How can I refactor converting this array into a Hash

No, it's all Ruby; It's a great language.

Arup Rakshit Over a year ago

Can you explain exactly how this works?? I mean it looks like it sets the key as a[1] (which is cut with slice! and removes it from the array) and has the values set to whatever is left over. I looked at the documentation here: ruby-doc.org/core-2.0/Enumerable.html#method-i-group_by But I still don't understand how it works

I will go with this answer. +1 to you.

@Senjai, that's what Ruby is doing.

You have ary = [a1, a2, a3], where a1 = [1,2,3], so a1[1] == a1.slice!(1) && a1 = [1,3]

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tokland · Accepted Answer · 2013-05-31 11:22:15Z

4

Functional approach using the abstraction Enumerable#map_by from Facets:

require 'facets'
records.map_by { |name, key, price| [key, [name, price]] }
#=> {"DM1210"=>[["Yonkers", "70.00 USD"], ... }

It's a pity that Ruby does not ship map_by within the core, it's a very useful (as it's unknown) variation of Enumerable#group_by (where you choose the grouping key and the value to accumulate).

edited May 31, 2013 at 11:22

answered May 31, 2013 at 9:19

tokland

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3 Comments

sawa Over a year ago

I just checked map_by. I think it is tricky to use it, particularly when the second element of an array is falesy. For example, [[:a, true], [:b, false]].map_by{|v1, v2| [v1, v2]} gives { :a => [true], :b => [[ :b, false ]] }, which is counter-intuitive.

tokland Over a year ago

@sawa, you are right, but that's not a problem of the abstraction per se, it's a bug in the implementation: it should do a more thorough check of the returned value or drop the "act like group_by" 'feature' altogether (it makes no sense, if I want something to act like group_by... I use group_by). Of course it should return {:a=>[true], :b=>[false]}. I'll open a ticket.

Facets has some very usable code it in, similar to Rails. I'd like to see the core team do some cherry picking and pull in more of the utility methods from both. Active Support has some great tools I often pull in though I seldom work in Rails.

Benoît · Accepted Answer · 2013-05-31 06:37:36Z

3

What about

result = trans_data.each_with_object({}) do |arr, hash|
  (hash[arr[1]] ||= []) << [arr[0], arr[2]]
end

answered May 31, 2013 at 6:37

Benoît

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4 Comments

I didn't even know each_with_object existed. so that passes in an empty hash every iteration? Or is it persistent.

Benoît Over a year ago

Let's say it keeps track of the object's object_id you passed. So you don't need to repass the object each time like in inject.

Ahh, I understand. Excellent. Thanks for teaching me something new and taking your time out to help others :)

Aha! I didn't know about each_with_object either. Updating!

Narfanator · Accepted Answer · 2013-05-31 07:17:30Z

2

Note: Accepted answer is best answer, but I'm really happy with the weird awesomeness I use and how I explain it:

arr = [["Yonkers", "DM1210", "70.00 USD"], ["Yonkers", "DM1182", "19.68 AUD"], 
["Nashua", "DM1182", "58.58 AUD"], ["Scranton", "DM1210", "68.76 USD"], 
["Camden", "DM1182", "54.64 USD"]]
 arr.each_with_object({}){|(a, b, c), hash| (hash[b] || hash[b]=[]).push [a,c]}

Props to Older God for each_with_object!

Explanation: There are two wacky things going on here. The first, the (a, b, c) magic, I think it works like this:

( 

  #This bit:
  arr.collect{|(a,b,c)| "#{a}#{b}#{c}"}

) - (

  #Is equivalent to this bit:
  (0..arr.size).collect {|i|
    (a,b,c) = arr[i] #=> (a,b,c) = ["Yonkers", "DM1210", "70.00 USD"]
    "#{a}#{b}#{c}"
  }

  #as you can see, they generate identical arrays:
) == []

Note that you can treat the parens as implicit in certain circumstances: arr.collect{|a, b, c| [a, b, c]} == arr

The second wacky thing:

(hash[b] || hash[b]=[]).push(...)

Remember that everything in Ruby is both an expression and a reference.

[

 (hash[:a] || "foo") == (nil || "foo"),
 (hash[:b]=[]) == [],
 (hash[:b]=[]) === hash[:b],
 (hash[:b] || "foo") == ([] || "foo"),

] == [true, true, true, true]

hash[b], when the key does not exist, evaluates to nil (which is falsey), so we evaluate and return the second half: hash[b]=[] which returns the value of the assignment, which is the array now referenced by hash[b], so we can push on to it, and hash[b] will [still be a] reference the updated array.

:D

PS - This is, I think, the first Ruby question I've ever answered, and it's the first time I've ever even thought of, let alone be able to, turn the comments into code, and oh my do I like it. Thank you for the puzzle!

edited May 31, 2013 at 7:17

answered May 31, 2013 at 6:32

Narfanator

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4 Comments

I love your use of short circuit eval here! Brilliant

What you are calling "wacky" aren't really wacky at all. They might have surprised you, but they're logical uses of the language. We used the || trick for assignment, or initialization then assignment, in Perl years ago. It's probably borrowed from shell scripting with a bit different syntax, or C.

It's totally inherited from Perl (I did Perl before I did Ruby), and it's one of the things that I love about ruby is that is has fewer "things" and those things are more similar: everything is an expression, everything is a reference. These kind of tricks are less readable when you first start programming, but as you start gaining the concepts, they're /absolutely/ natural.

(Well, and I think Perl probably got it from Lisp, which got it from... and gave it to...)

Sachin Singh · Accepted Answer · 2013-05-31 06:37:13Z

0

try this out

arr = [["Yonkers", "DM1210", "70.00 USD"], ["Yonkers", "DM1182", "19.68 AUD"], ["Nashua", "DM1182", "58.58 AUD"], ["Scranton", "DM1210", "68.76 USD"], ["Camden", "DM1182", "54.64 USD"]] 

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

 arr.each{|a| hash[a[1]].push([a[0],a[2]])}


   hash => {"DM1210"=>[["Yonkers", "70.00 USD"], ["Scranton", "68.76 USD"]], "DM1182"=>[["Yonkers", "19.68 AUD"], ["Nashua", "58.58 AUD"], ["Camden", "54.64 USD"]]}

edited May 31, 2013 at 6:37

answered May 31, 2013 at 6:32

Sachin Singh

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1 Comment