regular expression replace word in camelWords

Question

I want to replace a word in camelWords like: to replace "foo" with "bar" in text:

ifootest // not replace this foo
Ifootest // not replace this foo
IfooTest // << replace this foo
I foo Test // << replace this foo
I_foo_Test // << replace this foo

or to replace "Foo" with "Bar" in text:

IFootest // not replace
IFooTest // not replace
iFooTest // replace
i Foo Test //replace
I_Foo_Test // replace

the rule is: if I intput a word.

the character before the first character of the word should not be the same case of the the first character of the word.

the character after the last character of the word should not be the same case of the the last character of the word.

Then why you replace the last line of the second list of examples? — Francesco
– Francesco, Commented May 17, 2020 at 14:05

Stéphane Chazelas · Accepted Answer · 2020-05-17 16:04:43Z

1

You could do something like:

perl -pe 's/(?<![[:lower:]])foo(?![[:lower:]])/bar/g'

That is replace foo instances that are neither preceded nor followed by a lowercase letter using the negative look-behind and look-ahead operators.

That works with ASCII text only. Do work with your locale's charset, you can add a -Mopen=locale option. Or use -C to work with UTF-8 text.

That would need to be adapted for words like Foo/foO/FoO where the first or last character is an uppercase letter.

To make it work for arbitrary words, you could do something like:

WORD=FoO REPL=bar perl  -pe 's{
  (?(?=[[:lower:]])      # if following character is lowercase
      (?<![[:lower:]])|  # preceding must not be lower 
      (?<![[:upper:]])   # otherwise preceding must not be upper
  ) \Q$ENV{WORD}\E
  (?(?<=[[:lower:]])     # if preceding character is lowercase
      (?![[:lower:]])|   # following must not be lower 
      (?![[:upper:]])    # otherwise following must not be upper
  )}{$ENV{REPL}}gx'

edited May 17, 2020 at 16:04

answered May 17, 2020 at 15:37

Stéphane Chazelas

587k96 gold badges1.1k silver badges1.7k bronze badges

hello thank you, I use it every day, but in some siuation like I "Foo", I want to replace the Foo with bar in this way, it does not work.

John Chen
– John Chen

2020-06-25 08:13:02 +00:00
Commented Jun 25, 2020 at 8:13
1

@JohnChen, you mean that echo 'I "Foo"' | perl -pe 's/(?<![[:upper:]])Foo(?![[:lower:]])/bar/g' doesn't output I "bar"? I can't reproduce. It works for me. Same for the second approach with WORD=Foo

Stéphane Chazelas
– Stéphane Chazelas

2020-06-25 08:17:54 +00:00
Commented Jun 25, 2020 at 8:17

Add a comment |

bu5hman · Accepted Answer · 2020-05-17 18:21:34Z

This is probably about 1,000,000 times slower than the perl but here's an awk version just for the challenge of it. But anyway

awk -v gzin="Foo" -v gzout="Bar" '
  BEGIN {FS=gzin;
    cb=(substr(gzin,1,1)~/[a-z]/)?"[a-z]$":"[A-Z]$"
    ca=(substr(gzin,length(gzin)-1,1)~/[a-z]/)?"^[a-z]":"^[A-Z]"
  }
  {printf $1; for (f=2; f<=NF; f++) printf ("%s%s", ((($(f-1) ~ cb) || ( $(f) ~ ca ))?gzin:gzout), $f) ; 
  print ""}' file

And it even matches the comments

ifootest // not replace this foo
Ifootest // not replace this foo
IbarTest // << replace this bar
I bar Test // << replace this bar
I_bar_Test // << replace this bar

and with -v gzin="Foo" -v gzout="Bar"

IFootest // not replace
IFooTest // not replace
iBarTest // replace
i Bar Test //replace
I_Bar_Test // replace

Walkthrough

awk -v gzin="Foo" -v gzout="Bar" '

Load your match gzin and replacement gzout as variables

  BEGIN {FS=gzin;

Split on gzin

    cb=(substr(gzin,1,1)~/[a-z]/)?"[a-z]$":"[A-Z]$"

Test the case of the first character of gzin and set up a regex to match it

    ca=(substr(gzin,length(gzin)-1,1)~/[a-z]/)?"^[a-z]":"^[A-Z]"

Ditto the last character

  }
  {printf $1; for (f=2; f<=NF; f++) printf ("%s%s", ((($(f-1) ~ cb) || ( $(f) ~ ca ))?gzin:gzout), $f) ;

Iterate over the fields testing the previous and current fields and put the appropriate value between them

  print ""}' file

End each line

PS I think I hurt my brain

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