Python concatenate text files

问题:

I have a list of 20 file names, like [\'file1.txt\', \'file2.txt\', ...]. I want to write a Python script to concatenate these files into a new file. I could open each file by f = open(...), read line by line by calling f.readline(), and write each line into that new file. It doesn\'t seem very \"elegant\" to me, especially the part where I have to read//write line by line.

Is there a more \"elegant\" way to do this in Python?

回答1:

This should do it

For large files:

filenames = [\'file1.txt\', \'file2.txt\', ...]
with open(\'path/to/output/file\', \'w\') as outfile:
    for fname in filenames:
        with open(fname) as infile:
            for line in infile:
                outfile.write(line)

For small files:

filenames = [\'file1.txt\', \'file2.txt\', ...]
with open(\'path/to/output/file\', \'w\') as outfile:
    for fname in filenames:
        with open(fname) as infile:
            outfile.write(infile.read())

… and another interesting one that I thought of:

filenames = [\'file1.txt\', \'file2.txt\', ...]
with open(\'path/to/output/file\', \'w\') as outfile:
    for line in itertools.chain.from_iterable(itertools.imap(open, filnames)):
        outfile.write(line)

Sadly, this last method leaves a few open file descriptors, which the GC should take care of anyway. I just thought it was interesting

回答2:

Use shutil.copyfileobj. It should be more efficient.

with open(\'output_file.txt\',\'wb\') as wfd:
    for f in [\'seg1.txt\',\'seg2.txt\',\'seg3.txt\']:
        with open(f,\'rb\') as fd:
            shutil.copyfileobj(fd, wfd, 1024*1024*10)
            #10MB per writing chunk to avoid reading big file into memory.

回答3:

That\'s exactly what fileinput is for:

import fileinput
with open(outfilename, \'w\') as fout, fileinput.input(filenames) as fin:
    for line in fin:
        fout.write(line)

For this use case, it\'s really not much simpler than just iterating over the files manually, but in other cases, having a single iterator that iterates over all of the files as if they were a single file is very handy. (Also, the fact that fileinput closes each file as soon as it\'s done means there\'s no need to with or close each one, but that\'s just a one-line savings, not that big of a deal.)

There are some other nifty features in fileinput, like the ability to do in-place modifications of files just by filtering each line.

---

As noted in the comments, and discussed in another post, fileinput for Python 2.7 will not work as indicated. Here slight modification to make the code Python 2.7 compliant

with open(\'outfilename\', \'w\') as fout:
    fin = fileinput.input(filenames)
    for line in fin:
        fout.write(line)
    fin.close()

回答4:

I don\'t know about elegance, but this works:

    import glob
    import os
    for f in glob.glob(\"file*.txt\"):
         os.system(\"cat \"+f+\" >> OutFile.txt\")

回答5:

What\'s wrong with UNIX commands ? (given you\'re not working on Windows) :

ls | xargs cat | tee output.txt does the job ( you can call it from python with subprocess if you want)

回答6:

Check out the .read() method of the File object:

http://docs.python.org/2/tutorial/inputoutput.html#methods-of-file-objects

You could do something like:

concat = \"\"
for file in files:
    concat += open(file).read()

or a more \'elegant\' python-way:

concat = \'\'.join([open(f).read() for f in files])

which, according to this article: http://www.skymind.com/~ocrow/python_string/ would also be the fastest.

回答7:

If the files are not gigantic:

with open(\'newfile.txt\',\'wb\') as newf:
    for filename in list_of_files:
        with open(filename,\'rb\') as hf:
            newf.write(hf.read())
            # newf.write(\'\\n\\n\\n\')   if you want to introduce
            # some blank lines between the contents of the copied files

If the files are too big to be entirely read and held in RAM, the algorithm must be a little different to read each file to be copied in a loop by chunks of fixed length, using read(10000) for example.

回答8:

An alternative to @inspectorG4dget answer (best answer to date 29-03-2016). I tested with 3 files of 436MB.

@inspectorG4dget solution: 162 seconds

The following solution : 125 seconds

from subprocess import Popen
filenames = [\'file1.txt\', \'file2.txt\', \'file3.txt\']
fbatch = open(\'batch.bat\',\'w\')
str =\"type \"
for f in filenames:
    str+= f + \" \"
fbatch.write(str + \" > file4results.txt\")
fbatch.close()
p = Popen(\"batch.bat\", cwd=r\"Drive:\\Path\\to\\folder\")
stdout, stderr = p.communicate()

The idea is to create a batch file and execute it, taking advantage of \"old good technology\". Its semi-python but works faster. Works for windows.

回答9:

If you have a lot of files in the directory then glob2 might be a better option to generate a list of filenames rather than writing them by hand.

import glob2

filenames = glob2.glob(\'*.txt\')  # list of all .txt files in the directory

with open(\'outfile.txt\', \'w\') as f:
    for file in filenames:
        with open(file) as infile:
            f.write(infile.read()+\'\\n\')

回答10:

outfile.write(infile.read()) 2.1085190773010254s
shutil.copyfileobj(fd, wfd, 1024*1024*10) 0.60599684715271s

A simple benchmark shows that the shutil performs better.

回答11:

def concatFiles():
    path = \'input/\'
    files = os.listdir(path)
    for idx, infile in enumerate(files):
        print (\"File #\" + str(idx) + \"  \" + infile)
    concat = \'\'.join([open(path + f).read() for f in files])
    with open(\"output_concatFile.txt\", \"w\") as fo:
        fo.write(path + concat)

if __name__ == \"__main__\":
    concatFiles()