如何在Python中复制文件?
当前回答
复制文件是一个相对简单的操作,如下面的示例所示,但是您应该使用shutilstdlib模块。
def copyfileobj_example(source, dest, buffer_size=1024*1024):
"""
Copy a file from source to dest. source and dest
must be file-like objects, i.e. any object with a read or
write method, like for example StringIO.
"""
while True:
copy_buffer = source.read(buffer_size)
if not copy_buffer:
break
dest.write(copy_buffer)
如果要按文件名复制,可以执行以下操作:
def copyfile_example(source, dest):
# Beware, this example does not handle any edge cases!
with open(source, 'rb') as src, open(dest, 'wb') as dst:
copyfileobj_example(src, dst)
其他回答
Function | Copies metadata |
Copies permissions |
Uses file object | Destination may be directory |
---|---|---|---|---|
shutil.copy | No | Yes | No | Yes |
shutil.copyfile | No | No | No | No |
shutil.copy2 | Yes | Yes | No | Yes |
shutil.copyfileobj | No | No | Yes | No |
复制文件是一个相对简单的操作,如下面的示例所示,但是您应该使用shutilstdlib模块。
def copyfileobj_example(source, dest, buffer_size=1024*1024):
"""
Copy a file from source to dest. source and dest
must be file-like objects, i.e. any object with a read or
write method, like for example StringIO.
"""
while True:
copy_buffer = source.read(buffer_size)
if not copy_buffer:
break
dest.write(copy_buffer)
如果要按文件名复制,可以执行以下操作:
def copyfile_example(source, dest):
# Beware, this example does not handle any edge cases!
with open(source, 'rb') as src, open(dest, 'wb') as dst:
copyfileobj_example(src, dst)
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)
与公认的答案类似,如果您还想确保在目标路径中创建任何(不存在的)文件夹,以下代码块可能会很有用。
from os import path, makedirs
from shutil import copyfile
makedirs(path.dirname(path.abspath(destination_path)), exist_ok=True)
copyfile(source_path, destination_path)
正如公认的答案所指出的,这些行将覆盖目标路径上存在的任何文件,因此有时在该代码块之前添加:if not path.exists(destination_path):可能会很有用。
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。
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