Finding max of a column grouping by the time

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# 1  
Old 03-17-2014
Finding max of a column grouping by the time

I have the below text:
16:00 0.50
16:00 0.30
16:00 0.00
16:00 0.00
16:00 0.30
16:01 0.00
16:01 0.30

I want to find the max of the 2nd column grouping by the values in the 1st column using awk. So
16:00 0.50
16:01 0.30

I have tried
 awk -F" " '{ if(!($1 in max) || (max[$1]<$3)) max[$1]=$3 ; count[$1]++ } END { for(element in max) printf("%s %.2f\n", element, max[element] ) } ' file.txt

but all i get is
16:00 0.00
15:59 0.00
16:01 0.00

 cat file.txt | awk -F " " 'BEGIN{if(a[$1]<$2) { a[$1]=$3;}}END {for(i in a){print i " " a[i];}}'

does not work...

Any help much appreciated.
# 2  
Old 03-17-2014
A BEGIN rule is invoked even before the first record in a file is read. So never put any code over there which deals with input file records:
awk '
                if ( A[$1] < $2 )
                        A[$1] = $2
        END {
                for ( k in A )
                        print k, A[k]
' file

This User Gave Thanks to Yoda For This Post:
# 3  
Old 03-17-2014

Following is one more approach for same.

awk 'NR==FNR{a[$1]=$2;} ($1 in a){if(a[$1]<$2) {a[$1]=$2}} END{for(j in a){print j OFS a[j]}}' check_2nd_column_max_value12111 check_2nd_column_max_value12111

Output will be as follows.

16:00 0.50
16:01 0.30

NOTE: Where check_2nd_column_max_value12111 is the input file name.

R. Singh
# 4  
Old 03-18-2014
awk '{if($2 > a[$1]) {a[$1] = $2}} END {for(x in a) {print x, a[x]}}' file

sorry, I didn't see Yoda's solution...I have provided the same solution
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