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Top Forums Shell Programming and Scripting Need a quick and dirty solution
# 1  
Old 06-28-2012
Need a quick and dirty solution
I have a list of multiple versions of software. The list is formated as follows:

NAME VERSION

I simply need to pull out the highest version of each software, for example:

Original File
Code:
a v1.0
a v1.1
a v1.2
b v2.1
b v2.2
b v2.21
b v3.0

Output
Code:
a v1.2
b v3.0

# 2  
Old 06-28-2012
Code:
$ awk '(!A[$1]) || (A[$1] < $2) { A[$1]=$2 }; END { for (X in A) { print X, A[X] } }' <<EOF
a v1.0
a v1.1
a v1.2
b v2.1
b v2.2
b v2.21
b v3.0
EOF

a v1.2
b v3.0

$

# 3  
Old 06-28-2012
Quote:
Originally Posted by Corona688
Code:
$ awk '(!A[$1]) || (A[$1] < $2) { A[$1]=$2 }; END { for (X in A) { print X, A[X] } }' <<EOF

Given that the sample version data contains components of varying widths, e.g. 2.2 and 2.21, I don't think this code is sufficient.

The highlighted expression is a string comparison. v3.9 will be considered greater than v3.10. If the comparison were made using numeric substrings, 3.9 would still be greater than 3.10.

Regards,
Alister

---------- Post updated at 07:04 PM ---------- Previous update was at 07:00 PM ----------

Quick and dirty, as per the topic:
Code:
tr . '\t' < file | sort -k1,1 -k2.2b,2nr -k3,3nr | awk 'p!=$1 {p=$1; print}' | tr '\t' .

Assumes there is only the one dot in the version numbers, that there are no tabs in the file, and that your field delimiter is a space. If the delimiter is a tab and there are no spaces, change \t to a space.

Regards,
Alister
Last edited by alister; 06-28-2012 at 08:15 PM..
# 4  
Old 06-28-2012
Hi.

Using available (but non-standard) utility msort:
Code:
#!/usr/bin/env bash

# @(#) s1	Demonstrate msort hybrid sort.
# See: http://freecode.com/projects/msort

pe() { for _i;do printf "%s" "$_i";done; printf "\n"; }
pl() { pe;pe "-----" ;pe "$*"; }
db() { ( printf " db, ";for _i;do printf "%s" "$_i";done;printf "\n" ) >&2 ; }
db() { : ; }
C=$HOME/bin/context && [ -f $C ] && $C msort awk

FILE=${1-data1}
pl " Input data $FILE:"
cat $FILE

pl " Results:"
msort -q -l -n 2,2 -I -c hybrid $FILE |
tee f1 |
awk '!a[$1]++'

exit 0

producing:
Code:
% ./s1

Environment: LC_ALL = C, LANG = C
(Versions displayed with local utility "version")
OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution        : Debian GNU/Linux 5.0.8 (lenny) 
bash GNU bash 3.2.39
msort 8.44
awk GNU Awk 3.1.5

-----
 Input data data1:
d v4.0.9
d v4.0.10
a v1.0
a v1.1
a v1.2
b v2.1
b v2.2
b v2.21
b v3.0
c v3.10
c v3.9

-----
 Results:
d v4.0.10
c v3.10
b v3.0
a v1.2

The intermediate results are on file f1. The awk prints the first occurrence of an item, skipping the rest.

If msort is not in your repository, see link in script for binary or source to compile.

Best wishes ... cheers, drl
# 5  
Old 06-29-2012
Code:
awk -F"[ v]" 'a[$1]+0<$3+0{a[$1]=$3} END{for(i in a) print i" v"a[i]}' inputfile

# 6  
Old 06-29-2012
try below
Code:
for var in `awk -F " " '{print $1}' test_file_data | uniq`
do
  grep $var test_file_data | sort | tail -1
done

hope this would work out as simple..
Last edited by Franklin52; 06-29-2012 at 04:10 AM.. Reason: Please use code tags for data and code samples
# 7  
Old 06-29-2012
Code:
$ sort -nr infile | nawk '!x[$1]++'
b v3.0
a v1.2
$
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