基于文件名复制多文件

Question

基于文件名的前3个字符我想创建一个文件夹，然后复制到相关文件中。

我有一个脚本可以在第一时间运行，但是如果我多次运行它，我会得到一个错误 我相信我需要检查文件是否先存在，但是我一直无法得到它的工作。 或者从os.list

任何帮助将不胜感激过滤掉新创建的文件夹：

srcpath = 'C:\\temp\\Test' 
srcfiles = os.listdir(srcpath) 
destpath = 'C:\\temp\\Test' 

# extract the three letters from filenames 
destdirs = list(set([filename[0:3] for filename in srcfiles])) 


def create(destdirs, destpath): 
    full_path = os.path.join(destpath, destdirs) 
    if not os.path.exists(full_path): 
     os.mkdir(full_path) 
    return full_path 


def copy(filename, dirpath): 
    shutil.copy(os.path.join(srcpath, filename), dirpath) 

# create destination directories and store their names along with full paths 
targets = [ 
    (folder, create(folder, destpath)) for folder in destdirs 
] 

for destdirs, full_path in targets: 
    for filename in srcfiles: 
      if destdirs == filename[0:3]: 
       copy(filename, full_path) 

ERROR

Traceback (most recent call last): 
    File "C:/Users/Desktop/copy_only.py", line 45, in <module> 
    copy(filename, full_path) 
    File "C:/Users/Desktop/copy_only.py", line 35, in copy 
    shutil.copy(os.path.join(srcpath, filename), dirpath) 
    File "C:\Python27\lib\shutil.py", line 119, in copy 
    copyfile(src, dst) 
    File "C:\Python27\lib\shutil.py", line 82, in copyfile 
    with open(src, 'rb') as fsrc: 
IOError: [Errno 13] Permission denied: 'C:\\temp\\Test\\F12' 

Answer 1

它看起来像你试图打开一个目录C:\\temp\\Test\\F12作为filename的copy。 否则，请检查您是否有权打开/读取文件。