Find the (deep) complement of two hashes

Question

The complement is the mathematical term for what I'm looking for, but for context and possibly more targeted solution: I have hash A, which can have nested hashes (i.e. they're N-dimensional), and I apply to it a process (over which I have no control) which returns hash B, which is hash A with some elements removed. From there on, I am trying to find the elements in A which have been removed in B.

For example: (note that I use symbols for simplicity. Keys will always be symbols, but values won't.)

a = {:a => :b,
     :c => {:d => :e, :f => :g},
     :h => :i,
     :j => {:k => :l, :m => :n},
     :o => {:p => :q, :r => :s},
     :t => :u}

b = {:h => :i,
     :j => {:k => :l, :m => :n},
     :o => {:r => :s},
     :t => :u}

complement(a,b)
#=> {:a => :b,
#    :c => {:d => :e, :f => :g},
#    :o => {:p => :q}}

What is the best (ruby-esque) way of doing this?

Is the depth arbitrary? Are there potentially duplicate elements among the arrays? Do you care *where* the elements are or just that they were removed? — Andrew Marshall, Mar 10 '12 at 02:26
The depth is arbitrary (it's user input) although it shouldn't reasonably go beyond two or three levels. As for duplicates, this is hashes we're talking about, so keys *can't* have duplicates; values can, of course. Finally, last I checked, hashes aren't sorted. — Félix Saparelli, Mar 10 '12 at 02:30
Hashes are sorted by insertion order in 1.9, but the order isn't specified in 1.8. — Andrew Marshall, Mar 10 '12 at 02:34
And so you're trying to find whole removed key/value pairs then? In other words, not that the value is different? — Andrew Marshall, Mar 10 '12 at 02:36

Ismael Abreu · Accepted Answer · 2012-03-10T04:08:58.743

1

Came up with this

a = {a: "thing", b: [1,2,3], c:2}
b = {a: "thing", b: [1,2,3]}
c= {}
a.each do |k, v|
  c[k] = v unless b[k]
end
p c

EDIT: Now checking nested hashes. But yes, there should be some better ruby way to do this.

def check_deleted(a, b)
    c = Hash.new
    a.each do |k, v|
        if ! b.has_key? k
            c[k] = v
        elsif b[k].is_a? Hash
            c[k] = check_deleted(v, b[k])
        end
    end
    c
end
a = {a: "thing", b: [1,2,3], c:2, d: {e: 1, r:2}}
b = {a: "thing", b: [1,2,3], d: {r:2}}

p check_deleted(a,b) #=> {:c=>2, :d=>{:e=>1}}

edited Mar 10 '12 at 04:08

answered Mar 10 '12 at 02:27

Ismael Abreu

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That works... I'll accept it if no one comes and gives a ruby-esque-er solution :) – Félix Saparelli Mar 10 '12 at 03:04
Fair enough :) I'm pretty sure someone will answer with some mind blowing solution. – Ismael Abreu Mar 10 '12 at 03:08
`b.has_key? k` is better than `b[k]`. For readability but also if you had `{:somekey => nil}`. – Andrew Marshall Mar 10 '12 at 03:59
You are right. I actually used has_key? method when giving another try to this, using select instead of each just because it seamed more legible. Will edit – Ismael Abreu Mar 10 '12 at 04:06

Find the (deep) complement of two hashes

1 Answers1