How to copy files

Asked 2023-09-20 20:14:54 View 482,393

How do I copy a file in Python?

  • It would be nice if you specify a reason for copy. For many applications hard linking may be a viable alternative. And people looking for such solutions may also think about this "copy" opportunity (like me, but my answer here was downvoted). - anyone

Answers

shutil has many methods you can use. One of which is:

import shutil

shutil.copyfile(src, dst)

# 2nd option
shutil.copy(src, dst)  # dst can be a folder; use shutil.copy2() to preserve timestamp
  • Copy the contents of the file named src to a file named dst. Both src and dst need to be the entire filename of the files, including path.
  • The destination location must be writable; otherwise, an IOError exception will be raised.
  • If dst already exists, it will be replaced.
  • Special files such as character or block devices and pipes cannot be copied with this function.
  • With copy, src and dst are path names given as strs.

Another shutil method to look at is shutil.copy2(). It's similar but preserves more metadata (e.g. time stamps).

If you use os.path operations, use copy rather than copyfile. copyfile will only accept strings.

Answered   2023-09-20 20:14:54

  • In Python 3.8 this has received some significant speed boosts (~50% faster, depending on OS). - anyone
  • does the destination folder need to exist? - anyone
  • @KansaiRobot: Yes, otherwise you get an exception: FileNotFoundError: Directory does not exist: foo/ - anyone
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

Answered   2023-09-20 20:14:54

copy2(src,dst) is often more useful than copyfile(src,dst) because:

  • it allows dst to be a directory (instead of the complete target filename), in which case the basename of src is used for creating the new file;
  • it preserves the original modification and access info (mtime and atime) in the file metadata (however, this comes with a slight overhead).

Here is a short example:

import shutil
shutil.copy2('/src/dir/file.ext', '/dst/dir/newname.ext') # complete target filename given
shutil.copy2('/src/file.ext', '/dst/dir') # target filename is /dst/dir/file.ext

Answered   2023-09-20 20:14:54

In Python, you can copy the files using


import os
import shutil
import subprocess

1) Copying files using shutil module

shutil.copyfile signature

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

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

shutil.copy signature

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

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

shutil.copy2 signature

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

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

shutil.copyfileobj signature

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) Copying files using os module

os.popen signature

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 signature

os.system(command)


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

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

3) Copying files using subprocess module

subprocess.call signature

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 signature

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)

Answered   2023-09-20 20:14:54

  • Thanks for many options you listed. IMHO, "os.popen('cp source.txt destination.txt') " (with its likes) is bad design. Python is a platform-independent language, and with code like this you destroy this great feature. - anyone
  • Only the first example answers the question. Running shell commands isn't portable and also isn't really performing actions. - anyone

You can use one of the copy functions from the shutil package:

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Function              preserves     supports          accepts     copies other
                      permissions   directory dest.   file obj    metadata  
――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――――
shutil.copy              ✔             ✔                 ☐           ☐
shutil.copy2             ✔             ✔                 ☐           ✔
shutil.copyfile          ☐             ☐                 ☐           ☐
shutil.copyfileobj       ☐             ☐                 ✔           ☐
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Example:

import shutil
shutil.copy('/etc/hostname', '/var/tmp/testhostname')

Answered   2023-09-20 20:14:54

  • What does this add over jezrael answer from 2 years earlier? - anyone
  • @wim you have to compare my answer with jesrael's answer at the time I posted mine. So i added another column, used better column headers and less distracting table layout. I also added an example for the most common case. You are not the first one who asked this and I've answered this question several times already, but unfortunately, some moderator deleted the old comments here. After checking my old notes it even appears that you, wim, already had asked me this question in the past, and your question with my answer was deleted, too. - anyone
  • @wim I also added links into the Python documentation. The added column contains information about directory destinations, Jezrael's answer didn't include. - anyone
  • Oh dear, that's funny that I already asked. Yes I agree that the doc links and extra column are improvements, however, it may have been better to edit them into other answer. Looks like some users subsequently did so, and now we have two answers which aren't significantly different apart from the formatting - recommend deletion? - anyone

Copying a file is a relatively straightforward operation as shown by the examples below, but you should instead use the shutil stdlib module for that.

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)

If you want to copy by filename you could do something like this:

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)

Answered   2023-09-20 20:14:54

Use the shutil module.

copyfile(src, dst)

Copy the contents of the file named src to a file named dst. The destination location must be writable; otherwise, an IOError exception will be raised. If dst already exists, it will be replaced. Special files such as character or block devices and pipes cannot be copied with this function. src and dst are path names given as strings.

Take a look at filesys for all the file and directory handling functions available in standard Python modules.

Answered   2023-09-20 20:14:54

Directory and File copy example, from Tim Golden's Python Stuff:

import os
import shutil
import tempfile

filename1 = tempfile.mktemp (".txt")
open (filename1, "w").close ()
filename2 = filename1 + ".copy"
print filename1, "=>", filename2

shutil.copy (filename1, filename2)

if os.path.isfile (filename2): print "Success"

dirname1 = tempfile.mktemp (".dir")
os.mkdir (dirname1)
dirname2 = dirname1 + ".copy"
print dirname1, "=>", dirname2

shutil.copytree (dirname1, dirname2)

if os.path.isdir (dirname2): print "Success"

Answered   2023-09-20 20:14:54

For small files and using only Python built-ins, you can use the following one-liner:

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

This is not optimal way for applications where the file is too large or when memory is critical, thus Swati's answer should be preferred.

Answered   2023-09-20 20:14:54

  • not safe towards symlinks or hardlinks as it copies the linked file multiple times - anyone

Firstly, I made an exhaustive cheat sheet of the shutil methods for your reference.

shutil_methods =
{'copy':['shutil.copyfileobj',
          'shutil.copyfile',
          'shutil.copymode',
          'shutil.copystat',
          'shutil.copy',
          'shutil.copy2',
          'shutil.copytree',],
 'move':['shutil.rmtree',
         'shutil.move',],
 'exception': ['exception shutil.SameFileError',
                 'exception shutil.Error'],
 'others':['shutil.disk_usage',
             'shutil.chown',
             'shutil.which',
             'shutil.ignore_patterns',]
}

Secondly, explaining methods of copy in examples:

  1. shutil.copyfileobj(fsrc, fdst[, length]) manipulate opened objects

    In [3]: src = '~/Documents/Head+First+SQL.pdf'
    In [4]: dst = '~/desktop'
    In [5]: shutil.copyfileobj(src, dst)
    AttributeError: 'str' object has no attribute 'read'
    
    # Copy the file object
    In [7]: with open(src, 'rb') as f1,open(os.path.join(dst,'test.pdf'), 'wb') as f2:
        ...:      shutil.copyfileobj(f1, f2)
    In [8]: os.stat(os.path.join(dst,'test.pdf'))
    Out[8]: os.stat_result(st_mode=33188, st_ino=8598319475, st_dev=16777220, st_nlink=1, st_uid=501, st_gid=20, st_size=13507926, st_atime=1516067347, st_mtime=1516067335, st_ctime=1516067345)
    
  2. shutil.copyfile(src, dst, *, follow_symlinks=True) Copy and rename

    In [9]: shutil.copyfile(src, dst)
    IsADirectoryError: [Errno 21] Is a directory: ~/desktop'
    # So dst should be a filename instead of a directory name
    
  3. shutil.copy() Copy without preseving the metadata

    In [10]: shutil.copy(src, dst)
    Out[10]: ~/desktop/Head+First+SQL.pdf'
    
    # Check their metadata
    In [25]: os.stat(src)
    Out[25]: os.stat_result(st_mode=33188, st_ino=597749, st_dev=16777220, st_nlink=1, st_uid=501, st_gid=20, st_size=13507926, st_atime=1516066425, st_mtime=1493698739, st_ctime=1514871215)
    In [26]: os.stat(os.path.join(dst, 'Head+First+SQL.pdf'))
    Out[26]: os.stat_result(st_mode=33188, st_ino=8598313736, st_dev=16777220, st_nlink=1, st_uid=501, st_gid=20, st_size=13507926, st_atime=1516066427, st_mtime=1516066425, st_ctime=1516066425)
    # st_atime,st_mtime,st_ctime changed
    
  4. shutil.copy2() Copy with preserving the metadata

    In [30]: shutil.copy2(src, dst)
    Out[30]: ~/desktop/Head+First+SQL.pdf'
    In [31]: os.stat(src)
    Out[31]: os.stat_result(st_mode=33188, st_ino=597749, st_dev=16777220, st_nlink=1, st_uid=501, st_gid=20, st_size=13507926, st_atime=1516067055, st_mtime=1493698739, st_ctime=1514871215)
    In [32]: os.stat(os.path.join(dst, 'Head+First+SQL.pdf'))
    Out[32]: os.stat_result(st_mode=33188, st_ino=8598313736, st_dev=16777220, st_nlink=1, st_uid=501, st_gid=20, st_size=13507926, st_atime=1516067063, st_mtime=1493698739, st_ctime=1516067055)
    # Preserved st_mtime
    
  5. shutil.copytree()

    Recursively copy an entire directory tree rooted at src, returning the destination directory.

Answered   2023-09-20 20:14:54

As of Python 3.5 you can do the following for small files (ie: text files, small jpegs):

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 will overwrite whatever was at the destination's location

Answered   2023-09-20 20:14:54

  • Why only small files? - anyone
  • @Kevin my mistake I should have written down the why when it was fresh. I'll have to review again and update this as I can't remember anymore ¯_(ツ)_/¯ - anyone
  • @Kevin because this loads all the content in memory. - anyone

You could use os.system('cp nameoffilegeneratedbyprogram /otherdirectory/').

Or as I did it,

os.system('cp '+ rawfile + ' rawdata.dat')

where rawfile is the name that I had generated inside the program.

This is a Linux-only solution.

Answered   2023-09-20 20:14:54

  • this is not portable, and unnecessary since you can just use shutil. - anyone
  • Even when shutil is not available - subprocess.run() (without shell=True!) is the better alternative to os.system(). - anyone
  • shutil is more portable - anyone
  • subprocess.run() as suggested by @maxschlepzig is a big step forward, when calling external programs. For flexibility and security however, use the ['cp', rawfile, 'rawdata.dat'] form of passing the command line. (However, for copying, shutil and friends are recommended over calling an external program.) - anyone
  • try that with filenames with spaces in it. - anyone

There are two best ways to copy file in Python.

1. We can use the shutil module

Code Example:

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

There are other methods available also other than copyfile, like copy, copy2, etc, but copyfile is best in terms of performance,

2. We can use the OS module

Code Example:

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

Another method is by the use of a subprocess, but it is not preferable as it’s one of the call methods and is not secure.

Answered   2023-09-20 20:14:54

  • It ought to link to documentation (but not naked links). - anyone
  • Why do we need a new answer? - anyone
  • @PeterMortensen, added links. The answer would add value as other answers are either too broad or either too straight, this answer contains the helpful information that can be used easiy. - anyone

Use

open(destination, 'wb').write(open(source, 'rb').read())

Open the source file in read mode, and write to the destination file in write mode.

Answered   2023-09-20 20:14:54

  • All answers need explanation, even if it is one sentence. No explanation sets bad precedent and is not helpful in understanding the program. What if a complete Python noob came along and saw this, wanted to use it, but couldn't because they don't understand it? You want to be helpful to all in your answers. - anyone
  • Isn't that missing the .close() on all of those open(...)s? - anyone
  • No need of .close(), as we are NOT STORING the file pointer object anywhere(neither for the src file nor for the destination file). - anyone
  • AFAIK, it is undefined when the files are actually closed, @SundeepBorra. Using with (as in the example above) is recommended and not more complicated. Using read() on a raw file reads the entire file into memory, which may be too big. Use a standard function like from shutil so that you and whoever else is involved in the code does not need to worry about special cases. docs.python.org/3/library/io.html#io.BufferedReader - anyone
  • Same suboptimal memory-wasting approach as yellow01's answer. - anyone

Use subprocess.call to copy the file

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

Answered   2023-09-20 20:14:54

  • This depends on the platform, so i would not use is. - anyone
  • Such a call is unsecure. Please refere to the subproces docu about it. - anyone
  • this is not portable, and unnecessary since you can just use shutil. - anyone
  • Hmm why Python, then? - anyone
  • Maybe detect the operating system before starting (whether it's DOS or Unix, because those are the two most used) - anyone

For large files, I read the file line by line and read each line into an array. Then, once the array reached a certain size, append it to a new file.

for line in open("file.txt", "r"):
    list.append(line)
    if len(list) == 1000000: 
        output.writelines(list)
        del list[:]

Answered   2023-09-20 20:14:54

  • this seems a little redundant since the writer should handle buffering. for l in open('file.txt','r'): output.write(l) should work find; just setup the output stream buffer to your needs. or you can go by the bytes by looping over a try with output.write(read(n)); output.flush() where n is the number of bytes you'd like to write at a time. both of these also don't have an condition to check which is a bonus. - anyone
  • Yes, but I thought that maybe this could be easier to understand because it copies entire lines rather than parts of them (in case we don't know how many bytes each line is). - anyone
  • @owns To add to this question a year later, writelines() has shown slightly better performance over write() since we don't waste time consistently opening a new filestream, and instead write new lines as one large bytefeed. - anyone
  • looking at the source - writelines calls write, hg.python.org/cpython/file/c6880edaf6f3/Modules/_io/bytesio.c. Also, the file stream is already open, so write wouldn't need to reopen it every time. - anyone
  • This is awful. It does unnecessary work for no good reason. It doesn't work for arbitrary files. The copy isn't byte-identical if the input has unusual line endings on systems like Windows. Why do you think that this might be easier to understand than a call to a copy function in shutil? Even when ignoring shutil, a simple block read/write loop (using unbuffered IO) is straight forward, would be efficient and would make much more sense than this, and thus is surely easier to teach and understand. - anyone

In case you've come this far down. The answer is that you need the entire path and file name

import os

shutil.copy(os.path.join(old_dir, file), os.path.join(new_dir, file))

Answered   2023-09-20 20:14:54

  • if show importing of os, you should import shutil as well - anyone

Here is a simple way to do it, without any module. It's similar to this answer, but has the benefit to also work if it's a big file that doesn't fit in RAM:

with open('sourcefile', 'rb') as f, open('destfile', 'wb') as g:
    while True:
        block = f.read(16*1024*1024)  # work by blocks of 16 MB
        if not block:  # end of file
            break
        g.write(block)

Since we're writing a new file, it does not preserve the modification time, etc.
We can then use os.utime for this if needed.

Answered   2023-09-20 20:14:54

Similar to the accepted answer, the following code block might come in handy if you also want to make sure to create any (non-existent) folders in the path to the destination.

from os import path, makedirs
from shutil import copyfile
makedirs(path.dirname(path.abspath(destination_path)), exist_ok=True)
copyfile(source_path, destination_path)

As the accepted answers notes, these lines will overwrite any file which exists at the destination path, so sometimes it might be useful to also add: if not path.exists(destination_path): before this code block.

Answered   2023-09-20 20:14:54

I would like to propose a different solution.

def copy(source, destination):
   with open(source, 'rb') as file:
       myFile = file.read()
   with open(destination, 'wb') as file:
       file.write(myFile)

copy("foo.txt", "bar.txt")

The file is opened, and it's data is written to a new file of your choosing.

Answered   2023-09-20 20:14:54

  • in Linux, this is not safe towards symlinks or hardlinks, the linked file is rewritten multiple times - anyone

For the answer everyone recommends, if you prefer not to use the standard modules, or have completely removed them as I've done, preferring more core C methods over poorly written python methods

The way shutil works is symlink/hardlink safe, but is rather slow due to os.path.normpath() containing a while (nt, mac) or for (posix) loop, used in testing if src and dst are the same in shutil.copyfile()

This part is mostly unneeded if you know for certain src and dst will never be the same file, otherwise a faster C approach could potentially be used.
(note that just because a module may be C doesn't inherently mean it's faster, know that what you use is actually written well before you use it)

After that initial testing, copyfile() runs a for loop on a dynamic tuple of (src, dst), testing for special files (such as sockets or devices in posix).

Finally, if follow_symlinks is False, copyfile() tests if src is a symlink with os.path.islink(), which varies between nt.lstat() or posix.lstat() (os.lstat()) on Windows and Linux, or Carbon.File.ResolveAliasFile(s, 0)[2] on Mac.
If that test resolves True, the core code that copies a symlink/hardlink is:

        os.symlink(os.readlink(src), dst)

Hardlinks in posix are done with posix.link(), which shutil.copyfile() doesn't call, despite being callable through os.link().
(probably because the only way to check for hardlinks is to hashmap the os.lstat() (st_ino and st_dev specifically) of the first inode we know about, and assume that's the hardlink target)

Else, file copying is done via basic file buffers:

       with open(src, 'rb') as fsrc:
           with open(dst, 'wb') as fdst:
               copyfileobj(fsrc, fdst)

(similar to other answers here)

copyfileobj() is a bit special in that it's buffer safe, using a length argument to read the file buffer in chunks:

def copyfileobj(fsrc, fdst, length=16*1024):
    """copy data from file-like object fsrc to file-like object fdst"""
    while 1:
        buf = fsrc.read(length)
        if not buf:
            break
        fdst.write(buf)

Hope this answer helps shed some light on the mystery of file copying using core mechanisms in python. :)

Overall, shutil isn't too poorly written, especially after the second test in copyfile(), so it's not a horrible choice to use if you're lazy, but the initial tests will be a bit slow for a mass copy due to the minor bloat.

Answered   2023-09-20 20:14:54