Ruby Hash initialization (default value nil)

Question

I've been reading the Ruby docs, and looking at some other posts on the issue, but I am still wondering about this:

#counts each number in an array once
array = [1,1,2,5,3,2,5,3,3,3]
numbers = {}
array.each { |num| numbers[num] += 1 }

=> in `block in mode': undefined method `+' for nil:NilClass (NoMethodError)

In the Hash documentation the default value for a Hash is nil, which is why I am getting this error I assume. Is there a better way to insert each key/(value += 1) into the numbers array?

Suppose you execute numbers = Hash.new(0) and then numbers[num] += 1 when num is not a hash key. Since you gave a default value of 0 when creating the hash, when Ruby finds that num is not a hash key, it creates a new hash element h[num] = 0. It then executes h[num] += 1 # => 1 and Bob's your uncle. — Cary Swoveland
– Cary Swoveland, Commented Nov 6, 2013 at 4:14
In case you are wondering, when new is given a default argument, Ruby won't create a hash element unless an assignment is performed. If h = Hash.new(0), for example, h[6] will equal nil in the if statement if h[6] == 0, resulting in the condition being false, and no hash element with key = 6 will be created (which otherwise would cause the if condition to be evaluted as true). — Cary Swoveland
– Cary Swoveland, Commented Nov 6, 2013 at 5:13

squiguy · Accepted Answer · 2013-11-06 03:40:58Z

30

Try passing a default value to your new hash as such

numbers = Hash.new(0)

answered Nov 6, 2013 at 3:40

squiguy

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Roland Mai · Accepted Answer · 2013-11-06 03:43:10Z

2

You can explicitly do it this way as well:

array.each { |num| numbers[num] = (numbers[num] || 0) + 1 }

answered Nov 6, 2013 at 3:43

Roland Mai

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zolter · Accepted Answer · 2013-11-06 03:55:01Z

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Variant with inject and Hash.new(0)

  numbers = [1,1,2,5,3,2,5,3,3,3].inject(Hash.new(0)){|numbers, number| numbers[number] +=1; numbers}

answered Nov 6, 2013 at 3:55

zolter

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the Tin Man · Accepted Answer · 2013-11-06 03:59:24Z

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Aside from using the Hash default, you could also try something with group_by:

array = [1,1,2,5,3,2,5,3,3,3]
numbers = Hash[*array.group_by { |i| i }.flat_map { |k, v| [k , v.size] }]

There's probably a better way if you play around with it some.

edited Nov 6, 2013 at 3:59

the Tin Man

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answered Nov 6, 2013 at 3:55

Some Guy

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