sort a field in ascending order and delete the first and last number [closed]

2024/12/9 20:11:16

I have a data that looks like following:

a     10,5,3,66,50
b     2,10,1,88,5,8,9
c     4,60,10,39,55,22
d     1,604,3,503,235,45,60,7
e     20,59,33,2,6,45,36,34,22

I want to sort the data in second column in ascending order

a     3,5,10,50,66
b     1,2,5,8,9,10,88
c     4,10,22,39,55,60
....
....

Then delete the smallest and the largest value from it. So like this:

a     5,10,50
b     2,5,8,9,10
c     10,22,39,55
....
....

Any help would be appreciated!

Answer

Here you go:

awk '{l=split($2,a,",");asort(a);printf "%s\t",$1;for(i=2;i<l;i++) printf "%s"(i==l-1?RS:","),a[i]}' t
a       5,10,50
b       2,5,8,9,10
c       10,22,39,55
d       3,7,45,60,235,503
e       6,20,22,33,34,36,45

PS If I remember correct, you need gnu awk due to asort

How it works:

awk '{l=split($2,a,",")                      # Split the data into array "a" and set "l" to length of arrayasort(a)                                # Sort the array "a"printf "%s\t",$1                        # Print the first columnfor(i=2;i<l;i++)                        # Run a loop from second element to second last element in array "a"printf "%s"(i==l-1?RS:","),a[i]     # Print the element separated by "," except for last element, print a new line}'  file                                # Read the file
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