计算每行文件中的模式？

Question

My file looks like this:

id12 ack dko hhhh chfl dkl dll chfl
id14 slo ksol chfl dloo
id13 mse
id23 clos chfl dll alo

grep -c 'chfl' filename , gives me the number of occurrence of chfl , but I want to count occurrence of chfl per line. Like this:

id12 2
id14 1
id13 0
id23 1

Also how do I do the same with two patterns to match? Like chfl and dll ?

Answer 1

perl -lane 'undef $c;
            for(@F){$c++ if(/^chfl$/)};
            print "$F[0] ",$c?$c:"0"' your_file

Or simply:

perl -lane '$c=0;
            for(@F){$c++ if(/^chfl$/)};
            print "$F[0] $c"' your_file

Tested below:

> cat temp
id12 ack dko hhhh chfl dkl dll chfl
id14 slo ksol chfl dloo
id13 mse
id23 clos chfl dll alo
> perl -lane '$c=0;for(@F){$c++ if(/^chfl$/)};print "$F[0] $c"' temp
id12 2
id14 1
id13 0
id23 1
> 

Also in awk:(Logic here remains the same as above one in perl)

awk '{a=0;
     for(i=1;i<=NF;i++)if($i~/chfl/)a++;
     print $1,a}' your_file

Answer 2

A Perl version that copes with multiple strings.

#!/usr/bin/perl

use strict;
use warnings;
use 5.010;

die "Usage: $0 pattern [pattern ...] file\n" unless @ARGV > 1;

my @patterns;
until (@ARGV == 1) {
  push @patterns, shift;
}

my $re = '(' . join('|', map { "\Q$_\E" } @patterns) . ')';

my %match;
while (<>) {
  if (my @matches = /$re/g) {
    $match{$_}++ for @matches;
  }
}

say "$_: $match{$_}" for sort keys %match;

A couple of test runs:

$ ./cgrep chfl dll cgrep.txt 
chfl: 4
$ ./cgrep chfl dll cgrep.txt 
chfl: 4
dll: 2

Answer 3

How about:

my %res;
while(<DATA>) {
    chomp;
    my ($id,$rest) = $_ =~ /^(\S+)(.*)$/;
    $res{chfl}{$id} =()= $rest =~ /(chfl)/g;
    $res{dll}{$id} =()= $rest =~ /(dll)/g;
}
say Dumper\%res;

__DATA__
id12 ack dko hhhh chfl dkl dll chfl
id14 slo ksol chfl dloo
id13 mse
id23 clos chfl dll alo

output:

$VAR1 = {
          'dll' => {
                     'id13' => 0,
                     'id12' => 1,
                     'id23' => 1,
                     'id14' => 0
                   },
          'chfl' => {
                      'id13' => 0,
                      'id12' => 2,
                      'id23' => 1,
                      'id14' => 1
                    }
        };

Answer 4

Use this:

awk 'BEGIN {print "id\tchfl\tdll\n--------------------"}{c=d=i=0;while(i++<NF){if($i=="chfl")c++; if($i=="dll")d++}; print $1,c,d}' OFS="\t" file
id      chfl    dll
--------------------
id12    2       1
id14    1       0
id13    0       0
id23    1       1

Answer 5

bash one liner with grep:

while read line ; do echo $line | grep -o 'chfl' | wc -l  ; done < your_file

-o outputs every occurence on a new line and wc counts them.

Edit for multiple patterns:

patterns=(chfl dll)

while read line ; do
    for pattern in ${patterns[@]} ; do
        echo -ne $pattern"\t" ; echo $line | grep -o $pattern | wc -l 
    done
done < your_file

Answer 6

Another version of awk :

$ awk '{c1=gsub(var1,x);c2=gsub(var2,x);print $1,var1"="c1,var2"="c2}' var1="chfl" var2="dll"  file
id12 chfl=2 dll=1
id14 chfl=1 dll=0
id13 chfl=0 dll=0
id23 chfl=1 dll=1

Just pass the variables you want to count at the end of the file.

Answer 7

你可以用这个awk ，

awk '{d=c=0;for(i=1;i<=NF;i++){ if($i ~ /chfl/)c++; if($i ~ /dll/)d++;} print $1,c,d}' yourfile

Answer 8

perl -ne 'my $c=s/chfl//g||0;my $d=s/dll//g||0;s/ .*//s;print "$_ chfl $c dll $d\n"' file

Explanation:

• s///g in scalar context returns the number of substitutions made
• ||0 make sure the variable is set to zero if there are no matches
• s/ .*//s throws away everything from the 1st space from $_ , leaving the id only

It will produce the following output:

id12 chfl 2 dll 1
id14 chfl 1 dll 0
id13 chfl 0 dll 0    
id23 chfl 1 dll 1