unix sort on column without separator

Question

I'd like to sort a file content with a Unix script depending on a particular column :

ex : sort the following file on the 3rd column :

ax5aa 
aa3ya 
fg7ds 
pp0dd 
aa1bb

would result as

pp0dd
aa1bb
aa3ya
ax5aa
fg7ds

I have tried sort -k 3,3, but it just sort on the 3d group of word (separator=SPACE).

Is there any way to have unix sort behave the way I like, or should I use another tool?

Answer 1

$ sort --key=1.3,1.3 inputfile
pp0dd
aa1bb
aa3ya
ax5aa
fg7ds

man page of sort:

[...]

-k, --key=POS1[,POS2]

start a key at POS1 (origin 1), end it at POS2 (default end of line)

[...]

POS is F[.C][OPTS], where F is the field number and C the character position in the field; both are origin 1. If neither -t nor -b is in effect, characters in a field are counted from the beginning of the preceding whitespace. OPTS is one or more single-letter ordering options, which override global ordering options for that key. If no key is given, use the entire line as the key.

With --key=1.3,1.3, you said that there only one field (the entire line) and that you're comparing the third character position of this field.

Answer 2

use sed to create the columns before sorting

$ echo "ax5aa 
aa3ya 
fg7ds 
pp0dd 
aa1bb" | sed 's/\(.\)/\1 /g' | sort -t ' ' -k3,3 | tr -d ' '

pp0dd
aa1bb
aa3ya
ax5aa
fg7ds