Sort lines by number of words per line

Question

Given input:

hello: world foo bar baz
bar:
baz: bin boop bop fiz bang beep
bap: bim bam bop
boatkeeper: poughkeepsie

I would like to sort it into most words at the top, to least at the end, like so:

baz: bin boop bop fiz bang beep
hello: world foo bar baz
bap: bim bam bop
boatkeeper: poughkeepsie
bar:

How would I do this with sort or some other tool?

Answer 1

You could do something like:

awk '{print NF,$0}' file | sort -nr | cut -d' ' -f 2-

We use awk to prefix the number of fields to each line. We then sort by that number and remove it with cut.

Answer 2

In recent GNU awk one can use PROCINFO array to define many internal parameters including order in which array elements are printed (controlled by element "sorted_in"). Thus we can built and array indexed with the value of NF" "NR, which elements have value of $0 and print it in desired output, in your case that would be "@ind_num_desc":

awk '{a[NF" "NR]=$0}END{PROCINFO["sorted_in"]="@ind_num_desc"; for(i in a) print a[i]}' file

Answer 3

Perl one-liner:

print sort { split(' ',$a) <=> split(' ',$b) } <>;

If you want to break ties using alphabetical order:

print sort { split(' ',$a) <=> split(' ',$b) or $a cmp $b } <>;

Answer 4

Through python.

s = '''hello: world foo bar baz
bar:
baz: bin boop bop fiz bang beep
bap: bim bam bop'''.splitlines()
for i in sorted(s, key=lambda x: len(x.split()), reverse=True):
    print(i)

or

with open('/path/to/the/input/file') as f:
    m = f.readlines()
    for i in sorted(m, key=lambda x: len(x.split()), reverse=True):
        print(i, end="")