Ordering a string by the count of substrings?
How about
$ perl -F'-' -lpe '$_ = join "-", sort { length $a <=> length $b } @F' file
22-212-1234-11153782-0114232192380
14-234-6756-09867378-8807698823332
14-221-45323-43676256-238372635363
23-234-9983-62736373-8863345637388
and
$ perl -F'-' -lpe '$_ = join "-", sort { $a <=> $b } map length, @F' file
2-3-4-8-13
2-3-4-8-13
2-3-5-8-12
2-3-4-8-13
Thanks to Stéphane Chazelas for suggested improvements
GNU awk can sort, so the trickiest part is deciding how to separate the two desired outputs; this script generates both results, and you can decide if you'd like them somewhere other than hard-coded output files:
function compare_length(i1, v1, i2, v2) {
return (length(v1) - length(v2));
}
BEGIN {
PROCINFO["sorted_in"]="compare_length"
FS="-"
}
{
split($0, elements);
asort(elements, sorted_elements, "compare_length");
reordered="";
lengths="";
for (element in sorted_elements) {
reordered=(reordered == "" ? "" : reordered FS) sorted_elements[element];
lengths=(lengths == "" ? "" : lengths FS) length(sorted_elements[element]);
}
print reordered > "reordered.out";
print lengths > "lengths.out";
}
How far would this get you:
awk -F- ' # set "-" as the field separator
{
for (i=1; i<=NF; i++){
L = length($i) # for every single field, calc its length
T[L] = $i # and populate the T array with length as index
if (L>MX){ MX = L } # keep max length
}
$0 = "" # empty line
for (i=1; i<=MX; i++){
if (T[i]){
$0 = $0 OFS T[i] # append each non-zero T element to the line, separated by "-"
C = C OFS i # keep the field lengths in separate variable C
}
}
print substr ($0, 2) "\t" substr (C, 2) # print the line and the field lengths, eliminating each first char
C = MX = "" # reset working variables
split ("", T) # delete T array
}
' OFS=- file
22-212-1234-11153782-0114232192380 2-3-4-8-13
14-234-6756-09867378-8807698823332 2-3-4-8-13
14-221-45323-43676256-238372635363 2-3-5-8-12
23-234-9983-62736373-8863345637388 2-3-4-8-13
You may want to split the printout into two result files.