Statistics/Normal distribution: Difference between revisions

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immutable r = sqrt(-2 * uniform!"]["(0., 1.).log) * sigma;
immutable r = sqrt(-2 * uniform!"]["(0., 1.).log) * sigma;
immutable x = 2 * PI * uniform(0.0, 1.0);
immutable x = 2 * PI * uniform(0.0, 1.0);
state = [mu + r * x.sin, mu + r * x.cos]; // Heap alloc.
state = [mu + r * x.sin, mu + r * x.cos];
index = 0;
index = 0;
}
}
Line 96: Line 96:
void main() {
void main() {
const data = Normals(0.0, 0.5).take(100_000).array;
const data = Normals(0.0, 0.5).take(100_000).array;
writefln("Mean: %8.6f, SD: %8.6f\n", data.meanStdDev.tupleof);
writefln("Mean: %8.6f, SD: %8.6f\n", data.meanStdDev[]);
data.map!q{ max(0.0, min(0.9999, a / 3 + 0.5)) }.showHistogram01;
data.map!q{ max(0.0, min(0.9999, a / 3 + 0.5)) }.showHistogram01;
}</lang>
}</lang>

Revision as of 12:02, 11 April 2013

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

D

This uses the Box-Muller method as in the Go entry, and the module from the Statistics/Basic. A ziggurat-based normal generator for the Phobos standard library is in the works. <lang d>import std.stdio, std.random, std.math, std.range, std.algorithm, 

      statistics_basic;

struct Normals { 

   double mu, sigma;
   double[2] state;
   size_t index = state.length;
   enum empty = false;

   void popFront() pure nothrow { index++; }

   @property double front() {
       if (index >= state.length) {
           immutable r = sqrt(-2 * uniform!"]["(0., 1.).log) * sigma;
           immutable x = 2 * PI * uniform(0.0, 1.0);
           state = [mu + r * x.sin, mu + r * x.cos];
           index = 0;
       }
       return state[index];
   }

}

void main() { 

   const data = Normals(0.0, 0.5).take(100_000).array;
   writefln("Mean: %8.6f, SD: %8.6f\n", data.meanStdDev[]);
   data.map!q{ max(0.0, min(0.9999, a / 3 + 0.5)) }.showHistogram01;

}</lang>

Output:
Mean: 0.000528, SD: 0.502245

 0.0: *
 0.1: ******
 0.2: *****************
 0.3: ***********************************
 0.4: *************************************************
 0.5: **************************************************
 0.6: **********************************
 0.7: *****************
 0.8: ******
 0.9: *

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	▎
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56	███████████████████████▍
57	███████████████▉
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59	█████▍
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61	█▋
62	▊
63	▍
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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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#

Python

This uses the external matplotlib package as well as the built-in standardlib function random.gauss. <lang python>from __future__ import division import matplotlib.pyplot as plt import random

mean, stddev, size = 50, 4, 100000 data = [random.gauss(mean, stddev) for c in range(size)]

mn = sum(data) / size sd = (sum(x*x for x in data) / size 

     - (sum(data) / size) ** 2) ** 0.5

print("Sample mean = %g; Stddev = %g; max = %g; min = %g for %i values" 

     % (mn, sd, max(data), min(data), size))

plt.hist(data,bins=50)</lang>

Output:
Sample mean = 49.9822; Stddev = 4.00938; max = 66.8091; min = 33.5283 for 100000 values

REXX

The REXX language doesn't have any "higher math" functions like SIN/COS/LN/LOG/SQRT/POW/etc,
so we hoi polloi programmers have roll our own. <lang rexx>/*REXX pgm gens 10,000 normally distributed #s (Gaussian distribution).*/ parse arg n seed . /*allow specification of N | seed*/ if n== | n==',' then n=10000 /* N is the size of the array. */ if seed\== then call random ,,seed /*use seed for repeatable RANDOM#*/ call pi /*call subroutine to define pi. */ 

         do g=1  for n                /*generate N uniform random nums.*/
         #.g=sqrt(-2*ln(rand())) * cos(2*pi*rand())
         end   /*g*/

mn=#.1; mx=mn; s=0; ss=0; noise=n*.0005 /*calculate noise: .05%∙N.*/ 

         do j=1  for n;   _=#.j;   s=s+_;    ss=ss+_*_   /*sum & sum ²s*/
         mn=min(mn,#.j);  mx=max(mx,#.j)                 /*find min/max*/
         end   /*j*/

!.=0 say 'number of data points = ' n say ' minimum =' mn say ' maximum = ' mx say ' arithmetic mean = ' s/n say ' standard deviation = ' sqrt(ss/n - (s/n)**2) r=mx-mn /*used for scaling the histogram.*/ sd=50; sde=sd-4 /*screen depth, effective depth. */ sw=80; swe=sw-1 /*screen width, effective width. */ ?mn=; ?mx= 

         do i=1  for n
         ?=trunc( (#.i-mn)/r * sde)
         !.?=!.?+1
         if ?mn== then do; ?mn=!.j;  ?mx=!.j; end
         ?mn=min(?mn,!.?)
         ?mx=max(?mx,!.?)
         end   /*h*/

f=swe/?mx 

         do h=0  for sde;                     _=!.h
         if _>noise then say copies('─',trunc(_*f))   /*show histogram.*/
         end   /*h*/

exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────subroutines─────────────────────────*/ e: e=2.7182818284590452353602874713526624977572470936999595749669676277240766303535; return e pi: pi=3.1415926535897932384626433832795028841971693993751058209749445923078164062862; return pi rand: return random(0,1e5)/1e5 /*REXX gens uniform rand integers*/ r2r: return arg(1)//(2*pi()) sqrt: procedure;parse arg x; if x=0 then return 0; d=digits(); numeric digits 11; g=.sqrtGuess() 

       do j=0 while p>9; m.j=p; p=p%2+1; end; do k=j+5 to 0 by -1; if m.k>11 then numeric digits m.k
       g=.5*(g+x/g); end; numeric digits d; return g/1

.sqrtGuess: numeric form; m.=11; p=d+d%4+2 

       parse value format(x,2,1,,0) 'E0' with g 'E' _ .; return g*.5'E'_%2

cos: procedure; arg x; x=r2r(x); a=abs(x); numeric fuzz min(9,digits()-9); if a=pi() then return -1 

       if a=pi()/2|a=2*pi() then return 0;if a=pi()/3 then return .5;if a=2*pi()/3 then return -.5;return .sincos(1,1,-1)

.sincos:parse arg z,_,i; x=x*x; p=z; do k=2 by 2; _=-_*x/(k*(k+i)); z=z+_; if z=p then leave; p=z; end; return z ln: procedure; parse arg x,f; call e; ig=x>1.5; is=1-2*(ig\==1); ii=0; xx=x; return .ln_comp() .ln_comp: do while ig&xx>1.5|\ig&xx<.5;_=e;do k=-1;iz=xx*_**-is;if k>=0&(ig&iz<1|\ig&iz>.5) then leave;_=_*_;izz=iz;end 

       xx=izz;ii=ii+is*2**k;end;x=x*e**-ii-1;z=0;_=-1;p=z;do k=1;_=-_*x;z=z+_/k;if z=p then leave;p=z;end;return z+ii

</lang> output when using the default input 

number of data points =  10000
              minimum = -4.16421320
              maximum =  3.90674449
      arithmetic mean =  -0.00429983726
   standard deviation =  0.998221879

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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





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

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

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