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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