Populate hash with array keys and default value

Question

Stuck on a Code Wars Challenge: Complete the solution so that it takes an array of keys and a default value and returns a hash with all keys set to the default value.

My answer results in a parse error:

def solution([:keys, :default_value])
  return { :keys => " ", :default_value => " " }
end

Am I missing something to do with returning a hash key with all the keys set to the default value?

While I prefer Arup's solution, the "initialize a collection-add to it-return it" pattern is handled by Enumerable#inject in Ruby: keys.inject({}) { |h, k| h.merge({k => default_val}) } or keys.inject({}) { |h, k| h[k] = default_val ; h }. — Michael Kohl
– Michael Kohl, Commented Jan 12, 2014 at 16:24
@Michael, to get rid of that unsightly h at the end of the block: keys.each_with_object({}) { |k, h| h[k] = default_val }. — Cary Swoveland
– Cary Swoveland, Commented Jan 12, 2014 at 19:25
To replace it with an unsightly method name? I'd rather not. — Michael Kohl
– Michael Kohl, Commented Jan 13, 2014 at 0:54

Arup Rakshit · Accepted Answer · 2014-01-12 11:57:32Z

15

Do as below :

def solution(keys,default_val)
  Hash[keys.product([default_val])]
end

solution([:key1,:key2],12)  # => {:key1=>12, :key2=>12}

Read Array#product and Kernel#Hash.

edited Jan 12, 2014 at 11:57

answered Jan 12, 2014 at 11:41

Arup Rakshit

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7 Comments

HandDisco Over a year ago

Thanks. So using Hash before an array turns that array into a Hash? Can you give me a real world example of where the product method could be used - i mean, why would the same value be assigned to multiple keys? Is there another method i could have used other than product?

Arup Rakshit Over a year ago

@Abraham yes... then do Hash[keys.zip([default_val]* keys.size)].. But in this case it is not idiomatic.. so I use Array#product.

Darek Nędza Over a year ago

@Abraham For example: you have hash that contain number of letters of in a sentence ( {:a => 0, :b => 0...} ) and you count each letter, using each_char for example, you can do this hash1[:a] += 1. Without default value you have to check if hash contain key, if yes you can hash[:a]+=1, if not you must assign it via hash[:a] = 1. ps. []#* is interesting - it can take string as an argument while []#product will raise TypeError. [1,2,3].* '3' #=> "13233". You forgot to change 3 to Numeric, and you have string instead of array.

Denis de Bernardy Over a year ago

This forgets to dup the default value when possible, though.

Arup Rakshit Over a year ago

@Denis Is your advice for me ?

|

Denis de Bernardy · Accepted Answer · 2014-01-12 13:51:00Z

1

I'd advise amending your solution to this:

def solution(keys, default_value)
  hash = {}
  keys.each do |key|
    value = default_value.dup rescue default_value
    hash[key] = value
  end
  hash
end

The dup is to work around the nasty case where default_value is a string and you then do e.g.:

hash[:foo] << 'bar'

… with your version, this would modify multiple values in place instead of a single one.

answered Jan 12, 2014 at 13:51

Denis de Bernardy

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2 Comments

Cary Swoveland Over a year ago

Denis, I am sure you are aware that you could have written keys.each_with_object({}) do |key, hash| to get rid of hash = {} and hash at the end, but chose to do it this way. I'd be interested in your views on the stylistic difference.

Denis de Bernardy Over a year ago

I was merely fixing OP's function. Code I'd write myself would be more like Hash[keys.map{ |k| [k, val] }.

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