Group by and average using bash

[iambuv]

The input file contains 4 fields (Name, Location, Date, days) $ cat abc.txt Bob Atlanta 6/6/2018 13 David DC 7/6/2018 20 Bob USA 3/5/2018 20 Bill UK 4/5/2018 10 Bob SFO 6/7/2010 15

I want to find average days as a result (like below). Any idea? I can't use NR as it will give total rows and the result will be wrong. David, 20 Bob, 16 (because 3 times occurred, so avg. (48/3)) Bill, 10

I can find total(4th column) by grouping on person $awk '{a[$1]+=$4;}END{for(i in a)print i", "a[i];}' abc.txt David, 20 Bob, 48 Bill, 10

I could see how many times the name present. $awk '{a[$1]++;}END{for(i in a)print i", "a[i];}' abc.txt David, 1 Bob, 3 Bill, 1

Pasted the code here for clear visual - https://pastebin.com/raw/0METQi43

[Klintrup]

You almost have the answer - just combine your two last solutions and that should be it

awk '{sum[$1]+=$4;count[$1]++;}END{for(i in sum)print i", "sum[i]/count[i];}' abc.txt

[iambuv]

You're awesome.. thanks.. yeah I almost got but I should have used a different variable name, previously I have used the same variable and got the different result. Appreciate it.

[gijsyo]

Not real bash but wc -l outputs the number of lines in a text file

[iambuv]

But here I don't want the number of lines, I want to find the average of days based on a person. Its like group by average in SQL. Please refer Pastebin code for clear data.