Ruby multiple string replacement

str = "Hello☺ World☹"

Expected output is: 

"Hello:) World:("

I can do this: str.gsub("☺", ":)").gsub("☹", ":(")

Is there any other way so that I can do this in a single function call?. Something like:

str.gsub(['s1', 's2'], ['r1', 'r2'])
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You could do something like this:

replacements = [ ["☺", ":)"], ["☹", ":("] ]
replacements.each {|replacement| str.gsub!(replacement[0], replacement[1])}

There may be a more efficient solution, but this at least makes the code a bit cleaner

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最佳答案

Since Ruby 1.9.2, String#gsub accepts hash as a second parameter for replacement with matched keys. You can use a regular expression to match the substring that needs to be replaced and pass hash for values to be replaced.

Like this:

'hello'.gsub(/[eo]/, 'e' => 3, 'o' => '*')    #=> "h3ll*"
'(0) 123-123.123'.gsub(/[()-,. ]/, '')    #=> "0123123123"

In Ruby 1.8.7, you would achieve the same with a block:

dict = { 'e' => 3, 'o' => '*' }
'hello'.gsub /[eo]/ do |match|
dict[match.to_s]
end #=> "h3ll*"
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Set up a mapping table:

map = {'☺' => ':)', '☹' => ':(' }

Then build a regex:

re = Regexp.new(map.keys.map { |x| Regexp.escape(x) }.join('|'))

And finally, gsub:

s = str.gsub(re, map)

If you're stuck in 1.8 land, then:

s = str.gsub(re) { |m| map[m] }

You need the Regexp.escape in there in case anything you want to replace has a special meaning within a regex. Or, thanks to steenslag, you could use:

re = Regexp.union(map.keys)

and the quoting will be take care of for you.

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Another simple way, and yet easy to read is the following:

str = '12 ene 2013'
map = {'ene' => 'jan', 'abr'=>'apr', 'dic'=>'dec'}
map.each {|k,v| str.sub!(k,v)}
puts str # '12 jan 2013'
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Late to the party but if you wanted to replace certain chars with one, you could use a regex

string_to_replace.gsub(/_|,| /, '-')

In this example, gsub is replacing underscores(_), commas (,) or ( ) with a dash (-)

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You can also use tr to replace multiple characters in a string at once,

Eg., replace "h" to "m" and "l" to "t"

"hello".tr("hl", "mt")
=> "metto"

looks simple, neat and faster (not much difference though) than gsub 

puts Benchmark.measure {"hello".tr("hl", "mt") }
0.000000   0.000000   0.000000 (  0.000007)


puts Benchmark.measure{"hello".gsub(/[hl]/, 'h' => 'm', 'l' => 't') }
0.000000   0.000000   0.000000 (  0.000021)
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Riffing on naren's answer above, I'd go with

tr = {'a' => '1', 'b' => '2', 'z' => '26'}
mystring.gsub(/[#{tr.keys}]/, tr)

So 'zebraazzeebra'.gsub(/[#{tr.keys}]/, tr) returns "26e2r112626ee2r1"


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