Question

If I call the function anagrams below in irb, I get a non-empty hash container as expected. But if you comment out the print "No Key\n" line, the returned hash container is now empty. In fact for all elements in list the code in the elsif branch seems to execute. Either I’m going nuts or there is a nasty bug here:

def anagrams(list = ['cars', 'for', 'potatoes', 'racs', 'four','scar', 'creams', 'scream'])
        aHash = Hash.new()
        list.each { |el|
            aKey = el.downcase.chars.sort.to_a.hash
            if aHash.key?(aKey)
                # print "Has Key\n"
                aHash[aKey] << el
            elsif
                # print "No Key\n"
                aHash[aKey] = [el]
            end
        }

        return aHash
end

I have the following versions of ruby and irb installed:

ruby 1.9.2p290 (2011-07-09 revision 32553) [x86_64-linux]
irb 0.9.6(09/06/30)

Answer 1

Your problem is that you’re using an elsif where you mean else. This:

elsif
    print "No Key\n"
    aHash[aKey] = [el]

is misleading formatting, it is actually interpreted more like this:

elsif(print "No Key\n")
    aHash[aKey] = [el]

but print returns nil so the logic is like this:

elsif(nil)
    aHash[aKey] = [el]

and nil is false in a boolean context so aHash[aKey] = [el] never occurs. If you remove the print then you end up with this:

elsif(aHash[aKey] = [el])

and the assignment occurs; the assignment is also true in a boolean context (because an Array is) but the truthiness is irrelevant in this case.

You want to use else here:

if aHash.key?(aKey)
    aHash[aKey] << el
else
    aHash[aKey] = [el]
end

Even better would be to use a Hash with an Array (via a block) as its default value:

aHash = Hash.new { |h, k| h[k] = [ ] }

and then you don’t need the if at all, you can just do this:

list.each do |el|
    aKey = el.downcase.chars.sort.to_a.hash
    aHash[aKey] << el
end

And you can use anything as a key in a Ruby Hash so you don’t even need to .to_a.hash, you can simply use the Array itself as the key; furthermore, sort will give you an array so you don’t even need the to_a:

list.each { |el| aHash[el.downcase.chars.sort] << el }

Someone will probably complain about the return at the end of your method so I’ll do it: you don’t need the return at the end of your method, just say aHash and it will be the method’s return value:

def anagrams(list = ['cars', 'for', 'potatoes', 'racs', 'four','scar', 'creams', 'scream'])
    aHash = Hash.new { |h, k| h[k] = [ ] }
    list.each { |el| aHash[el.downcase.chars.sort] << el }
    aHash
end

You could also use each_with_object to compress it even more:

def anagrams(list = ['cars', 'for', 'potatoes', 'racs', 'four','scar', 'creams', 'scream'])
    list.each_with_object(Hash.new { |h, k| h[k] = [ ] }) do |el, h|
        h[el.downcase.chars.sort] << el
    end
end

but I’d probably do it like this to cut down on the noise:

def anagrams(list = ['cars', 'for', 'potatoes', 'racs', 'four','scar', 'creams', 'scream'])
    h = Hash.new { |h, k| h[k] = [ ] }
    list.each_with_object(h) { |el, h| h[el.downcase.chars.sort] << el }
end