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Top Forums Shell Programming and Scripting Finding most repeated entry in a column and giving the count Post 302677305 by necro98 on Thursday 26th of July 2012 02:31:43 AM
Old 07-26-2012
Quote:
Originally Posted by summer_cherry
perl

Code:
open $fh,"<", "a";
while(<$fh>){
    chomp;
    my @tmp = split(",",$_);
    $hash{$tmp[1]}->{'CNT'}++;
    $hash{$tmp[1]}->{'CONTENT'}=$hash{$tmp[1]}->{'CONTENT'}."\n".$_;
}
close $fh;
my $key = (sort {$hash{$b}->{'CNT'} cmp $hash{$a}->{'CNT'}} keys %hash)[0];
print $key,"=",$hash{$key}->{'CNT'},"\n";
print $hash{$key}->{'CONTENT'};

awk:

Code:
awk -F"," '{
    cnt[$2]++
    content[$2]=sprintf("%s\n%s",content[$2],$0)
}
END{
    for(i in cnt){
        if(ind ==""){
            ind=i
            max=cnt[i]
        }
        else{
            if(cnt[i]>=max){
                ind=i
                max=cnt[i]
            }
        }
    }
    print ind"="cnt[ind]
    print content[ind]
}' a

Thanks very much f this ,it worked

In addition Some of the lines in the same file contain the letter C: with a value
Here the value is 0

1,00: This , is a good script c:0

I want to output of the lines with top 3 highest value for c:

1,00: This , is a nice script c:9999
1,00: This , is a good script c:9998
1,00: This , is a cool script c:9000
1,00: This , is a fun script c:12

So the output should be

1,00: This , is a nice script c:9999
1,00: This , is a good script c:9998
1,00: This , is a cool script c:9000

---------- Post updated at 01:31 AM ---------- Previous update was at 12:30 AM ----------

Hi summer , Please can you help with the above
 

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bup-margin(1)						      General Commands Manual						     bup-margin(1)

NAME
bup-margin - figure out your deduplication safety margin SYNOPSIS
bup margin [options...] DESCRIPTION
bup margin iterates through all objects in your bup repository, calculating the largest number of prefix bits shared between any two entries. This number, n, identifies the longest subset of SHA-1 you could use and still encounter a collision between your object ids. For example, one system that was tested had a collection of 11 million objects (70 GB), and bup margin returned 45. That means a 46-bit hash would be sufficient to avoid all collisions among that set of objects; each object in that repository could be uniquely identified by its first 46 bits. The number of bits needed seems to increase by about 1 or 2 for every doubling of the number of objects. Since SHA-1 hashes have 160 bits, that leaves 115 bits of margin. Of course, because SHA-1 hashes are essentially random, it's theoretically possible to use many more bits with far fewer objects. If you're paranoid about the possibility of SHA-1 collisions, you can monitor your repository by running bup margin occasionally to see if you're getting dangerously close to 160 bits. OPTIONS
--predict Guess the offset into each index file where a particular object will appear, and report the maximum deviation of the correct answer from the guess. This is potentially useful for tuning an interpolation search algorithm. --ignore-midx don't use .midx files, use only .idx files. This is only really useful when used with --predict. EXAMPLE
$ bup margin Reading indexes: 100.00% (1612581/1612581), done. 40 40 matching prefix bits 1.94 bits per doubling 120 bits (61.86 doublings) remaining 4.19338e+18 times larger is possible Everyone on earth could have 625878182 data sets like yours, all in one repository, and we would expect 1 object collision. $ bup margin --predict PackIdxList: using 1 index. Reading indexes: 100.00% (1612581/1612581), done. 915 of 1612581 (0.057%) SEE ALSO
bup-midx(1), bup-save(1) BUP
Part of the bup(1) suite. AUTHORS
Avery Pennarun <apenwarr@gmail.com>. Bup unknown- bup-margin(1)
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