计算两个模式之间的重复记录

Question

我想计算两个模式之间的重复记录，并将它们打印在模式一和二之后。

INPUT

  

群集1   一个
  一个
  2个
  2个
  2个
  集群2
  2个
  一个
  一个
  2个
  集群3
  一个
  集群4
  2个
  2个
  一个

输出

  

群集1 2 3
  集群2 2 2
  群集3 1 0
  群集4 1 2

我尝试过awk，但它只计算Cluster之间的行，而不是重复行。

我试过这段代码; .....

cut -d _ -f2 norm_nonnorm_merged-95-count-new.clstr | sort | uniq -c | awk '$1>1&&$0=$2 " - " $1 " times"'

打印

one - 6 times
two - 7 times

也试过

awk ' />Cluster/ { if ( f!="") {print f-1}; f=0;f1=1 } f1 { f++ } ' norm_nonnorm_merged-95.clstr

打印

5
4
1
3

Answer 1

更长形式的perl脚本解决方案：

use strict;
use warnings;

my %allkeys;
my @clusters;

while (<DATA>) {
    chomp;
    if (/^Cluster/) {
        push @clusters, [$_, {}];
    } else {
        $clusters[-1][1]{$_}++;
        $allkeys{$_}++;
    }
}

my @sortedkeys = sort keys %allkeys;
for my $cluster (@clusters) {
    print "$cluster->[0] " . join(' ', map {$cluster->[1]{$_} // 0} @sortedkeys) . "\n";
}


__DATA__
Cluster 1
one
one
two
two
two
Cluster 2
two
one
one
two
Cluster 3
one
Cluster 4
two
two
one

输出：

Cluster 1 2 3
Cluster 2 2 2
Cluster 3 1 0
Cluster 4 1 2

Answer 2

这适用于Python：

txt='''\
Cluster 1
one
one
two
two
two
Cluster 2
two
one
one
two
Cluster 3
one
Cluster 4
two
two
one
'''

import re

for t in re.findall(r'(Cluster \d)(.*?)(?=Cluster|$)', txt, re.S):
    print t[0],t[1].count('one'), t[1].count('two')

打印：

Cluster 1 2 3
Cluster 2 2 2
Cluster 3 1 0
Cluster 4 1 2

---

（未经测试......）

import re

fn1='one.txt'
fn2='two.txt'
fn3='one-two.txt'

with open(fn1, 'w') as f1, open(fn2, 'w') as f2, open(fn3, 'w') as f3:
    for t in re.findall(r'(Cluster \d+)(.*?)(?=Cluster|$)', txt, re.S):
        one=t[1].count('one')
        two=t[1].count('two')
        out= '{} {} {}\n'.format(t[0], one, two)
        if one and two:
            f3.write(out)
        elif one:
            f1.write(out)
        elif two:
            f2.write(out)