Replace one digit by two digit using sed


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# 1  
Replace one digit by two digit using sed

Folks,

Is there a simple way to replace one digit by two digit using sed.

Example,
mydigit1918_2006_8_8_lag1.csv should be
mydigit1918_2006_08_08_lag01.csv.

I tried this way, but doesn't work.
echo mydigit1989_2006_8_8_lag1.csv|sed 's/[[:digit:]]/0[[:digit:]]/'

Thank you,
# 2  
Apparently your sed doesn't understand the [[:digit:]] class, but if it did, it would replace the first digit -- the 1 in 1989 -- and replace it with a literal string "[[:digit:]]". You need to supply more context; and, apparently, use the old-fashioned [0-9] construct for matching a digit.

Code:
echo mydigit1989_2006_8_8_lag1.csv | sed 's/_\([0-9]\)/_0\1/g'

The regular expression matches a single digit after an underscore, and captures it into a backreference so you can refer to it as "\1" in the substitution part. (That's what the backslashed parentheses are for. Your sed might want them without backslashes, or something.) The /g at the end says to do this as many times as possible ("globally"), not just on the first occurrence.
# 3  
Code:
echo mydigit1918_2006_8_8_lag1.csv | sed -e 's/\([^0-9]\)\([0-9]\)\([^0-9]\)/\10\2\3/' -e 's/\([^0-9]\)\([0-9]\)\([^0-9]\)/\10\2\3/g'

# 4  
Looks simple, but..

Quote:
Originally Posted by era
Apparently your sed doesn't understand the [[:digit:]] class, but if it did, it would replace the first digit -- the 1 in 1989 -- and replace it with a literal string "[[:digit:]]". You need to supply more context; and, apparently, use the old-fashioned [0-9] construct for matching a digit.

Code:
echo mydigit1989_2006_8_8_lag1.csv | sed 's/_\([0-9]\)/_0\1/g'

The regular expression matches a single digit after an underscore, and captures it into a backreference so you can refer to it as "\1" in the substitution part. (That's what the backslashed parentheses are for. Your sed might want them without backslashes, or something.) The /g at the end says to do this as many times as possible ("globally"), not just on the first occurrence.
Thanks Era,

Your approach looks simple, but not quite right. Shamrock's method below looks long, but it gives me a right answer.
# 5  
Ah yes, it gets the 2006 too. It's possible to fix that with a different regex but you already have something which works for you.
# 6  
regexp can be powerful, but can get confusing for the faint hearted
Code:
# echo "mydigit1918_2006_08_08_lag01.csv" | awk -F"_" '{$3+=0;$4+=0}1' OFS="_"
mydigit1918_2006_8_8_lag01.csv

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