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How to select the largest number from column 2 according to the value of column 1 bash script -sed-awk

I have a file like this:

1 12:00

1 12:34

1 01:01

1 08:06

2 09:56

2 06:52
...

and I want to select from each value of the first column the largest value of the second.

New file:

1 12:34

2 09:56 
...

How can i do this? Thanks in advance!

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4 answers
awk '
  { if ($2>values[$1]) values[$1]=$2; }
  END {
    for (i in values) {
      print i,values[i]
    }
  }
' file
+4
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This might work for you:

sort -k1,1n -k2,2r file | sort -uk1,1
+2
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perl -nale '$F[1]=~s/://;$h{$F[0]}=$F[1]if($F[1]>$h{$F[0]}); 
     END{for(sort keys(%h)){($x=$h{$_})=~s/^(..)(..)$/\1:\2/;print"$_ $x"}}' file

Look it up

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Bash

# build arrays with 1.column value in its name
# use all digits in the row as index, the row as value
while read a b ; do
  eval "array$a[$a${b//:/}]=\"$a $b\""
done < "$infile"

# select the last element of each array
for name in ${!array*}; do
  last='${'${name}'[-1]}'
  eval "echo ${last}"
done
0
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