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Full Discussion: Count and print the number of occurences
Top Forums Shell Programming and Scripting Count and print the number of occurences Post 302835113 by arch on Sunday 21st of July 2013 11:33:46 PM
Old 07-22-2013
Count and print the number of occurences
I have some files as shown below

Code:
GLL  ALM  654-656  654  656 
SEM  LYG  655-657  655  657  
SEM  LYG  655-657  655  657  
ALM  LEG  656-658  656  658  
ALM  LEG  656-658  656  658  
ALM  LEG  656-658  656  658  
LEG  LEG  658-660  658  660 
LEG  LEG  658-660  658  660

The value of GLL is 654. The value of ALM is 656. In the same way, 4th column represents the values of first column. 5th column represents the values of second column.

I tried the following program to count the occurrences of each number in the fourth and fifth column.

Code:
 for i in folder1/*.pdb;
do
awk '
BEGIN {
    path=sprintf("%s", "/home/arch/Desktop/folder2/")
}
!s[1":"$4":"$5]++{sU[$4]++;tot++} 
!s[2":"$4":"$5]++{sU[$5]++;tot++} 
END { 
    sub(/.*\//,"",FILENAME)
    for (x in sU) 
        print x, sU[x], sU[$1] > path FILENAME;       
}'  $i;
 done

The above program prints as follows

Code:
660 1 
654 1 
655 1 
656 2 
657 1 
658 2

Desired Output:-

Code:
 660 LEG 1
 654 GLL 1
 655 SEM 1
 656 ALM 2
 657 LYG 1
 658 LEG 2

your suggestions would be appreciated!!
Last edited by arch; 07-22-2013 at 02:24 AM.. Reason: code tags
 

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LEARN ABOUT DEBIAN

bup-margin

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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