Complex mapping of array to object in Ruby

Question

I have an array of strings:

["username", "String", "password", "String"]

And I want to convert this array to a list of Field objects:

class Field
    attr_reader :name, :type
    def initialize(name, type)
        @name = name
        @type = type
    end
end

So I need to map "username", "String" => Field.new("username", "String") and so on. The length of the array will always be a multiple of two.

Does anyone know if this is possible using a map style method call?

Magnar · Accepted Answer · 2009-08-17 12:43:52Z

4

A bracketed Hash call does exactly what you need. Given

a = ["username", "String", "password", "String"]

Then:

fields = Hash[*a].map { |name, type| Field.new name, type }

edited Aug 17, 2009 at 12:43

answered Aug 17, 2009 at 12:32

Magnar

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2 Comments

BefittingTheorem Over a year ago

Thanks Magnar, an interesting solution that I didn't know existed. I'm going to go with sepp2k's answer as it is more generic. As far as I can see this will only handle the case where there is two object mapping to one, or am I wrong? Could you map say four strings to one object using this method?

Magnar Over a year ago

The Hash-bracket constructor creates a hash of key=>value pairs out of every other value in the array. So it's perfect for your example, but no, it won't work for larger constructors.

sepp2k · Accepted Answer · 2009-08-17 12:58:02Z

2

1.8.6:

require 'enumerator'
result = []
arr = ["username", "String", "password", "String"]
arr.each_slice(2) {|name, type| result << Field.new(name, type) }

Or Magnar's solution which is a bit shorter.

For 1.8.7+ you can do:

arr.each_slice(2).map {|name, type| Field.new(name, type) }

answered Aug 17, 2009 at 12:58

sepp2k

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1 Comment

BefittingTheorem Over a year ago

Excellent that's what I'm looking for. I'm choosing this as the answer as it is more scalable that the Hash method for mapping multiple items to an object.

Matt Grande · Accepted Answer · 2009-08-17 12:27:07Z

2

Take a look at each_slice. It should do what you need.

answered Aug 17, 2009 at 12:27

Matt Grande

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4 Comments

Dave Ray Over a year ago

1.9 only, FWIW. Too bad there isn't a map_slice, which is what he's really after.

Dave Ray Over a year ago

Maybe 1.8.7 too, I'm not sure.

sepp2k Over a year ago

each_slice is in 1.8.6 too, if you require 'enumerator' (stdlib)

Matt Grande Over a year ago

It's 1.8.6. Definitely not 1.9, the docs I linked are for 1.8.7

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