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I want to replace a character in string when a particular condition is satisfied. So , I went through the API doc of Ruby and found the gsub , gsub! etc for similar purpose. When I implemented that in my program I didn't got any error but din't got the desired output too.

The code which I was trying is this : 

name.each_char  { |c|

if name[c] == "a"
    name.sub( name[c] , c )
    puts "matched....   "
end

So , for example , I have a string called huzefa and want to replace all the letters with its index numbers . So , what is the way to do it ? Please explain in detail by giving a simple example.

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  • what is the reason to downvote ? please explain for clearification . Commented Oct 25, 2015 at 13:15
  • 1
    may be they think its simple to be asked; or the title smells duplication or there may be other reasons. or the first downvote might have triggered/influenced the second downvote. (I didnot downvote though, rather upvoted) Commented Oct 25, 2015 at 13:32
  • @illusionist : thank you for your edit. And yes that may be the reason. Commented Oct 25, 2015 at 13:36
  • Converting each character of a string to the string representation of its index depends only on the length of the string: len = 'huzefa'.size; (0...len).map(&:to_s).join # "012345". The result is the same for every other six-character string. Commented Oct 25, 2015 at 16:26
  • Your suggestion is very proper for the case if my requirement was to convert all . But i just want to map only those char to index which satisfies my particular condition . But , still i got to know something new from your answer . Thank you. :) Commented Oct 25, 2015 at 16:43

3 Answers 3

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3

You could pass block to gsub and do whatever you want when match happend.

To do it inplace you could use gsub! method.

name = "Amanda"
new_name = name.gsub("a") do |letter|
  puts "I've met letter: " + letter
  "*"
end
# I've met letter: a
# I've met letter: a
# => "Am*nd*"

If you want to work with indexes, you could do something like this:

new_name = name.chars.map.with_index do |c, i|
  if i.odd?
    "*"
  else
    c
  end
end.join
#=> => "A*a*d*"

Here c and i are passed to the block. c is a character and i is an index.

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9 Comments

Thank you , your second suggestion worked for me. If this is not stupid to ask , how does that end.join works ? . Thanx :)
Good answer. A small suggestion: whenever the characters of a string are to be food for an enumerable, use each_char, rather than chars, to avoid the creation of an unneeded temporary array. On a different subject, you've had those eyeglasses for some time. I'm compelled to tell you that they've fallen out of style.
Suppose you want to return a string that is the same as 'cat' except with the letter "a" capitalized. If you write 'cat'.chars.map { |c| c=='a' ? 'A' : c }.join #=> "cAt", the array arr = 'cat'.chars #=> ["c", "a", "t"] is first constructed, then arr.map { |c| c=='a' ? 'A' : c }.join #=> "cAt" is executed. This works because the class Array includes Enumerable and implements Array#each. (cont...)
By contrast, if you write 'cat'.each_char.map { |c| c=='a' ? 'A' : c }.join #=> "cAt", the first step is to create an enumerator enum = 'cat'.each_char #=> #<Enumerator: "cat":each_char>, then enum.map { |c| c=='a' ? 'A' : c }.join #=> "cAt". This works because enum is an instance of the class Enumerator, which includes Enumerable and implements the method each. (We can see the elements enum will pass to Enumerable#map by converting enum to an array: enum.to_a #=> ["c", "a", "t"].) (cont...)
Think of an enumerator as a rule that acts on its receiver (here a string). Being a rule, it requires less memory than does chars, which creates a temporary array (if the string is more than a few characters anyway). On the other hand, suppose we wanted to return a string that shuffles the letters in "cat". Here we must write "cat".chars.shuffle.join #=> "atc" because shuffle is an instance method of Array. If we tried using each_char we get "cat".each_char.shuffle.join #=> undefined method 'shuffle' for #<Enumerator: "cat":each_char>.
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if name=huzefa and you want to replace 'a' with its index..

name.split(//).map.with_index{|x,y| (x=='a')? y : x}.join

to result in #> "huzef5"

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Part of string before index + new char + part of string after index:

a = "hello"

idx = 3

new_char = "z"

a[0..(idx-1)] + new_char + a[(idx+1)..-1]

# => "helzo"
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