group by 'last' value in bash

Question

I have a two-column file:

1,112
1,123
2,123
2,124
2,144
3,158
4,123
4,158
5,123

I need to know last column2 value for each column1:

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Answer 1

Couple of solutions:

1) With tac to reverse input file and sort

$ tac ip.txt | sort -u -t, -k1,1n
1,123
2,144
3,158
4,158
5,123

2) With perl

$ perl -F, -ne '$h{$F[0]} = $_; END{print $h{$_} foreach (sort {$a <=> $b} keys %h)}' ip.txt 
1,123
2,144
3,158
4,158
5,123

Input lines split on , and hash variable keeps updating based on first field, effectively throwing away previous lines if first field matches. At end, the hash variable is printed based on sorted keys

Thanks @choroba for pointing out that numeric sort is needed in both cases

Answer 2

This is pretty similar to @Sundeep's solution but here it goes:

$ tac file|uniq -w 1|tac
1,123
2,144
3,158
4,158
5,123

ie. reverse the record order with cat, uniq outputs based on the first character only and then the order is reversed again.

Answer 3

With GNU bash:

declare -A array   # associative array

# read from file
while IFS=, read a b; do array[$a]="$b"; done < file

# print array
for i in "${!array[@]}"; do echo "$i,${array[$i]}"; done

Output:

1,123
2,144
3,158
4,158
5,123

Answer 4

You can use awk delimit on , to store each $2 in an array using key as $1:

awk 'BEGIN{FS=OFS=","} {seen[$1]=$2} END{for (i in seen) print i, seen[i]}' file.csv

1,123
2,144
3,158
4,158
5,123