your directory 


os.chdir('folder2') 
print(os.getcwd()) 
COUNT = a

# Function to increment count 
# to make the files sorted. 
def increment(): 
    global COUNT 
    COUNT = COUNT + 1


for f in os.listdir(): 
    f_name, f_ext = os.path.splitext(f) 
    f_name = "foldername" + str(COUNT) 
    increment() 

    new_name = '{} {}'.format(f_name, f_ext) 
    os.rename(f, new_name)

【讨论】：

【解决方案3】：

解决了...

# Change this based on your path
PATH1 = "/path/to/folder1"
PATH1 += "/"

PATH2 = "/path/to/folder2"
PATH2 += "/"

TAG = "file"

files_in_raw1 = os.listdir(PATH1)
files_in_raw2 = os.listdir(PATH2)

unique_files_in_raw1 = []
unique_files_in_raw2 = []


for files1, files2 in zip(files_in_raw1, files_in_raw2):
    # i[:4] gets the file name "file". This var will change based on the tag that needs substitution.
    tag1 = list(os.path.splitext(files1[:4]))[0]
    tag2 = list(os.path.splitext(files2[:4]))[0]
    if tag1 not in unique_files_in_raw1:
        unique_files_in_raw1.append(tag1)
    if tag2 not in unique_files_in_raw2:
        unique_files_in_raw2.append(tag2)

unique_files_in_raw1.sort()
unique_files_in_raw2.sort()

COUNT = len(unique_files_in_raw1)
START = COUNT + 1
END = START + len(unique_files_in_raw2)

continue_naming_files = []
for x in range(START, END):
    continue_naming_files.append(TAG + "0" + str(x))

new_name = []
for file in files_in_raw2:
    for name in continue_naming_files:
        new_file = re.sub(file[:11], name, file)
        new_name.append(new_file)

for i, j in zip(files_in_raw2, new_name):
    os.symlink(PATH2 + i, PATH2 + j)

【讨论】：