Count if there is more than 1 occurrence in a 2d array in Ruby

Question

I've got a sorted array:

array = [[4, 13], [1, 12], [3, 8], [2, 8], [0, 3]]

Which shows me a position (array[n][0]) and the number of occurrences of that position (array[n][1]).

I need to test to see if more than one item in the array has the same number of occurrences as the last item.

I thought I might be able to do it with this:

array.detect {|i| i[1] == array.last[1] }.length

But it returns 2 for the above array, and seems to also return 2 for the following array:

array = [[4, 13], [1, 12], [3, 8], [2, 3], [0, 3]]

When I run it without length it always returns the first occurrence.

Is there a way to get this to count the occurrences?

EDIT:

Sorry, will ask my follow up question in a new question.

Answer 1

Try using find_all instead of detect. detect returns the first match. In your first example, that's array[3], which is another array of length 2. That's why it's returning 2 (it should always be returning 2 or nil for your arrays). find_all will return an array of the matches (instead of the first match itself), and its length will be the value you want.

Answer 2

Here's a clearer way of doing what you want, with none of the associated worries.

array.count{|i| i[1] == array.last[1]}