Sort by frequency without deleting lines?

Question

How can I sort by frequency without deleting any?

one
two
one

to

one
one
two

The output of the standard sort-by-frequency :%!sort | uniq -c | sort -nr is

   2 one
   1 two

It has deleted all the copies of the line (bc of uniq), and the spaces and numbers are not desired in the output.

So how can I sort by frequency and keep all lines in VIM?

Answer 1

The output of your current pipeline is always in the format

[zero or more spaces][number of repetitions][space][actual line contents]
                     ^^^^^^^^^^^^^^^^^^^^^^^       ^^^^^^^^^^^^^^^^^^^^^^
                         First field: $1         Remaining fields: $2, $3...

We can thus set up a for-loop that prints the actual line contents $1 times.

If the file is as simple as you make it look (each line only contains one word), use

:%!sort | uniq -c | sort -nr | awk '{for(i=$1;i>0;i--)print $2}'

Otherwise (this is the general answer), use

:%!sort | uniq -c | sort -nr | awk '{n=$1;sub(/ *[0-9]+ +/,"");for(i=n;i>0;i--)print}'

Answer 2

Here's a mostly vimscript way. Start with your pipeline:

:%!sort | uniq -c | sort -nr

Now, either

• use a :global to delete the number and the line, then repeatedly paste

:global/./execute 'normal! 0"adawdd' | let @a = str2nr(@a) | execute 'normal!' @a.'P'

• somewhat shorter, but may break depending on the number format (e.g., 010 becomes 8)

:global/./execute 'normal! ^"adiwdd' | execute 'normal!' @a.'P'

• use a substitute with \= to compute the replacement

:%substitute/^\s*\(\d\) \(.*\)/\=repeat([submatch(2)], submatch(1)->str2nr())->join("\n")