将文件移动到文件夹或制作重命名的副本(如果它存在于目标文件夹中)

时间:2015-03-10 09:38:54

标签: python file-copying

我为学校写了一段代码:

import os

source = "/home/pi/lab"
dest = os.environ["HOME"]

for file in os.listdir(source):
   if file.endswith(".c")
      shutil.move(file,dest+"/c")
   elif file.endswith(".cpp")
      shutil.move(file,dest+"/cpp")
   elif file.endswith(".sh")
      shutil.move(file,dest+"/sh")

此代码正在执行的操作是在源目录中查找文件,然后如果找到某个扩展名,则将文件移动到该目录。这部分有效。如果文件已经存在于同名的目标文件夹中,则在文件名的末尾添加1,在扩展名之前添加test1.c,如果它们是多个副本,请执行" 1 ++"。 像这样:test2.ctest3.cos.isfile(filename) 我尝试使用{{1}}但这仅查看源目录。而且我得到了真或假。

2 个答案:

答案 0 :(得分:0)

要测试文件是否存在于目标文件夹中,您应该os.path.join dest文件夹的文件夹

import os                                                                       
import shutil                                                                      
source = "/home/pi/lab"
dest = os.environ["HOME"]                                                                                

# Avoid using the reserved word 'file' for a variable - renamed it to 'filename' instead
for filename in os.listdir(source):                                             
    # os.path.splitext does exactly what its name suggests - split the name and extension of the file including the '.'
    name, extension = os.path.splitext(filename)                                
    if extension == ".c":                                                       
        dest_filename = os.path.join(dest, filename)                            
        if not os.path.isfile(dest_filename):                                      
            # We copy the file as is
            shutil.copy(os.path.join(source, filename) , dest)                  
        else:                       
            # We rename the file with a number in the name incrementing the number until we find one that is not used. 
            # This should be moved to a separate function to avoid code duplication when handling the different file extensions                                        
            i = 0                                                                  
            dest_filename = os.path.join(dest, "%s%d%s" % (name, i,     extension)) 
            while os.path.isfile(dest_filename):                                   
                i += 1                                                                                                                      
                dest_filename = os.path.join(dest, "%s%d%s" % (name, i, extension))
            shutil.copy(os.path.join(source, filename), dest_filename)    
    elif extension == ".cpp"
        ...
        # Handle other extensions

如果您想使用globre将重命名逻辑放在单独的函数中,这是一种方式:

import glob
import re
...
def rename_file(source_filename, source_ext):                                   
    filename_pattern = os.path.join(dest, "%s[0-9]*%s"                          
                                    % (source_filename, source_ext)) 
    # Contains file such as 'a1.c', 'a2.c', etc...
    existing_files = glob.glob(filename_pattern)
    regex = re.compile("%s([0-9]*)%s" % (source_filename, source_ext))          
    # Retrieve the max of the index used for this file using regex
    max_index = max([int(match.group(1))                                        
                     for match in map(regex.search, existing_files)
                     if match])                                                   
    source_full_path = os.path.join(source, "%s%s"                              
                                    % (source_filename, source_ext))            
    # Rebuild the destination filename with the max index + 1 
    dest_full_path = os.path.join(dest, "%s%d%s"                                
                                  % (source_filename,                           
                                     (max_index + 1),                           
                                     source_ext))                               
    shutil.copy(source_full_path, dest_full_path)

 ...
 # If the file already exists i.e. replace the while loop in the else statement
 rename_file(name, extension)

答案 1 :(得分:0)

我不测试代码。但是这样的事情应该可以胜任: -

i = 0
filename = "a.txt"
while True:
    if os.isfile(filename):
        i+= 1
    break
if i:
    fname, ext = filename.split('.')
    filename = fname + str(i) + '.' + ext