a = "foobarfoobarhmm"
I want the output as `"fooBARfoobarhmm"
ie only the first occurrence of "bar" should be replaced with "BAR".
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3 Answers
Use #sub:
a.sub('bar', "BAR")
Comments
String#sub is what you need, as Yossi already said. But I'd use a Regexp instead, since it's faster:
a = 'foobarfoobarhmm'
output = a.sub(/foo/, 'BAR')
7 Comments
Niels B.
I just did a benchmark and regex takes 50% longer than just using a string as parameter for sub.
tbuehlmann
I checked it, and the Regexp version is faster. I used MRI 1.9.2, 1.9.3, 2.0.0 and even JRuby, all of them being faster on the Regexp version. The benchmark code: gist.github.com/tbuehlmann/5574713 Wanna provide your benchmark?
Niels B.
That's not very intuitive. I would expect a regex to be slower than a specialised method.
ma11hew28
In my opinion (IMO), when you don't need a regex, the string version is more readable.
Brenden
@NielsB. RegEx engines vary greatly on their level of optimization. String parsing has a wide spectrum of efficiency from horrible to impressive. RegEx can be on both ends.
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to replace the first occurrence, just do this:
str = "Hello World"
str['Hello'] = 'Goodbye'
# the result is 'Goodbye World'
you can even use regular expressions:
str = "I have 20 dollars"
str[/\d+/] = 500.to_s
# will give 'I have 500 dollars'
2 Comments
Cary Swoveland
500.to_s is sometimes written "500". :-)
Nafaa Boutefer
I wrote it like that intentionally. to indicate that you need to assign an explicit string. Because the conversion is not automatic here. If you do this instead
str[/\d+/] = 500 you'll get an error TypeError: no implicit conversion of Fixnum into String .