如何使用txt列表复制文件以定义文件名的开头

import shutil

# 读取txt文件并存储为列表
file_list = []
with open('file_list.txt', 'r') as file:
    file_list = file.read().splitlines()

# 源文件目录
source_directory = '/path/to/source/directory/'

# 目标文件目录
destination_directory = '/path/to/destination/directory/'

# 复制文件并定义文件名开头
for file_path in file_list:
    file_name = file_path.split('/')[-1]  # 获取文件名
    new_file_name = 'prefix_' + file_name  # 定义文件名开头
    source_file = source_directory + file_path
    destination_file = destination_directory + new_file_name
    shutil.copyfile(source_file, destination_file)