How do I calculate the average values in objects located in an array?

Question

Let's say I have an array like this:

[
  {
    "player_id"         => 1,
    "number_of_matches" => 2,
    "goals"             => 5
  },
  {
    "player_id"         => 2,
    "number_of_matches" => 4,
    "goals"             => 10
  }
]

I want to have the average goals per match among all the players, not the average for each individual player, but the total average.

I have in mind doing it with .each and storing each of the individual averages, and at the end add them all and divide by the number of players I have. However, I am looking for a Ruby/ one-liner way of doing this.

You might want to fix your array/hash so that it's actually valid Ruby. — Dominik Honnef, Mar 09 '12 at 18:45
Sorry, I get a JSON and I map it to a hash. Let me edit that. — Nobita, Mar 09 '12 at 18:46
One-liners are interesting, but often overrated, IMO. I think asking for an *elegant* and *clean* solution is better than asking for a one-liner. — Andrew Marshall, Mar 09 '12 at 19:02
@Andrew: Agreed, especially because this doesn't seem to be a problem that can be solved elegantly in one libe. — Niklas B., Mar 09 '12 at 19:04
@AndrewMarshall You are right. I will take that into account when formulating this kind of questions to avoid confusion in answers. — Nobita, Mar 09 '12 at 19:05

tokland · Accepted Answer · 2016-06-07T18:35:19.497

17

As requested, a one-liner:

avg = xs.map { |x| x["goals"].to_f / x["number_of_matches"] }.reduce(:+) / xs.size

A more readable snippet:

goals, matches = xs.map { |x| [x["goals"], x["number_of_matches"]] }.transpose 
avg = goals.reduce(:+).to_f / matches.reduce(:+) if goals

edited Jun 07 '16 at 18:35

answered Mar 09 '12 at 18:52

tokland

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Kyle: Doing this in one line will require code repetition or inaccurate results. – Niklas B. Mar 09 '12 at 19:01
Niklas: code repetition? Please explain how my solution is repeating code? It makes one pass over the array and produces the result. – Kyle Mar 09 '12 at 19:18
@Kyle: I said *or inaccurate results*, like your solution, which sums up the individual division errors. If at all, you'd have to use `Rational`. – Niklas B. Mar 09 '12 at 19:21
one-liner solution will lead to inaccurate result on big numbers, please look at my variant – megas Mar 09 '12 at 19:28
2

@Kyle: Just replace every line break with a semicolon. Voilà: one-liner. – Jörg W Mittag Mar 10 '12 at 01:35
@JörgWMittag I answered with my one liner solution. – Kyle Mar 10 '12 at 01:48

score 1 · Answer 2 · answered Mar 09 '12 at 19:10

1

A slight modification to tokland's answer.

items.map{|e| e.values_at("goals", "number_of_matches")}.transpose.map{|e| e.inject(:+)}.instance_eval{|goals, matches| goals.to_f/matches}

answered Mar 09 '12 at 19:10

sawa

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Heh, nice trick with `instance_eval` :) I'd rather not see that in production code, though :P – Niklas B. Mar 09 '12 at 19:11

Kyle · Answer 3 · 2012-03-09T19:26:55.533

0

a = [{player_id:1 , match_num:2, goals: 5}, {player_id:2 , match_num:4, goals: 10}]

a.reduce(0){|avg, p| avg += p[:goals].to_f/p[:match_num]}/a.size

Edit: renamed keys and block args to reduce char count. For those who care.

First, your keys need to use => if your going to use strings as keys.

reduce will iterate over the array and sum the individual averages for each player and finally we divide that result by the number of total players. The '0' in the parenthesis is your starting number for reduce.

edited Mar 09 '12 at 19:26

answered Mar 09 '12 at 18:52

Kyle

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`arr.map { |p| p[:goals].to_f / p[:number_of_matches] }.reduce(:+) / arr.size` would be a bit shorter (and not overflow the code div). – Niklas B. Mar 09 '12 at 19:05
Out of 93 characters in your one-liner, only 3 are spaces, and a few more around operators would make it far more readable. – Andrew Marshall Mar 09 '12 at 19:05
Niklas: you are mapping and then reducing, thus iterating over the array twice when only one pass is required. – Kyle Mar 09 '12 at 19:14
1

@Kyle: Yes, but a one-liner that does not fit the screen is not a one-liner. Usually code should wrap at about column 80. If performance were a concern, one would probably not choose Ruby for the job. Also, both solutions are `O(n)`, so it's not a "real" algorithmic difference. – Niklas B. Mar 09 '12 at 19:21

megas · Answer 4 · 2012-03-10T11:26:12.030

0

To make string shorter, lets rename "number_of_matches" to "matches"

a = [
  {"player_id":1 , "matches":2, "goals": 5}, 
  {"player_id":2 , "matches":4, "goals": 10}
]

a.reduce([0,0]){|sum,h|[sum.first+h["goals"],sum.last+h["matches"]]}.reduce{|sum,m|sum.to_f/m}
#=> 2.5

edited Mar 10 '12 at 11:26

answered Mar 09 '12 at 19:20

megas

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How do I calculate the average values in objects located in an array?

4 Answers4