Convert Ruby Array into average

Question

Say I have an array as such:

array = [["male", 59], ["female", 31], ["unknown gender", 38]]

The number inside the array is the total number of users by gender. What I need to do is get each element of the array and find the average for each male, female and unknown and put them into a new array that looks like this:

new_array = [["male", 46], ["female", 24], ["unknown gender", 30]]

Where the 2nd value inside the tuple is the average.

I have tried setting it as such:

new_array = []
sum = 0
array.each do |k,v|
sum += v
new_array << [k,(v/sum)*100]
end

My new array however turns out to be [["male", 100], ["female", 0], ["unknown gender", 0]]

Ive tried a few different things and cannot figure out what Im doing wrong. Any help would be great.

I don't see how you would get the second array from the first. Does the first actually have more subarrays? — Mark Thomas
– Mark Thomas, Commented Nov 3, 2014 at 0:12
I added an edit above, the numbers inside the first array are the total number of users by gender, I need to create a new array with the average % of the demographics — jthedudeoflife
– jthedudeoflife, Commented Nov 3, 2014 at 0:16

August · Accepted Answer · 2014-11-03 00:26:05Z

1

There are two issues here.

The first is that you are doing integer division at (v/sum). The result of this is probably not what you expect, so you should turn v into a float: (v.to_f/sum).

The second problem is that you are using the sum variable before it actually contains the sum.

Putting these two together:

array = [["male", 59], ["female", 31], ["unknown gender", 38]] 
new_array = []
sum = 0
# calculating sum before it's used in division below.
array.each { |_, v| sum += v }

array.each do |k, v|
  new_array << [k, ((v.to_f / sum) * 100).round]
end
# => [["male", 46], ["female", 24], ["unknown gender", 30]]

answered Nov 3, 2014 at 0:26

August

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2 Comments

jthedudeoflife Over a year ago

Thats it, thank you. My goal was trying to make this into a single iteration, Im learning about Big O notation so Im trying to write this in the best possible way, unfortunately it looks like there was no other way then creating 2 separate iterations. Thank you again.

spickermann Over a year ago

Iteration two times over the same array means O(2n). What is still in O(n) complexity. That is fine, especially because it is unlikely that a list of genders will get huge.

ruby_newbie · Accepted Answer · 2014-11-03 00:22:38Z

0

You need to do a few things. what is happening now is that sum is 59. v/59 is one on the first iteration. On the second iteration 31/90 == 0 because you are not dividing the sum by the number of iterations. What you really want is 31/45 because 45 is 90/2(2 being the number of iterations)

1) you need to divide the sum by the number of iterations to get the current average.

2) your syntax of

array.each do |k,v|

is misleading because you are not dealing with keys and values but with array indexes.

answered Nov 3, 2014 at 0:22

ruby_newbie

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Comments

spickermann · Accepted Answer · 2014-11-03 01:37:05Z

0

I would prefer to use more specialised Ruby methods like inject or map instead of the simple each:

array = [['male', 59], ['female', 31], ['unknown gender', 38]] 

total = array.inject(0) { |sum, (_, n)| sum + n }

new_array = array.map { |k, v| [k, ((v.to_f / total) * 100).round] }
#=> [["male", 46], ["female", 24], ["unknown gender", 30]]

Furthermore the ((v.to_f / total) * 100).round part can be extracted into its own method like this:

def percentage(value, total)
  ((value.to_f / total) * 100).round
end

What makes the map more readable:

new_array = array.map { |k, v| [k, percentage(v, total)] }

edited Nov 3, 2014 at 1:37

answered Nov 3, 2014 at 1:05

spickermann

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Comments

Cary Swoveland · Accepted Answer · 2014-11-03 06:58:08Z

0

If you have a choice, I suggest you consider using hashes rather than an arrays and symbols rather than strings:

hash = { male: 59, female: 31, unknown_gender: 38 }
  #=> {:male=>59, :female=>31, :unknown_gender=>38}
tot = hash.values.reduce(:+)
  #=> 128
percents = hash.each_with_object({}) { |(k,v),h| h[k] = (100.0*v/tot).round }
  #=> {:male=>46, :female=>24, :unknown_gender=>30}

answered Nov 3, 2014 at 6:58

Cary Swoveland

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