i want to convert in ruby
[[1, 1], [2, 3], [3, 5], [4, 1], [1, 2], [2, 3], [3, 5], [4, 1]]
into
[{1=>1}, {2=>3}, {3=>5}, {4=>1}, {1=>2}, {2=>3}, {3=>5}, {4=>1}]
and after this to obtain sum of all different keys:
{1=>3,2=>6,3=>10,4=>2}
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do you need the second step or just want to achieve the last stepPritesh Jain– Pritesh Jain2012-09-06 12:56:29 +00:00Commented Sep 6, 2012 at 12:56
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no , the second step was just to make task more clearNicolae Rotaru– Nicolae Rotaru2012-09-07 08:42:16 +00:00Commented Sep 7, 2012 at 8:42
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4 Answers
For the second question
sum = Hash.new(0)
original_array.each{|x, y| sum[x] += y}
sum # => {1 => 3, 2 => 6, 3 => 10, 4 => 2}
3 Comments
Joshua Cheek
Unless I know I'm the only user of the hash, I don't like doing things like this, it returns a hash which behaves in ways most people aren't expecting. e.g. it's standard practice to check existence with
Hash#[], and this will then return a truthy value.
sawa
@JoshuaCheek That is not the standard way to check existence. That will fail when you have nil values. The standard way is to use Hash#key?, and my code works well with it.
Joshua Cheek
The ratio in which I have seen
Hash#[] to Hash#key? is probably on the order of 25:1, and that only because I'm grouping Hash#has_key? in with it. In those cases, it was used because false was a legitimate hash value (or the hash contents weren't known). Personally, as soon as I use a default value or block, I stop thinking of it as a hash and start thinking of it as an interface (which I occasionally replace later with a custom object).Functional approach:
xs = [[1, 1], [2, 3], [3, 5], [4, 1], [1, 2], [2, 3], [3, 5], [4, 1]]
Hash[xs.group_by(&:first).map do |k, pairs|
[k, pairs.map { |x, y| y }.inject(:+)]
end]
#=> {1=>3, 2=>6, 3=>10, 4=>2}
Using Facets is much simpler thanks to the abstractions map_by (a variation of group_by) and mash (map + Hash):
require 'facets'
xs.map_by { |k, v| [k, v] }.mash { |k, vs| [k, vs.inject(:+)] }
#=> {1=>3, 2=>6, 3=>10, 4=>2}
Comments
You don't need the intermediate form.
arrays = [[1, 1], [2, 3], [3, 5], [4, 1], [1, 2], [2, 3], [3, 5], [4, 1]]
aggregate = arrays.each_with_object Hash.new do |(key, value), hash|
hash[key] = hash.fetch(key, 0) + value
end
aggregate # => {1=>3, 2=>6, 3=>10, 4=>2}
3 Comments
willcodejavaforfood
Hmm. My first questions as a Java developer, does this involve any form of virgin sacrifice, or does it work out of the box?
Joshua Cheek
@willcodejavaforfood not really sure what you mean. If you'd like I can explain anything that isn't apparent.
willcodejavaforfood
@Joshua Cheek - It does look like magic to me at first glance, but having stared at it for a while it's beginning to make sense :)
arr= [[1, 1], [2, 3], [3, 5], [4, 1], [1, 2], [2, 3], [3, 5], [4, 1]]
final = Hash.new(0)
second_step = arr.inject([]) do |arr,inner|
arr << Hash[*inner]
final[inner.first] += inner.last
arr
end
second_step
#=> [{1=>1}, {2=>3}, {3=>5}, {4=>1}, {1=>2}, {2=>3}, {3=>5}, {4=>1}]
final
#=> {1=>3, 2=>6, 3=>10, 4=>2}
if you directly only need the last step
arr.inject(Hash.new(0)){|hash,inner| hash[inner.first] += inner.last;hash}
=> {1=>3, 2=>6, 3=>10, 4=>2}
