From: Glenn <glenn_ritz@...>
Date: 2008-10-01T07:30:04+09:00
Subject: Re: Mode method for Array

--0-1364420513-1222813867=:73318
Content-Type: text/plain; charset=iso-8859-1
Content-Transfer-Encoding: quoted-printable

Hi,=0A=0AI wrote 2 ways.  I don't know if either way is good or not.  Any f=
eedback is welcome.=0A=0AThe first way makes a hash of the array, with the =
unique values in the array as the keys, and the number of times the keys oc=
cur as the values.  Then I create a new hash out of that first hash with th=
e frequencies as the keys of the hash and the elements that had that freque=
ncy as the values.  Then I pick out the value of the highest key.=0A=0AThe =
second way creates that frequency hash, then iterates over the hash and cre=
ates an array with the elements that have the highest frequency.=0A=0A clas=
s Array  =0A  def hash_of_frequency=0A    h =3D Hash.new(0)=0A    each_with=
_index do |e, i|=0A      e =3D e.to_f if e !=3D nil=0A      h[e] =3D h[e] +=
=3D 1 =0A    end=0A    h=0A  end=0A=0A  def get_mode=0A    h =3D hash_of_fr=
equency.frequency_in_key=0A    h[h.keys.max].sort=0A  end=0Aend=0A=0Aclass =
Hash=0A  def frequency_in_key=0A    h =3D Hash.new { |k, v| k[v] =3D [] }=
=0A    each { |k, v| h[v] << k if k !=3D nil }=0A    h=0A  end=0A=0A  def g=
et_mode=0A    a =3D []=0A    max_value =3D values.max=0A    each { |k, v| a=
 << k if v =3D=3D max_value }=0A    a=0A  end=0Aend=0A=0A=0A[3, 1, 1, 55, 5=
5].hash_of_frequency.get_mode.inspect ## returns [1.0, 55.0]=0A[3, 1, 1, 55=
, 55].get_mode.inspect ## returns[1.0, 55.0]=0A=0A=0A=0A----- Original Mess=
age ----=0AFrom: Eust=E1quio 'TaQ' Rangel <eustaquiorangel@gmail.com>=0ATo:=
 ruby-talk ML <ruby-talk@ruby-lang.org>=0ASent: Tuesday, September 30, 2008=
 6:14:35 PM=0ASubject: Re: Mode method for Array=0A=0A> I'd like to write a=
 get_mode method for the Array class.  The method would return an array of =
the most frequently occurring element or elements.=0A> So [3, 1, 1, 55, 55]=
.get_mode would return [1, 55].=0A> I have a way to do this but I don't kno=
w if it's the best way.  I was wondering if anyone had any suggestions?=0A=
=0AWhat is your way? Maybe we can have some idea of what parameters you are=
 using=0Ato the the most frequently elements. Using something like=0A=0Airb=
(main):001:0> [3,1,1,55,55].inject(Hash.new(0)){|memo,item| memo[item] +=3D=
 1;=0Amemo}.sort_by {|e| e[1]}.reverse=0A=3D> [[55, 2], [1, 2], [3, 1]]=0A=
=0Acan return you some elements ordered by frequency.
--0-1364420513-1222813867=:73318--