I have an array of integers, and I am attempting to multiply every second element by 2 moving backwards in the array. However, it does not seem to accept the * operator on the array item, results in the error message: undefined method `*' for nil:NilClass (NoMethodError)
index = array.length
while index > 0
array[index] * 2
index - 2
end
I have checked the class of the array items and it is indeed an integer, so I am unsure why this does not work.
Thanks.
If you fix the indexing as others have suggested, and change the third and fourth lines to actually modify the array and index variables, then your code works as intended for any array with an even number of elements.
index = array.length - 1
while index > 0
array[index] *= 2
index -= 2
end
# array now =='s [4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]
Alternatively, you could perform the operation in one line using map.with_index by taking advantage of the fact that every second element of an array will have an index that is an odd number.
array.map.with_index { |n, i| i.odd? ? n * 2 : n }
# Or if you prefer,
array.map.with_index { |n, i| if i.odd? then n * 2 else n end }
#=> [4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]
You can iterate reversed array with indexes, and then reverse result Array#reverse
i % 3 is used to multiply each 3rd element.
Error undefined method *' for nil:NilClass (NoMethodError) in your example appears because array.length is 16(elements in array), but indexes starts from 0, so last index is 15, and in the first iteration you are trying to access array[16] that does not exist, so it is nil
array = [4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2]
result = array.reverse.each_with_index.map do |x, i|
(i % 3).zero? ? x * 2 : x
end.reverse
# p result
#=> [8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4]
UPDATE: Sekalf Nroc's answer is good with odd? if you need each 2nd element to multiply, but if you had for example 15 elements, you could see unexpected results, so reverse still needed to walk array backwards, by combining those approaches, you can do something like this:
array = [4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 99]
p array.map.with_index { |n, i| i.odd? ? n * 2 : n } # WRONG RESULT
#=>[4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 99]
p array.reverse.map.with_index { |n, i| i.even? ? n * 2 : n }.reverse
#=>[8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 198]
Another approach, that uses Numeric#step. It works good with all the examples even with another step(array3)
array1 = [4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2]
array2 = [2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2]
array3 = [2, 4, 4, 2, 4, 4, 2, 4, 4, 2, 4, 4, 2, 4, 4, 2]
(array1.size - 1).step(0, -2).each { |i| array1[i] *= 2 }
(array2.size - 1).step(0, -2).each { |i| array2[i] *= 2 }
(array3.size - 1).step(0, -3).each { |i| array3[i] *= 2 }
p array1, array2, array3
#=>[4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]
#=>[4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]
#=>[4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]
You could use a Range, which provides the Range#step method. For example, given the array:
ary = [4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2]
This is a way for "multiply every second element by 2 moving backwards in the array" using Array#reverse_each:
(1...ary.size).step(2).each { |i| ary.size.odd? ? ary[i] *=2 : ary[i-1] *=2 }
#=> [4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]
Note that the first part (1...ary.size).step(2).to_a returns the odd indexes:
#=> [1, 3, 5, 7, 9, 11, 13, 15]
Also note the three points ... in range.
For ary = [2, 4, 2] #=> [2, 8, 2]
For ary = [2, 4, 2, 4] #=> [4, 4, 4, 4]
[2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2]
array[index] * 2andindex - 2expressions, you need to assign them to something:array[index] *= 2,array[index] = array[index] * 2,index = index - 2,index -= 2, ...array.length - 1(remember, arrays are 0-indexed). Also you are not doing anything witharray[index] * 2(either print it, or assign to a variable)[4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2], what did you expect the output to be? I had interpreted the phrase "every second element" to mean the result should be[4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4]. Is that right?