CODEWITHSUNDEEP
bashawk
Ash
Ash

Reputation: 355

pick unique lowest row in unix

I have a 2 fields in below file.

Field 1 is name and field 2 is transaction date.

I want unique name with lowest transaction date

cat abc.lst
John_20130201
David_20130202
Scott_20130203
Li_20130201
John_20130202
Scott_20130201
David_20130201
Li_20130204
Torres_20121231

output desired

John_20130201
Li_20130201
Scott_20130201
David_20130201
Torres_20121231

Upvotes: 1

Views: 113

Answers (5)

TrueY
TrueY

Reputation: 7610

A pure bash solution (not ordered by name):

declare -A hash
while IFS=_ read nam val; do
  [[ -z "${hash[$nam]}" || $val < "${hash[$nam]}" ]] && hash[$nam]=$val
done <abc.lst

for i in ${!hash[*]}; do echo ${i}_${hash[$i]}; done

Output:

David_20130201
Li_20130201
John_20130201
Scott_20130201
Torres_20121231

Upvotes: 0

jaypal singh
jaypal singh

Reputation: 77105

sort -t_ -nk2 abc.lst | awk -F_ '!a[$1]++'

or save a pipe and do this - 

awk -F_ '!a[$1]++' <(sort -t_ -nk2 abc.lst)

Upvotes: 4

Arkku
Arkku

Reputation: 42139

awk -F_ '{ l = lowest[$1]; if (!l || $2 < l) { lowest[$1] = $2 } }
         END { for (name in lowest) print name FS lowest[name] }'

(if output order doesn't matter)

Upvotes: 0

Kent
Kent

Reputation: 195079

this one-liner selects those entries. but doesn't sort the output:

awk -F_ '$1 in a{a[$1]=$2>a[$1]?a[$1]:$2;next}{a[$1]=$2}
END{for(x in a)print x"_"a[x]}' file

Upvotes: 0

ramonovski
ramonovski

Reputation: 414

Use the sort command:

cat abc.lst | sort -u -t_ -k1,1

And read about it on man sort

Upvotes: -2

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