Unexpected keyword-else - ruby

Question

I have a method which returns finds out which one of the given numbers differs from the others in evenness and returns the index of it (+1).
def iq_test(numbers)
new_array = numbers.split(" ").collect {|n| n.to_i}
x = new_array.select(&:even?)
y = new_array.select(&:odd?)
if x.count > y.count
new_array.split.each_with_index do |value, index|
"#{index + 1}".to_i if value % 3 != 0
else y.count > x.count
"#{index + 1}".to_i if value % 2 == 0
end
end
end
For example iq_test("2 4 7 8 10") should return 3.
However, I am receiving
syntax error, unexpected keyword_else, expecting keyword_end
and I can't find out where I am not closing some code off.

Answer 1

This is going to be part code review as well as answer. Let's start with the first part:
new_array = numbers.split(" ").collect {|n| n.to_i}
x = new_array.select(&:even?)
y = new_array.select(&:odd?)
Spend time to think of good variable names. I wouldn't use "new_array" as a variable name, because it doesn't provide any clue as to what it contains. Do this kind of thing too many times, and a program becomes incomprehensible.
x and y are really evens and odds, aren't they? Those would be better variable names.
A common Ruby idiom is to use plural names for arrays, singular names for everything else.
The split by default splits on whitespace, so the (" ") is unnecessary.
Be careful with indentation.
Your selects are fine, however there is a shortcut: Enumerable's partition.
This is really my own style, but I use map when processing all values in an array, and collect only when doing something like extracting attributes from an array of objects. (In other words, I use map much more often).
note that (&:to_i) is a bit of a shortcut for {|n| n.to_i}
Rewritten considering the above, this part might look like this:
numbers = input.split.map(&:to_i)
evens, odds = numbers.partition(&:even?)
Now let's look at the rest:
if x.count > y.count
new_array.split.each_with_index do |value, index|
"#{index + 1}".to_i if value % 3 != 0
else y.count > x.count
"#{index + 1}".to_i if value % 2 == 0
end
end
And let's consider the error message you got: unexpected keyword else; expected end. This has everything you need to know to answer the question (and you'll find that most error messages do, if you think about them). It says it found an else where it expected an end. And that's exactly the problem, you need to put end before the else to close the do/end block. Also, your else part is missing your iteration logic.
Other notes:
Again, be careful with indentation. Your ends do not line up with what they're ending. Proper alignment can help catch these kinds of errors. Use of an IDE like Rubymine or a sophisticated text editor with Ruby support can help as well.
else clauses are standalone, you don't put a condition after them. Perhaps you meant elsif as Holger commented.
Using string interpolation ("#{}") converts expressions to a string. Here you're converting index + 1 to a string, and then back to an integer with .to_i which cancels it out, so to speak. Simply index + 1 will do.
Array#index can be used to determine the index of a value.
It's not clear if you want all indices in case there is more than one.
Here's a version considering the above:
if evens.count > odds.count
odds.map{|n| numbers.index(n) + 1}
elsif odds.count > evens.count
evens.map{|n| numbers.index(n) + 1}
end
If you like this kind of thing, bring your working code to http://codereview.stackexchange.com/!

Answer 2

This specific error is because you do not have a closing end for your each_with_index block. To fix this error you need:
def iq_test(numbers)
new_array = numbers.split(" ").collect {|n| n.to_i}
x = new_array.select(&:even?)
y = new_array.select(&:odd?)
if x.count > y.count
new_array.split.each_with_index do |value, index|
"#{index + 1}".to_i if value % 3 != 0
end #end of block
elsif y.count > x.count #in order for this to have a condition, it must be an elsif
"#{index + 1}".to_i if value % 2 == 0
end #end of if statement
#end - remove this extra end
end #end of function

Answer 3

Code
def odd_one_out(str)
evens, odds = str.split.partition { |s| s.to_i.even? }
case [evens.size <=> 1, odds.size <=> 1]
when [0, 1] then evens.first
when [1, 0] then odds.first
else nil
end
end
Examples
odd_one_out "0 3 4 6" #=> "3"
odd_one_out "1 5 7 8" #=> "8"
odd_one_out "0 2 4 6" #=> nil
odd_one_out "1 5 7 9" #=> nil
odd_one_out "0 2 3 5" #=> nil
odd_one_out "3" #=> nil
odd_one_out "8" #=> nil
odd_one_out "" #=> nil
Explanation
See Integer#<=>.
Suppose
str = "0 3 4 6"
then
a = str.split
#=> ["0", "3", "4", "6"]
evens, odds = a.partition { |s| s.to_i.even? }
#=> [["0", "4", "6"], ["3"]]
evens
#=> ["0", "4", "6"]
odds
#=> ["3"]
b = [evens.size <=> 1, odds.size <=> 1]
#=> [1, 0]
b == [0, 1]
#=> false
b == [1, 0]
#=> true
c = odds.first
#=> "3"
The case statement, and therefore the method, returns "3".