Map an array modifying only elements matching a certain condition

Question

In Ruby, what is the most expressive way to map an array in such a way that certain elements are modified and the others left untouched?

This is a straight-forward way to do it:

old_a = ["a", "b", "c"]                         # ["a", "b", "c"]
new_a = old_a.map { |x| (x=="b" ? x+"!" : x) }  # ["a", "b!", "c"]

Omitting the "leave-alone" case of course if not enough:

new_a = old_a.map { |x| x+"!" if x=="b" }       # [nil, "b!", nil]

What I would like is something like this:

new_a = old_a.map_modifying_only_elements_where (Proc.new {|x| x == "b"}) 
        do |y|
          y + "!"
        end
# ["a", "b!", "c"]

Is there some nice way to do this in Ruby (or maybe Rails has some kind of convenience method that I haven't found yet)?

---

Thanks everybody for replying. While you collectively convinced me that it's best to just use map with the ternary operator, some of you posted very interesting answers!

Answer 1

Because arrays are pointers, this also works:

a = ["hello", "to", "you", "dude"]
a.select {|i| i.length <= 3 }.each {|i| i << "!" }

puts a.inspect
# => ["hello", "to!", "you!", "dude"]

In the loop, make sure you use a method that alters the object rather than creating a new object. E.g. upcase! compared to upcase.

The exact procedure depends on what exactly you are trying to achieve. It's hard to nail a definite answer with foo-bar examples.

Answer 2

old_a.map! { |a| a == "b" ? a + "!" : a }

gives

=> ["a", "b!", "c"]

map! modifies the receiver in place, so old_a is now that returned array.