BASH: Using cut on space delimited file: Treating two spaces as one

I need to convert file full of lines like this:

# 2007  4 29 10  1 17.98 blah  other   stuff

into lines formatted like this


The original line is space delimited, and when a one's place digit number appears (such as 4) it gets listed as ' 4'. When I convert it, I need to be able to change it to '04'. Thus there are spaces that delimit the file, AND spaces that are placeholders for leading zeros.

I need to write a shell script to make that conversion. I tried using the cut command because each character stays in the same exact place, so the 7th char is always a delimiting space and the 8th char is always the ten's digit, or a space that should be a leading zero. However I soon discovered that it treats two spaces as one, which totally throws off the count (Since sometimes I have ' 4' and sometimes I will have '14'.

So: I need a way to read and convert this file, either using cut, or some other method (awk?) that will allow me to do this. Either a way to modify my current code (below) or another approach that would work a lot better would be much appreciated.

Just for reference, my present code is below:

while read LINE
    #IF line starts with '#', then
    if [[ $LINE == "#"* ]]; then

        # 2008 12 26 11 26 20.36
        # 2007  5 10  1  8 10.52

        #GET 4 digit year
        LINEyear=$(echo $LINE | cut -c3-6)

        #GET 2 digit month
        if [ $(echo $LINE | cut -c8-8) == " " ]; then
            LINEmonth=0$(echo $LINE | cut -c8-9)                
            LINEmonth=$(echo $LINE | cut -c8-9)

        #GET 2 digit day
        if [ $(echo $LINE | cut -c11-11) == " " ]; then
            LINEday=0$(echo $LINE | cut -c11-12)
            LINEday=$(echo $LINE | cut -c11-12)

        #GET hour, min, sec, (Removed to save space)

        echo $LINEnew



You can solve this in just one line of awk:

% awk '/^#/ {printf "%04d.%02d.%02d.%02d.%02d.%02d\n", $2, $3, $4, $5, $6, $7}' ~/stuff 



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