Finding max value of a specific date awk

Question

I have a file with several rows and with each row containing the following data-

name 20150801|1 20150802|4  20150803|6  20150804|7  20150805|7  20150806|8  2015         


        

Answer 1

As a quick&dirty solution, we can perform this in following Unix commands:

yourdatafile=
yourdate=

cat $yourdatafile | sed 's/|/_/g' | awk -F "${yourdate}_" '{print $1" "$2}' | sed 's/[0-9]*_[0-9]*//g' | awk '{print $1" "$2}' |sort -k 2n | tail -n 1

With following sample data:

$ cat $yourdatafile
Alice 20150801|44 20150802|21  20150803|7  20150804|76  20150805|71
Bob 20150801|31 20150802|5 20150803|21 20150804|133 20150805|71

and yourdate=20150803 we get:

$ cat $yourdatafile | sed 's/|/_/g' | awk -F "${yourdate}_" '{print $1" "$2}' | sed 's/[0-9]*_[0-9]*//g' | awk '{print $1" "$2}' |sort -k 2n | tail -n 1
Bob 21

and for yourdate=20150802 we get:

$ cat $yourdatafile | sed 's/|/_/g' | awk -F "${yourdate}_" '{print $2" "$1}' | sed 's/[0-9]*_[0-9]*//g' | awk '{print $2" "$1}' | sort -k 2n | tail -n 1
Alice 21

The drawback is that only one line is printed the highest value of a day was achieved by more than one name as can be seen with:

$ yourdate=20150805; cat $yourdatafile | sed 's/|/_/g' | awk -F "${yourdate}_" '{print $2" "$1}' | sed 's/[0-9]*_[0-9]*//g' | awk '{print $2" "$1}' | sort -k 2n | tail -n 1
Bob 71

I hope that helps anyway.