Assign to an array and replace emerged nil values

Question

Greetings!

When assigning a value to an array as in the following, how could I replace the nils by 0?

array = [1,2,3]
array         


        

Answer 1

a.select { |i| i }

This answer is too short so I am adding a few more words.

Answer 2

To change the array after assignment:

array.map! { |x| x || 0 }

Note that this also converts false to 0.

If you want to use zeros during assignment, it's a little messy:

i = 10
a = [1, 2, 3]
a += ([0] * (i - a.size)) << 2
# => [1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 2]

Answer 3

nil.to_i is 0, if all the numbers are integers then below should work. I think It is also the shortest answer here.

array.map!(&:to_i)

Answer 4

Another approach would be to define your own function for adding a value to the array.

class Array
  def addpad(index,newval)
    concat(Array.new(index-size,0)) if index > size
    self[index] = newval
  end
end

a = [1,2,3]
a.addpad(10,2)
a => [1,2,3,0,0,0,0,0,0,0,2]

Answer 5

To change the array in place

array.map!{|x|x ?x:0}

If the array can contain false you'll need to use this instead

array.map!{|x|x.nil? ? 0:x}

Answer 6

There is no built-in function to replace nil in an array, so yes, map is the way to go. If a shorter version would make you happier, you could do:

array.map {|e| e ? e : 0}