unix - breakdown of how many lines with number of character occurrences - linux

Question

Is there an inbuilt command to do this or has anyone had any luck with a script that does it?
I am looking to get counts of how many lines had how many occurrences of a specfic character. (sorted descending by the number of occurrences)
For example, with this sample file:
gkdjpgfdpgdp
fdkj
pgdppp
ppp
gfjkl
Suggested input (for the 'p' character)
bash/perl some_script_name "p" samplefile
Desired output:
occs count
4 1
3 2
0 2
Update:
How would you write a solution that worked off a 2 character string such as 'gd' not a just a specific character such as p?

Answer 1

$ sed 's/[^p]//g' input.txt | awk '{print length}' | sort -nr | uniq -c | awk 'BEGIN{print "occs", "count"}{print $2,$1}' | column -t
occs count
4 1
3 2
0 2

Answer 2

You could give the desired character as the field separator for awk, and do this:
awk -F 'p' '{ print NF-1 }' |
sort -k1nr |
uniq -c |
awk -v OFS="\t" 'BEGIN { print "occs", "count" } { print $2, $1 }'
For your sample data, it produces:
occs count
4 1
3 2
0 2
If you want to count occurrences of multi-character strings, just give the desired string as the separator, e.g., awk -F 'gd' ... or awk -F 'pp' ....

Answer 3

#!/usr/bin/env perl
use strict; use warnings;
my $seq = shift #ARGV;
die unless defined $seq;
my %freq;
while ( my $line = <> ) {
last unless $line =~ /\S/;
my $occurances = () = $line =~ /(\Q$seq\E)/g;
$freq{ $occurances } += 1;
}
for my $occurances ( sort { $b <=> $a} keys %freq ) {
print "$occurances:\t$freq{$occurances}\n";
}
If you want short, you can always use:
#!/usr/bin/env perl
$x=shift;/\S/&&++$f{$a=()=/(\Q$x\E)/g}while<>
;print"$_:\t$f{$_}\n"for sort{$b<=>$a}keys%f;
or, perl -e '$x=shift;/\S/&&++$f{$a=()=/(\Q$x\E)/g}while<>;print"$_:\t$f{$_}\n"for sort{$b<=>$a}keys%f' inputfile, but now I am getting silly.

Answer 4

Pure Bash:
declare -a count
while read ; do
cnt=${REPLY//[^p]/} # remove non-p characters
((count[${#cnt}]++)) # use length as array index
done < "$infile"
for idx in ${!count[*]} # iterate over existing indices
do echo -e "$idx ${count[idx]}"
done | sort -nr
Output as desired:
4 1
3 2
0 2

Answer 5

Can to it in one gawk process (well, with a sort coprocess)
gawk -F p -v OFS='\t' '
{ count[NF-1]++ }
END {
print "occs", "count"
coproc = "sort -rn"
for (n in count)
print n, count[n] |& coproc
close(coproc, "to")
while ((coproc |& getline) > 0)
print
close(coproc)
}
'

Answer 6

Shortest solution so far:
perl -nE'say tr/p//' | sort -nr | uniq -c |
awk 'BEGIN{print "occs","count"}{print $2,$1}' |
column -t
For multiple characters, use a regex pattern:
perl -ple'$_ = () = /pg/g' | sort -nr | uniq -c |
awk 'BEGIN{print "occs","count"}{print $2,$1}' |
column -t
This one handles overlapping matches (e.g. it finds 3 "pp" in "pppp" instead of 2):
perl -ple'$_ = () = /(?=pp)/g' | sort -nr | uniq -c |
awk 'BEGIN{print "occs","count"}{print $2,$1}' |
column -t
Original cryptic but short pure-Perl version:
perl -nE'
++$c{ () = /pg/g };
}{
say "occs\tcount";
say "$_\t$c{$_}" for sort { $b <=> $a } keys %c;
'