基於文件名複製多文件

Question

基於文件名的前3個字符我想創建一個文件夾，然後複製到相關文件中。

我有一個腳本可以在第一時間運行，但是如果我多次運行它，我會得到一個錯誤 我相信我需要檢查文件是否先存在，但是我一直無法得到它的工作。 或者從os.list

任何幫助將不勝感激過濾掉新創建的文件夾：

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

它看起來像你試圖打開一個目錄C:\\temp\\Test\\F12作爲filename的copy。 否則，請檢查您是否有權打開/讀取文件。