Ruby group by range, array of numbers?

Question

I'm trying to create a histogram from an array of numbers in the range [0,1].

Is there a way to use group_by to separate the array into N groups/bins by numeric interval (or some other fun Ruby one-liner)?

This is my current, boring, solution:

# values == array containing floating point numbers in the range [0,1]

n = 10

# EDITED from Array.new(n, 0) to Array.new(n, []), thanks emaillenin!
histogram = Array.new(n, [])
values.each do |val|
  histogram[(val * n).ceil - 1].push(val)
end

this code wouldn't work. histogram[anything] is a Fixnum and you cannot call push on it — Raj
– Raj, Commented Jul 25, 2014 at 17:43
@emaillenin oops you're correct! I meant that to be an array of arrays, my bad! — cjhin
– cjhin, Commented Jul 25, 2014 at 17:46
Should be Array.new(n) { [ ] } to avoid all slots referencing the same array. — tadman
– tadman, Commented Jul 25, 2014 at 17:50
Why does it have to be a one-liner? That constraint often results in code that is hard to read or understand. — the Tin Man
– the Tin Man, Commented Jul 25, 2014 at 19:32

DiegoSalazar · Accepted Answer · 2014-07-25 17:49:35Z

3

Not sure what you're trying to do but maybe this helps?

values = [0.0, 0.1, 0.2, 0.3]
values.group_by { |v| (v * 10).ceil - 1 }

That returns a hash:

{-1=>[0.0], 0=>[0.1], 1=>[0.2], 2=>[0.3]}

answered Jul 25, 2014 at 17:49

DiegoSalazar

13.5k2 gold badges42 silver badges58 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Daniël Knippers Over a year ago

Keep in mind that this can yield up to 11 buckets when both 0.0 and 1.0 are present in addition to the other values (0.x).

tadman Over a year ago

group_by does this very cleanly.

Cary Swoveland · Accepted Answer · 2014-07-25 23:33:33Z

This is one way to do it.

Code

def freq_by_bin(nbr_bins, *values)
  nbr_bins.times.to_a.product([0]).to_h.tap { |h|
    values.each { |v| h.update({ (v*nbr_bins).to_i=>1 }) { |_,o,_| o+1 } } }
end

Example

values =  [0.30, 0.25, 0.63, 0.94, 0.08, 0.94, 0.01,
           0.41, 0.28, 0.69, 0.61, 0.12, 0.66]
freq_by_bin(10, *values)
  #=> {0=>2, 1=>1, 2=>2, 3=>1, 4=>1,
  #    5=>0, 6=>4, 7=>0, 8=>0, 9=>2}

def histogram(nbr_bins, *values)
  h = freq_by_bin(nbr_bins, *values)
  puts "\nfreq"
  h.values.max.downto(0) do |n|
    print "%2d|" % n
    puts nbr_bins.times.with_object('   ') { |i,row|
           row << ((h[i]==n) ? ' X ' : '   ') }
  end
  puts "   __"+"___"*nbr_bins
  puts nbr_bins.times.each_with_object('      ') { |i,row| row << "%2d " % i }
end

histogram(10, *values)

freq
 4|                      X          
 3|                                 
 2|    X     X                    X 
 1|       X     X  X                
 0|                   X     X  X    
   ________________________________
       0  1  2  3  4  5  6  7  8  9

Notes

There are several ways to construct the hash whose elements are bin=>freq. Using Enumerable#group_by, which you mentioned and @diego used is one. I've used the form of Hash#update (aka Hash#merge!) that takes a block.
I used Object#tap merely to avoid the need to create a temporary (non-block) variable for the initialized hash.

Collectives™ on Stack Overflow

Ruby group by range, array of numbers?

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related