Recursively truncate each file in directory to 20 lines

Question

I want to share my codebase for others to get a glimpse without seeing all the details.

I just want them to see the directory structure and the first N lines of every file.

How can I do this using bash or some other standard scripting language like python?

NOTE: If some other tool can do this, it would also work

This seems to have nothing to do with Git or GitHub (even if you intend to use the result with them later, they're not relevant to how you would do this in code). — torek, Oct 31 '19 at 01:12
pluse uno for more interesting than average Q ;-) Good luck. — shellter, Oct 31 '19 at 01:16

score -1 · Answer 1 · answered Oct 31 '19 at 01:39

Something like this will do the trick:

import os

for r, d, files in os.walk('.') :
    for f in files :
        name = os.path.join( r, f )

        with open( name ) as fin :
            text = fin.readlines()[:20]

        with open( name, 'w' ) as fout :
            fout.write( ''.join( text ) )

score -1 · Answer 2 · answered Oct 31 '19 at 01:52

-1

One way using sed:

find . -type f -exec sed -i.bak '21,$d' {} \;

answered Oct 31 '19 at 01:52

Steve

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@oguzismail: Should also work with BSD/OSX sed, not just GNU sed. – Steve Oct 31 '19 at 02:52
2

`find -type f -exec sed -si 20q {} +`. – jthill Oct 31 '19 at 03:09

Recursively truncate each file in directory to 20 lines

2 Answers2