I want to create a fixed size array with a default number of elements already filled from another array, so lets say that I have this method:
def fixed_array(size, other)
array = Array.new(size)
other.each_with_index { |x, i| array[i] = x }
array
end
So then I can use the method like:
fixed_array(5, [1, 2, 3])
And I will get
[1, 2, 3, nil, nil]
Is there an easier way to do that in ruby? Like expanding the current size of the array I already have with nil objects?
def fixed_array(size, other)
Array.new(size) { |i| other[i] }
end
fixed_array(5, [1, 2, 3])
# => [1, 2, 3, nil, nil]
5.times.collect{|i| other[i]}
=> [1, 2, 3, nil, nil]
Is there an easier way to do that in ruby? Like expanding the current size of the array I already have with nil objects?
Yes, you can expand your current array by setting the last element via Array#[]=:
a = [1, 2, 3]
a[4] = nil # index is zero based
a
# => [1, 2, 3, nil, nil]
A method could look like this:
def grow(ary, size)
ary[size-1] = nil if ary.size < size
ary
end
Note that this will modify the passed array.
a = [1, 2, 3]
b = a.dup
Array.new(5){b.shift} # => [1, 2, 3, nil, nil]
Or
a = [1, 2, 3]
b = Array.new(5)
b[0...a.length] = a
b # => [1, 2, 3, nil, nil]
Or
Array.new(5).zip([1, 2, 3]).map(&:last) # => [1, 2, 3, nil, nil]
Or
Array.new(5).zip([1, 2, 3]).transpose.last # => [1, 2, 3, nil, nil]
You can also do the following:
(assuming other = [1,2,3])
(other+[nil]*5).first(5)
=> [1, 2, 3, nil, nil]
if other is [], you get
(other+[nil]*5).first(5)
=> [nil, nil, nil, nil]
this answer uses the fill method
def fixed_array(size, other, default_element=nil)
_other = other
_other.fill(default_element, other.size..size-1)
end
How about this? So you don't create a new array.
def fixed_array(size, other)
(size - other.size).times { other << nil }
end
