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Sorting a file based on the absolute value of a column

 
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# 1  
Old 08-08-2012
Sorting a file based on the absolute value of a column
I would like to sort a tab delimited text file based on the absolute value of its second column. How do I go about doing that? Thanks!

Example input:
Code:
A -12
B 0
C -6
D 7

Output:
Code:
A -12
D 7
C -6
B 0

# 2  
Old 08-08-2012
There might be more efficient ways, but this will work and is straight forward:

Code:
awk '{print $2 < 0 ? -1 * $2 : $2, $0}'  input-file-name | sort -k 1rn,1 | sed -E 's/^[0-9]+ //'

This User Gave Thanks to agama For This Post:
evelibertine
# 3  
Old 08-08-2012
Hi, try:

Code:
awk '{print $2*$2,$0}' infile | sort -rn | cut -d " " -f2-

These 3 Users Gave Thanks to Scrutinizer For This Post:
evelibertine praveen_reddy vgersh99
# 4  
Old 08-08-2012
Code:
sort -k2n data.txt

This User Gave Thanks to praveen_reddy For This Post:
evelibertine
# 5  
Old 08-08-2012
Try this:
Code:
$ cat test.txt
A       -12
B       0
C       -6
D       7

$ sed 's/\t\([^-]*[0-9]\{1,\}\)/\t-\1-/' test.txt | sort -n -t$'\t' -k2,3  | sed 's/\t-\([0-9]\{1,\}\)-/\t\1/'
A       -12
D       7
C       -6
B       0

This User Gave Thanks to spacebar For This Post:
evelibertine
 
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