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S-123-P Bash Pocket Ref.        2010    Cengage Learning    $55
E-P234  Python Pocket Ref.      2012    Cengage Learning    $45
55-MNP  Unix System Programming 2001    Sybex               $230

I need to replace the numbers after the $ not including the $ with *, so the output need to be:

S-123-P Bash Pocket Ref.        2010    Cengage Learning    $**
E-P234  Python Pocket Ref.      2012    Cengage Learning    $**
55-MNP  Unix System Programming 2001    Sybex               $***

I've been able to replace the last digit or last 2, but not every digit after the $. I've tried sed and awk gsub but nothing I try seems to work.

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  • 3
    Just an advice: if you want realy anonymize the price then it could be better to change any price to the same count of stars. ($1 -> $** as the $123 -> $**). Commented Nov 15, 2020 at 20:16

3 Answers 3

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6

You can use sed and advantage of not touching the Tabs/Spaces or fields intention:

sed -E ':a s/(\$\**)[^*]/\1*/; ta' infile

replace every ($<zero-or-more-*>)[<non-*-character>] with $<zero-or-more-*><plus-additional-*-added> (\1*; \1 is the back-reference to the first matched group in sed defining by (...)) until all <non-*-character>s replaced with *s.

A bit complex, but in case you wanted to force the changes only on the last field, you could use the command in this way:

sed -E ':a s/(\$\**)[^*]([^$]*)$/\1*\2/; ta' infile
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  • Do you have any idea why awk '{ print gensub ("$[0-9]+","$*", "g"); } does not work? The $ seems to confuse awk. Commented Nov 15, 2020 at 19:48
  • you have a awk answer also here you can take a look. your awk problem is because you are not setting the gensub substitution to any field to update the changes as well as you need double escape the dollar \\$, so gawk '{$NF=gensub("\\$[0-9]+","$*", "g", $NF); print }' infile, but as you see it does replace all digits with single * Commented Nov 15, 2020 at 20:00
  • OK, thanks. I thought I had tried double escaping but apparently not. Commented Nov 15, 2020 at 20:23
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With awk / gsub - assuming that we may replace any decimal digit in the last field, provided it begins with $ i.e. we don't need to handle things like 123$45 -> 123$**:

awk '$NF ~ /\$[0-9]+/ {gsub(/[0-9]/,"*",$NF)} 1' file
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  • to keep the fields intention you can force the awk to use filed separator as regex for example with awk -F'(\t)' or awk -F'\t|' and change the OFS to tab for sure, or if it's space it's the same -F' |' or ... Commented Nov 15, 2020 at 20:25
  • or even you can define the filed separator to $ to be able to handle the things like 123$45 you mentioned awk -F'$' '$NF ~ /[0-9]+/ {gsub(/[0-9]/,"*",$NF)} 1' OFS='$' infile Commented Nov 15, 2020 at 20:33
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Perl gives a nice way to do this if you have it available:

perl -lpe 's/\$([0-9]*)/"\$" . "*" x length($1)/e'

This takes advantage of the ability to have the replacement portion of a regex substitution be a Perl expression rather than a fixed string (the /e flag) to produce the appropriate number of asterisks on demand, instead of requiring a "repeat until done" loop.

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  • That should be perl -lpe 's/\$([0-9]*)/"\$" . "*" x length($1)/e' to handle an input like a $123 bar, where the dollar sum is not at the end. The way you posted it, it would change that to a $*******, replacing the space and bar, too. Commented Nov 17, 2020 at 15:43
  • @ilkkachu yeah, that works too. Problem spec was unclear, and no test cases covered that :) Commented Nov 18, 2020 at 1:29

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