Statistics/Normal distribution

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Revision as of 20:03, 7 July 2012 by Ulrie (talk | contribs)
Statistics/Normal distribution is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The Normal (or Gaussian) distribution is a freqently used distribution in statistics. While most programming languages provide a uniformly distributed random number generator, one can derive normally distributed random numbers from a uniform generator.

The task
  1. Take a uniform random number generator and create a large (you decide how large) set of numbers that follow a normal (Gaussian) distribution. Calculate the dataset's mean and stddev, and show the histogram here.
  2. Mention any native language support for the generation of normally distributed random numbers.
Reference

C++

showing features of C++11 here <lang C++>#include <random>

  1. include <map>
  2. include <string>
  3. include <iostream>
  4. include <cmath>
  5. include <iomanip>

int main( ) {

  std::random_device myseed ;
  std::mt19937 engine ( myseed( ) ) ;
  std::normal_distribution<> normDistri ( 2 , 3 ) ;
  std::map<int , int> normalFreq ;
  int sum = 0 ; //holds the sum of the randomly created numbers
  double mean = 0.0 ;
  double stddev = 0.0 ;
  for ( int i = 1 ; i < 10001 ; i++ ) 
     ++normalFreq[ normDistri ( engine ) ] ;
  for ( auto MapIt : normalFreq ) {
     sum += MapIt.first * MapIt.second ;
  }
  mean = sum / 10000 ;
  stddev = sqrt( sum / 10000 ) ;
  std::cout << "The mean of the distribution is " << mean << " , the " ;
  std::cout << "standard deviation " << stddev << " !\n" ;
  std::cout << "And now the histogram:\n" ;
  for ( auto MapIt : normalFreq ) {
     std::cout << std::left << std::setw( 4 ) << MapIt.first << 

std::string( MapIt.second / 100 , '*' ) << std::endl ;

  }
  return 0 ;

}</lang> Output:

The mean of the distribution is 1 , the standard deviation 1 !
And now the histogram:
-10 
-9  
-8  
-7  
-6  
-5  
-4  *
-3  **
-2  ****
-1  ******
0   *********************
1   ************
2   ************
3   ***********
4   *********
5   ******
6   ****
7   **
8   *
9   
10  
11  
12  
13  

Go

Box-Muller method shown here. Go has a normally distributed random function in the standard library, as shown in the Go Random numbers solution. It uses the ziggurat method. <lang go>package main

import (

   "fmt"
   "math"
   "math/rand"
   "strings"

)

// Box-Muller func norm2() (s, c float64) {

   r := math.Sqrt(-2 * math.Log(rand.Float64()))
   s, c = math.Sincos(2 * math.Pi * rand.Float64())
   return s * r, c * r

}

func main() {

   const (
       n     = 10000
       bins  = 12
       sig   = 3
       scale = 100
   )
   var sum, sumSq float64
   h := make([]int, bins)
   for i, accum := 0, func(v float64) {
       sum += v
       sumSq += v * v
       b := int((v + sig) * bins / sig / 2)
       if b >= 0 && b < bins {
           h[b]++
       }
   }; i < n/2; i++ {
       v1, v2 := norm2()
       accum(v1)
       accum(v2)
   }
   m := sum / n
   fmt.Println("mean:", m)
   fmt.Println("stddev:", math.Sqrt(sumSq/float64(n)-m*m))
   for _, p := range h {
       fmt.Println(strings.Repeat("*", p/scale))
   }

}</lang> Output:

mean: -0.0034970888831523488
stddev: 1.0040682925006286

*
****
*********
***************
*******************
******************
**************
*********
****
*

Mathematica

<lang Mathematica>x:= RandomReal[1] SampleNormal[n_] := (Print[#//Length, " numbers, Mean : ", #//Mean, ", StandardDeviation : ", #//StandardDeviation];

   Histogram[#, BarOrigin -> Left,Axes -> False])& [(Table[(-2*Log[x])^0.5*Cos[2*Pi*x], {n} ]]

Invocation: SampleNormal[ 10000 ] ->10000 numbers, Mean : -0.0122308, StandardDeviation : 1.00646 </lang>

MATLAB / Octave

<lang Matlab> N = 100000;

 x = randn(N,1);
 mean(x)
 std(x) 
 [nn,xx] = hist(x,100);
 bar(xx,nn);</lang>

Liberty BASIC

Uses LB Statistics/Basic <lang lb>call sample 100000

end

sub sample n

   dim dat( n)
   for i =1 to n
       dat( i) =normalDist( 1, 0.2)
   next i
   '// show mean, standard deviation. Find max, min.
   mx  =-1000
   mn  = 1000
   sum =0
   sSq =0
   for i =1 to n
       d =dat( i)
       mx =max( mx, d)
       mn =min( mn, d)
       sum =sum +d
       sSq =sSq +d^2
   next i
   print n; " data terms used."
   mean =sum / n
   print "Largest term was "; mx; " & smallest was "; mn
   range =mx -mn
   print "Mean ="; mean
   print "Stddev ="; ( sSq /n -mean^2)^0.5
   '// show histogram
   nBins =50
   dim bins( nBins)
   for i =1 to n
       z =int( ( dat( i) -mn) /range *nBins)
       bins( z) =bins( z) +1
   next i
   for b =0 to nBins -1
       for j =1 to int( nBins *bins( b)) /n *30)
           print "#";
       next j
       print
   next b
   print

end sub

function normalDist( m, s) ' Box Muller method

   u =rnd( 1)
   v =rnd( 1)
   normalDist =( -2 *log( u))^0.5 *cos( 2 *3.14159265 *v)

end function</lang>

100000 data terms used.
Largest term was 4.12950792 & smallest was -4.37934139
Mean =-0.26785425e-2
Stddev =1.00097669


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PARI/GP

Works with: PARI/GP version 2.4.3 and above

<lang parigp>rnormal()={ my(u1=random(1.),u2=random(1.); sqrt(-2*log(u1))*cos(2*Pi*u1) \\ Could easily be extended with a second normal at very little cost. }; mean(v)={

 sum(i=1,#v,v[i])/#v

}; stdev(v,mu="")={

 if(mu=="",mu=mean(v));
 sqrt(sum(i=1,#v,(v[i]-mu)^2))/#v

}; histogram(v,bins=16,low=0,high=1)={

 my(u=vector(bins),width=(high-low)/bins);
 for(i=1,#v,u[(v[i]-low)\width+1]++);
 u

}; show(n)={

 my(v=vector(n,i,rnormal()),m=mean(v),s=stdev(v,m),h,sz=ceil(n/300));
 h=histogram(v,,vecmin(v)-.1,vecmax(v)+.1);
 for(i=1,#h,for(j=1,h[i]\sz,print1("#"));print());

}; show(10^4)</lang>

For versions before 2.4.3, define <lang parigp>rreal()={

 my(pr=32*ceil(default(realprecision)*log(10)/log(4294967296))); \\ Current precision
 random(2^pr)*1.>>pr

};</lang> and use rreal() in place of random(1.).

A PARI implementation: <lang C>GEN rnormal(long prec) { pari_sp ltop = avma; GEN u1, u2, left, right, ret; u1 = randomr(prec); u2 = randomr(prec); left = sqrtr_abs(shiftr(mplog(u1), 1)); right = mpcos(mulrr(shiftr(mppi(prec), 1), u2));

ret = mulrr(left, right); ret = gerepileupto(ltop, ret); return ret; }</lang> Use mpsincos and caching to generate two values at nearly the same cost.

Perl 6

<lang perl6>constant τ = 2 * pi;

sub normdist ($m, $σ) {

   gather loop {
       my $r = sqrt -2 * log rand;
       my $Θ = τ * rand;
       take $r * $_($Θ) * $σ + $m for &cos, &sin;
   }

}

sub MAIN ($size = 100000, $mean = 50, $stddev = 4) {

   my @dataset = normdist($mean,$stddev)[^$size];
   my $m = [+](@dataset) / $size;
   say (:$m);
   my $σ = sqrt [+](@dataset X** 2) / $size - $mean**2;
   say (:$σ);
   (my %hash){.round}++ for @dataset;
   my $scale = 180 * $stddev / $size;
   constant @subbar = < ⎸ ▏ ▎ ▍ ▌ ▋ ▊ ▉ █ >;
   for %hash.keys».Int.minmax(+*) -> $i {
       my $x = (%hash{$i} // 0) * $scale;
       my $full = floor $x;
       my $part = 8 * ($x - $full);
       say $i, "\t", '█' x $full, @subbar[$part];
   }

}</lang>

Output:
"m" => 50.006107405837142e0
"σ" => 4.0814435639885254e0
33	⎸
34	⎸
35	⎸
36	▏
37	▎
38	▊
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40	███⎸
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56	███████████████████████▍
57	███████████████▉
58	█████████▉
59	█████▍
60	███▍
61	█▋
62	▊
63	▍
64	▏
65	⎸
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PureBasic

<lang purebasic>Procedure.f randomf(resolution = 2147483647)

 ProcedureReturn Random(resolution) / resolution

EndProcedure

Procedure.f normalDist() ;Box Muller method

  ProcedureReturn Sqr(-2 * Log(randomf())) * Cos(2 * #PI * randomf())

EndProcedure

Procedure sample(n, nBins = 50)

 Protected i, maxBinValue, binNumber
 Protected.f d, mean, sum, sumSq, mx, mn, range
 
 Dim dat.f(n)
 For i = 1 To n
   dat(i) = normalDist()
 Next
 
 ;show mean, standard deviation, find max & min.
 mx  = -1000
 mn  =  1000
 sum = 0
 sumSq = 0
 For i = 1 To n
   d = dat(i)
   If d > mx: mx = d: EndIf
   If d < mn: mn = d: EndIf
   sum + d
   sumSq + d * d
 Next
 
 PrintN(Str(n) + " data terms used.")
 PrintN("Largest term was " + StrF(mx) + " & smallest was " + StrF(mn))
 mean = sum / n
 PrintN("Mean = " + StrF(mean))
 PrintN("Stddev = " + StrF((sumSq / n) - Sqr(mean * mean)))
 
 ;show histogram
 range = mx - mn
 Dim bins(nBins)
 For i = 1 To n
   binNumber = Int(nBins * (dat(i) - mn) / range)
   bins(binNumber) + 1
 Next
  
 maxBinValue = 1
 For i = 0 To nBins
   If bins(i) > maxBinValue
     maxBinValue = bins(i)
   EndIf
 Next
 
 #normalizedMaxValue = 70
 For binNumber = 0 To nBins
   tickMarks = Round(bins(binNumber) * #normalizedMaxValue / maxBinValue, #PB_Round_Nearest)
   PrintN(ReplaceString(Space(tickMarks), " ", "#"))
 Next
 PrintN("")

EndProcedure

If OpenConsole()

 sample(100000)
 
 Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
 CloseConsole()

EndIf</lang> Sample output:

100000 data terms used.
Largest term was 4.5352029800 & smallest was -4.5405135155
Mean = 0.0012346541
Stddev = 0.9959455132





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SAS

<lang sas>data test; n=100000; twopi=2*constant('pi'); do i=1 to n; u=ranuni(0); v=ranuni(0); r=sqrt(-2*log(u)); x=r*cos(twopi*v); y=r*sin(twopi*v); z=rannor(0); output; end; keep x y z;

proc means mean stddev;

proc univariate; histogram /normal;

run;

/* Variable Mean Std Dev


x -0.0052720 0.9988467 y 0.000023995 1.0019996 z 0.0012857 1.0056536

  • /</lang>

Tcl

<lang tcl>package require Tcl 8.5

  1. Uses the Box-Muller transform to compute a pair of normal random numbers

proc tcl::mathfunc::nrand {mean stddev} {

   variable savednormalrandom
   if {[info exists savednormalrandom]} {

return [expr {$savednormalrandom*$stddev + $mean}][unset savednormalrandom]

   }
   set r [expr {sqrt(-2*log(rand()))}]
   set theta [expr {2*3.1415927*rand()}]
   set savednormalrandom [expr {$r*sin($theta)}]
   expr {$r*cos($theta)*$stddev + $mean}

} proc stats {size {slotfactor 10}} {

   set sum 0.0
   set sum2 0.0
   for {set i 0} {$i < $size} {incr i} {

set r [expr { nrand(0.5, 0.2) }]

incr histo([expr {int(floor($r*$slotfactor))}]) set sum [expr {$sum + $r}] set sum2 [expr {$sum2 + $r**2}]

   }
   set mean [expr {$sum / $size}]
   set stddev [expr {sqrt($sum2/$size - $mean**2)}]
   puts "$size numbers"
   puts "Mean:   $mean"
   puts "StdDev: $stddev"
   foreach i [lsort -integer [array names histo]] {

puts [string repeat "*" [expr {$histo($i)*350/int($size)}]]

   }

}

stats 100 puts "" stats 1000 puts "" stats 10000 puts "" stats 100000 20</lang> Sample output:

100 numbers
Mean:   0.49355955990390254
StdDev: 0.19651396178121985
***
*******
**************
***********************************
********************************************************
******************************************************************
*************************************************************************
******************************************
**************************************
**************

1000 numbers
Mean:   0.5066940614105869
StdDev: 0.2016794788065389


*
*****
**************
****************************
**********************************************************
****************************************************************
*************************************************************
******************************************************
***********************************
************
*********
*

10000 numbers
Mean:   0.49980964730768285
StdDev: 0.1968441612522318

*
*****
***************
*******************************
*****************************************************
******************************************************************
*******************************************************************
****************************************************
*********************************
***************
*****
*



100000 numbers
Mean:   0.49960438950922254
StdDev: 0.20060211160998606





*
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***
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******
***
**
*







The blank lines in the output are where the number of samples is too small to even merit a single unit on the histogram.