代码十几秒的事,为什么要手动去改一天呢!直接放源码~
重命名指定目录下所有文件夹
import os
import re
def extract_first_three_numbers_from_folder_name(folder_name):
numbers = re.findall(r'\d+', folder_name)
return numbers[:3]
def rename_folders_with_prefixes(directory):
# 获取目标目录下的所有文件夹
folder_names = [name for name in os.listdir(directory) if os.path.isdir(os.path.join(directory, name))]
for folder_name in folder_names:
# 提取文件夹名称中的前三个数字部分
numbers = extract_first_three_numbers_from_folder_name(folder_name)
# 只有在数字数量满足条件时才进行重命名
if len(numbers) == 3:
# 在每个数字前添加相应的字符
prefixed_numbers = [f'V{numbers[0]}', f'D{numbers[1]}', f'Q{numbers[2]}']
# 构建新的文件夹名称
new_folder_name = "_".join(prefixed_numbers)
# 构建完整的路径
old_path = os.path.join(directory, folder_name)
new_path = os.path.join(directory, new_folder_name)
# 重命名文件夹
os.rename(old_path, new_path)
print(f"已将文件夹 {folder_name} 重命名为 {new_folder_name}")
else:
print(f"文件夹 {folder_name} 的数字数量不足三个,未进行重命名")
# 替换为你的目标目录路径
target_directory = r'G:\jet_data\Jet_Forward_Underwater\V3456_D3_Q7'
rename_folders_with_prefixes(target_directory)
重命名指定目录下所有文件
import os
import re
def extract_numbers_from_file_names(directory):
# 获取目录下的所有文件
file_names = [name for name in os.listdir(directory) if os.path.isfile(os.path.join(directory, name))]
# 使用正则表达式提取文件名中的数字
numbers_list = []
pattern = re.compile(r'\d+')
for file_name in file_names:
matches = pattern.findall(file_name)
numbers_list.append(matches)
return numbers_list
def rename_files_with_prefix_suffix(directory):
# 获取目录下的所有文件
file_names = [name for name in os.listdir(directory) if os.path.isfile(os.path.join(directory, name))]
# 使用正则表达式提取文件名中的数字
numbers_list = extract_numbers_from_file_names(directory)
# 遍历文件并进行重命名
for old_name, numbers in zip(file_names, numbers_list):
# 判断提取的数字个数是否小于 6
if len(numbers) < 6:
continue
# 获取文件名和后缀
file_name, file_extension = os.path.splitext(old_name)
# 获取数字列表的最后一个数字
last_number = numbers[-1]
# 截取最后一个数字的后六位数字
last_six_digits = last_number[-6:]
# 构造新文件名
new_name = f'V{numbers[0]}_D{numbers[1]}_Q{numbers[2]}_{last_six_digits}{file_extension}'
# 构造新文件的完整路径
old_path = os.path.join(directory, old_name)
new_path = os.path.join(directory, new_name)
# 重命名文件
os.rename(old_path, new_path)
# 替换为你的目标目录路径
target_directory = r'G:\jet_data\test\V3_D5_Q4'
# 调用重命名函数
rename_files_with_prefix_suffix(target_directory)
print("文件重命名完成。")
重命名指定目录下所有文件夹内的所有文件
import os
import re
def extract_numbers_from_file_names(directory):
# 获取目录下的所有文件
file_names = [name for name in os.listdir(directory) if os.path.isfile(os.path.join(directory, name))]
# 使用正则表达式提取文件名中的数字
numbers_list = []
pattern = re.compile(r'\d+')
for file_name in file_names:
matches = pattern.findall(file_name)
numbers_list.append(matches)
return numbers_list
def rename_files_with_prefix_suffix_in_subdirectories(root_directory):
# 获取根目录下的所有文件夹
subdirectories = [d for d in os.listdir(root_directory) if os.path.isdir(os.path.join(root_directory, d))]
# 遍历每个文件夹
for subdirectory in subdirectories:
subdirectory_path = os.path.join(root_directory, subdirectory)
# 获取目录下的所有文件
file_names = [name for name in os.listdir(subdirectory_path) if os.path.isfile(os.path.join(subdirectory_path, name))]
# 使用正则表达式提取文件名中的数字
numbers_list = extract_numbers_from_file_names(subdirectory_path)
# 遍历文件并进行重命名
for old_name, numbers in zip(file_names, numbers_list):
# 判断提取的数字个数是否小于 6
if len(numbers) < 6:
continue
# 获取文件名和后缀
file_name, file_extension = os.path.splitext(old_name)
# 获取数字列表的最后一个数字
last_number = numbers[-1]
# 截取最后一个数字的后六位数字
last_six_digits = last_number[-6:]
# 构造新文件名
new_name = f'V{numbers[0]}_D{numbers[1]}_Q{numbers[2]}_{last_six_digits}{file_extension}'
# 构造新文件的完整路径
old_path = os.path.join(subdirectory_path, old_name)
new_path = os.path.join(subdirectory_path, new_name)
# 重命名文件
os.rename(old_path, new_path)
# 替换为你的目标根目录路径
root_directory = r'G:\jet_data\Jet_Forward_Collapse'
# 调用重命名函数
rename_files_with_prefix_suffix_in_subdirectories(root_directory)
print("文件重命名完成。")