Assign to an array and replace emerged nil values
Greetings!
When assigning a value to an array as in the following, how could I replace the nil
s by 0
?
array = [1,2,3]
array[10] = 2
array # => [1, 2, 3, nil, nil, nil, nil, nil, nil, nil, 2]
If not possible when assigning, how would I do it the best way afterwards? I thought of array.map { |e| e.nil? ? 0 : e }
, but w开发者_C百科ell…
Thanks!
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]
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}
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)
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}
a.select { |i| i }
This answer is too short so I am adding a few more words.
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]
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