How to parse an input according to column's variables in bash

Question

I need help to parse an input like this.

192.168.0.168: 1
192.168.0.158: 0
192.168.0.198: 0
192.168.0.148: 0
192.168.0.158: 1
192.168.0.168: 0

If there is 1 in second column of any ip, I want to delete the row which have 0 in second column and same ip in first column. So my output should be like this.

192.168.0.168: 1
192.168.0.198: 0
192.168.0.148: 0
192.168.0.158: 1

I guess It can be done by using awk, sed etc. but I have no idea how to do that. I hope I could explain my question correctly. Thanks...

Birei · Accepted Answer · 2012-08-29 07:30:13Z

2

One way:

awk '
    { 
        ips[ $1 ] = ( ips[ $1 ] == 1 ) ? 1 : $2 
    } 
    END { 
        for ( ip in ips ) { 
            print ip, ips[ ip ] 
        } 
    }
' infile

That yields (output could be unordered):

192.168.0.168: 1
192.168.0.198: 0
192.168.0.148: 0
192.168.0.158: 1

answered Aug 29, 2012 at 7:30

Birei

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Comments

potong · Accepted Answer · 2012-08-29 07:30:51Z

2

This might work for you (GNU sed):

cat -n file | 
sort -k2,2 -k3,3nr | 
sed ':a;$!N;/^\s*\S*\s*\(\S*\)\s*1\s*\n.*\1/s/\n.*0\s*//;ta;P;D' | 
sort -n | 
sed 's/^\s*\S*\s*//'

answered Aug 29, 2012 at 7:30

potong

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Comments

choroba · Accepted Answer · 2012-08-29 07:38:01Z

1

Perl solution:

perl -nae '$h{ $F[0] } += $F[1]
           }{
           print "$k ", $v ? 1 : 0, "\n" while ($k, $v) = each %h'

answered Aug 29, 2012 at 7:38

choroba

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Costi Ciudatu · Accepted Answer · 2012-08-29 07:52:50Z

1

A couple of sorts should do:

sort file -r | sort -u -k1,1

The former sort makes sure the lines are ordered so that lines with 1 on the second column will come first for every IP.

The latter sort will keep only the first entry for each IP: -u -> unique, -k1,1 -> first column only.

answered Aug 29, 2012 at 7:52

Costi Ciudatu

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1 Comment

Thor Over a year ago

To preserve line order you can use same trick that potong's answer uses. First number the lines and with cat -n, and post sort on first column: cat -n infile | sort -k2 -r | sort -u -k2,2 | sort -n | cut -f2-.

Vijay · Accepted Answer · 2012-08-29 07:54:57Z

1

awk '{a[$1]+=$2}END{for(i in a)print i,a[i]}' your_file

answered Aug 29, 2012 at 7:54

Vijay

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Zoltan Toth · Accepted Answer · 2012-08-29 10:07:53Z

1

Functional approach (haskell programming language):

-- function that having the two sublists with '0' and '1' ips,
-- filters and puts into   the '1' 
-- sublist all the '0' ips that are not included in '1'

fil [] result = result
fil (x: xs) result | (init x `elem` (map init result)) == False = fil xs (x:result)
            | otherwise = fil xs result


-- function that filters '0' and '1' sublists
getsublist alist character = filter (\x-> (last x) == character) alist



> let a = ["192.168.0.168: 1", "192.168.0.158: 0", "192.168.0.198: 0", "192.168.0.148: 0", "192.168.0.158: 1", "192.168.0.168: 0"]

> let b = getsublist a '0'

> let c = getsublist a '1'

> fil b c

Output:

["192.168.0.148: 0","192.168.0.198: 0","192.168.0.168: 1","192.168.0.158: 1"]

edited Aug 29, 2012 at 10:07

Zoltan Toth

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answered Aug 29, 2012 at 9:06

Igor Mandric

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