如何在Python中复制文件?


当前回答

对于小文件和仅使用Python内置程序,可以使用以下一行代码:

with open(source, 'rb') as src, open(dest, 'wb') as dst: dst.write(src.read())

对于文件太大或内存很关键的应用程序来说,这不是最佳方式,因此应该首选Swati的答案。

其他回答

在Python中,可以使用

梭式组件操作系统模块子流程模块


import os
import shutil
import subprocess

1) 使用shutil模块复制文件

shutil.copyfile签名

shutil.copyfile(src_file, dest_file, *, follow_symlinks=True)

# example    
shutil.copyfile('source.txt', 'destination.txt')

shutil.copy签名

shutil.copy(src_file, dest_file, *, follow_symlinks=True)

# example
shutil.copy('source.txt', 'destination.txt')

shutil.copy2签名

shutil.copy2(src_file, dest_file, *, follow_symlinks=True)

# example
shutil.copy2('source.txt', 'destination.txt')  

shutil.copyfileobj签名

shutil.copyfileobj(src_file_object, dest_file_object[, length])

# example
file_src = 'source.txt'  
f_src = open(file_src, 'rb')

file_dest = 'destination.txt'  
f_dest = open(file_dest, 'wb')

shutil.copyfileobj(f_src, f_dest)  

2) 使用os模块复制文件

os.popen签名

os.popen(cmd[, mode[, bufsize]])

# example
# In Unix/Linux
os.popen('cp source.txt destination.txt') 

# In Windows
os.popen('copy source.txt destination.txt')

os.system签名

os.system(command)


# In Linux/Unix
os.system('cp source.txt destination.txt')  

# In Windows
os.system('copy source.txt destination.txt')

3) 使用子流程模块复制文件

subprocess.call签名

subprocess.call(args, *, stdin=None, stdout=None, stderr=None, shell=False)

# example (WARNING: setting `shell=True` might be a security-risk)
# In Linux/Unix
status = subprocess.call('cp source.txt destination.txt', shell=True) 

# In Windows
status = subprocess.call('copy source.txt destination.txt', shell=True)

subprocess.check_output签名

subprocess.check_output(args, *, stdin=None, stderr=None, shell=False, universal_newlines=False)

# example (WARNING: setting `shell=True` might be a security-risk)
# In Linux/Unix
status = subprocess.check_output('cp source.txt destination.txt', shell=True)

# In Windows
status = subprocess.check_output('copy source.txt destination.txt', shell=True)

有两种在Python中复制文件的最佳方法。

1.我们可以使用梭动模块

代码示例:

import shutil
shutil.copyfile('/path/to/file', '/path/to/new/file')

除了copyfile,还有其他可用的方法,如copy、copy2等,但copyfile在性能方面是最好的,

2.我们可以使用OS模块

代码示例:

import os
os.system('cp /path/to/file /path/to/new/file')

另一种方法是使用子流程,但它不是优选的,因为它是调用方法之一,不安全。

使用subprocess.call复制文件

from subprocess import call
call("cp -p <file> <file>", shell=True)

这是一个利用“shutil.copyfileobj”的答案,它非常高效。我在不久前创建的一个工具中使用了它。我最初没有写这篇文章,但我稍微修改了一下。

def copyFile(src, dst, buffer_size=10485760, perserveFileDate=True):
    '''
    @param src:    Source File
    @param dst:    Destination File (not file path)
    @param buffer_size:    Buffer size to use during copy
    @param perserveFileDate:    Preserve the original file date
    '''
    #    Check to make sure destination directory exists. If it doesn't create the directory
    dstParent, dstFileName = os.path.split(dst)
    if(not(os.path.exists(dstParent))):
        os.makedirs(dstParent)

    # Optimize the buffer for small files
    buffer_size = min(buffer_size,os.path.getsize(src))
    if(buffer_size == 0):
        buffer_size = 1024

    if shutil._samefile(src, dst):
        raise shutil.Error("`%s` and `%s` are the same file" % (src, dst))
    for fn in [src, dst]:
        try:
            st = os.stat(fn)
        except OSError:
            # File most likely does not exist
            pass
        else:
            # XXX What about other special files? (sockets, devices...)
            if shutil.stat.S_ISFIFO(st.st_mode):
                raise shutil.SpecialFileError("`%s` is a named pipe" % fn)
    with open(src, 'rb') as fsrc:
        with open(dst, 'wb') as fdst:
            shutil.copyfileobj(fsrc, fdst, buffer_size)

    if(perserveFileDate):
        shutil.copystat(src, dst)

从Python 3.5开始,您可以对小文件(例如:文本文件、小jpegs)执行以下操作:

from pathlib import Path

source = Path('../path/to/my/file.txt')
destination = Path('../path/where/i/want/to/store/it.txt')
destination.write_bytes(source.read_bytes())

write_bytes将覆盖目标位置的任何内容