Add leading 0 in sed substitution - sed

Question

I have input data:
foo 24
foobar 5 bar
bar foo 125
and I'd like to have output:
foo 024
foobar 005 bar
bar foo 125
So I can use this sed substitutions:
s,\([a-z ]\+\)\([0-9]\)\([a-z ]*\),\100\2\3,
s,\([a-z ]\+\)\([0-9][0-9]\)\([a-z ]*\),\10\2\3,
But, can I make one substitution, that will do the same? Something like:
if (one digit) then two leading 0
elif (two digits) then one leading 0
Regards.

Answer 1

I doubt that the "if - else" logic can be incorporated in one substitution command without saving the intermediate data (length of the match for instance). It doesn't mean you can't do it easily, though. For instance:
$ N=5
$ sed -r ":r;s/\b[0-9]{1,$(($N-1))}\b/0&/g;tr" infile
foo 00024
foobar 00005 bar
bar foo 00125
It uses recursion, adding one zero to all numbers that are shorter than $N digits in a loop that ends when no more substitutions can be made. The r label basically says: try to do substitution, then goto r if found something to substitute. See more on flow control in sed here.

Answer 2

Use two substitute commands: the first one will search for one digit and will insert two zeroes just before, and the second one will search for a number with two digits and will insert one zero just before. GNU sed is needed because I use the word boundary command to search for digits (\b).
sed -e 's/\b[0-9]\b/00&/g; s/\b[0-9]\{2\}\b/0&/g' infile
EDIT to add a test:
Content of infile:
foo 24 9
foo 645 bar 5 bar
bar foo 125
Run previous command with following output:
foo 024 009
foo 645 bar 005 bar
bar foo 125

Answer 3

Add the max number of leading zeros first, then take this number of characters from the end:
echo 55 | sed -e 's:^:0000000:' -e 's:0\+\(.\{8\}\)$:\1:'
00000055

Answer 4

You seem to have the sed options covered, here's one way with awk:
BEGIN { RS="[ \n]"; ORS=OFS="" }
/^[0-9]+$/ { $0 = sprintf("%03d", $0) }
{ print $0, RT }

Answer 5

I find the following sed approach to pad an integer number with zeroes to 5 (n) digits quite straighforward:
sed -e "s/\<\([0-9]\{1,4\}\)\>/0000\1/; s/\<0*\([0-9]\{5\}\)\>/\1/"
If there is at least one, at most 4 (n-1) digits, add 4 (n-1) zeroes in
front
If there is any number of zeroes followed by 5 (n) digits after the first transformation, keep just these last 5 (n) digits
When there happen to be more than 5 (n) digits, this approach behaves the usual way -- nothing is padded or trimmed.
Input:
0
1
12
123
1234
12345
123456
1234567
Output:
00000
00001
00012
00123
01234
12345
123456
1234567

Answer 6

This might work for you (GNU sed):
echo '1.23 12,345 1 12 123 1234 1' |
sed 's/\(^\|\s\)\([0-9]\(\s\|$\)\)/\100\2/g;s/\(^\|\s\)\([0-9][0-9]\(\s\|$\)\)/\10\2/g'
1.23 12,345 001 012 123 1234 001
or perhaps a little easier on the eye:
sed -r 's/(^|\s)([0-9](\s|$))/\100\2/g;s/(^|\s)([0-9][0-9](\s|$))/\10\2/g'