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'''Many big data or scientific programs use boxplots to show distributions of data. In addition, sometimes saving large arrays for boxplots can be impractical and use extreme amounts of RAM. It can be useful to save large arrays as arrays with five numbers to save memory.
For example, the '''R''' programming language implements Tukey's [[wp:Five-number summary|five-number summary]] as the '''[https://stat.ethz.ch/R-manual/R-devel/library/stats/html/fivenum.html fivenum]''' function.
;Task: Given an array of numbers, compute the five-number summary.
;Note: While these five numbers can be used to draw a [[wp:Box plot|boxplot]], statistical packages will typically need extra data. Moreover, while there is a consensus about the "box" of the boxplot, there are variations among statistical packages for the whiskers.
C
{{trans|Kotlin}}
#include <stdio.h>
#include <stdlib.h>
double median(double *x, int start, int end_inclusive) {
int size = end_inclusive - start + 1;
if (size <= 0) {
printf("Array slice cannot be empty\n");
exit(1);
}
int m = start + size / 2;
if (size % 2) return x[m];
return (x[m - 1] + x[m]) / 2.0;
}
int compare (const void *a, const void *b) {
double aa = *(double*)a;
double bb = *(double*)b;
if (aa > bb) return 1;
if (aa < bb) return -1;
return 0;
}
int fivenum(double *x, double *result, int x_len) {
int i, m, lower_end;
for (i = 0; i < x_len; i++) {
if (x[i] != x[i]) {
printf("Unable to deal with arrays containing NaN\n\n");
return 1;
}
}
qsort(x, x_len, sizeof(double), compare);
result[0] = x[0];
result[2] = median(x, 0, x_len - 1);
result[4] = x[x_len - 1];
m = x_len / 2;
lower_end = (x_len % 2) ? m : m - 1;
result[1] = median(x, 0, lower_end);
result[3] = median(x, m, x_len - 1);
return 0;
}
int show(double *result, int places) {
int i;
char f[7];
sprintf(f, "%%.%dlf", places);
printf("[");
for (i = 0; i < 5; i++) {
printf(f, result[i]);
if (i < 4) printf(", ");
}
printf("]\n\n");
}
int main() {
double result[5];
double x1[11] = {15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0};
if (!fivenum(x1, result, 11)) show(result, 1);
double x2[6] = {36.0, 40.0, 7.0, 39.0, 41.0, 15.0};
if (!fivenum(x2, result, 6)) show(result, 1);
double x3[20] = {
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
};
if (!fivenum(x3, result, 20)) show(result, 9);
return 0;
}
{{out}}
[6.0, 25.5, 40.0, 42.5, 49.0]
[7.0, 15.0, 37.5, 40.0, 41.0]
[-1.950595940, -0.676741205, 0.233247060, 0.746070945, 1.731315070]
C++
{{trans|D}}
#include <algorithm>
#include <iostream>
#include <ostream>
#include <vector>
/////////////////////////////////////////////////////////////////////////////
// The following is taken from https://cpplove.blogspot.com/2012/07/printing-tuples.html
// Define a type which holds an unsigned integer value
template<std::size_t> struct int_ {};
template <class Tuple, size_t Pos>
std::ostream& print_tuple(std::ostream& out, const Tuple& t, int_<Pos>) {
out << std::get< std::tuple_size<Tuple>::value - Pos >(t) << ", ";
return print_tuple(out, t, int_<Pos - 1>());
}
template <class Tuple>
std::ostream& print_tuple(std::ostream& out, const Tuple& t, int_<1>) {
return out << std::get<std::tuple_size<Tuple>::value - 1>(t);
}
template <class... Args>
std::ostream& operator<<(std::ostream& out, const std::tuple<Args...>& t) {
out << '(';
print_tuple(out, t, int_<sizeof...(Args)>());
return out << ')';
}
/////////////////////////////////////////////////////////////////////////////
template <class RI>
double median(RI beg, RI end) {
if (beg == end) throw std::runtime_error("Range cannot be empty");
auto len = end - beg;
auto m = len / 2;
if (len % 2 == 1) {
return *(beg + m);
}
return (beg[m - 1] + beg[m]) / 2.0;
}
template <class C>
auto fivenum(C& c) {
std::sort(c.begin(), c.end());
auto cbeg = c.cbegin();
auto cend = c.cend();
auto len = cend - cbeg;
auto m = len / 2;
auto lower = (len % 2 == 1) ? m : m - 1;
double r2 = median(cbeg, cbeg + lower + 1);
double r3 = median(cbeg, cend);
double r4 = median(cbeg + lower + 1, cend);
return std::make_tuple(*cbeg, r2, r3, r4, *(cend - 1));
}
int main() {
using namespace std;
vector<vector<double>> cs = {
{ 15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0 },
{ 36.0, 40.0, 7.0, 39.0, 41.0, 15.0 },
{
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
}
};
for (auto & c : cs) {
cout << fivenum(c) << endl;
}
return 0;
}
{{out}}
(6, 25.5, 40, 43, 49)
(7, 15, 37.5, 40, 41)
(-1.9506, -0.676741, 0.233247, 0.746071, 1.73132)
C#
{{trans|Java}}
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Fivenum {
public static class Helper {
public static string AsString<T>(this ICollection<T> c, string format = "{0}") {
StringBuilder sb = new StringBuilder("[");
int count = 0;
foreach (var t in c) {
if (count++ > 0) {
sb.Append(", ");
}
sb.AppendFormat(format, t);
}
return sb.Append("]").ToString();
}
}
class Program {
static double Median(double[] x, int start, int endInclusive) {
int size = endInclusive - start + 1;
if (size <= 0) throw new ArgumentException("Array slice cannot be empty");
int m = start + size / 2;
return (size % 2 == 1) ? x[m] : (x[m - 1] + x[m]) / 2.0;
}
static double[] Fivenum(double[] x) {
foreach (var d in x) {
if (Double.IsNaN(d)) {
throw new ArgumentException("Unable to deal with arrays containing NaN");
}
}
double[] result = new double[5];
Array.Sort(x);
result[0] = x.First();
result[2] = Median(x, 0, x.Length - 1);
result[4] = x.Last();
int m = x.Length / 2;
int lowerEnd = (x.Length % 2 == 1) ? m : m - 1;
result[1] = Median(x, 0, lowerEnd);
result[3] = Median(x, m, x.Length - 1);
return result;
}
static void Main(string[] args) {
double[][] x1 = new double[][]{
new double[]{ 15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0},
new double[]{ 36.0, 40.0, 7.0, 39.0, 41.0, 15.0},
new double[]{
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
},
};
foreach(var x in x1) {
var result = Fivenum(x);
Console.WriteLine(result.AsString("{0:F8}"));
}
}
}
}
{{out}}
[6.00000000, 25.50000000, 40.00000000, 42.50000000, 49.00000000]
[7.00000000, 15.00000000, 37.50000000, 40.00000000, 41.00000000]
[-1.95059594, -0.67674121, 0.23324706, 0.74607095, 1.73131507]
D
{{trans|Java}}
import std.algorithm;
import std.exception;
import std.math;
import std.stdio;
double median(double[] x) {
enforce(x.length >= 0, "Array slice cannot be empty");
int m = x.length / 2;
if (x.length % 2 == 1) {
return x[m];
}
return (x[m-1] + x[m]) / 2.0;
}
double[] fivenum(double[] x) {
foreach (d; x) {
enforce(!d.isNaN, "Unable to deal with arrays containing NaN");
}
double[] result;
result.length = 5;
x.sort;
result[0] = x[0];
result[2] = median(x);
result[4] = x[$-1];
int m = x.length / 2;
int lower = (x.length % 2 == 1) ? m : m - 1;
result[1] = median(x[0..lower+1]);
result[3] = median(x[lower+1..$]);
return result;
}
void main() {
double[][] x1 = [
[15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0],
[36.0, 40.0, 7.0, 39.0, 41.0, 15.0],
[
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
]
];
foreach(x; x1) {
writeln(fivenum(x));
}
}
{{out}}
[6, 25.5, 40, 43, 49]
[7, 15, 37.5, 40, 41]
[-1.9506, -0.676741, 0.233247, 0.746071, 1.73132]
=={{header|F#|F sharp}}== {{trans|C#}}
open System
// Take from https://stackoverflow.com/a/1175123
let rec last = function
| hd :: [] -> hd
| _ :: tl -> last tl
| _ -> failwith "Empty list."
let median x =
for e in x do
if Double.IsNaN(e) then failwith "unable to deal with lists containing NaN"
let size = List.length(x)
if size <= 0 then failwith "Array slice cannot be empty"
let m = size / 2
if size % 2 = 1 then x.[m]
else (x.[m - 1] + x.[m]) / 2.0
let fivenum x =
let x2 = List.sort(x)
let m = List.length(x2) / 2
let lowerEnd = if List.length(x2) % 2 = 1 then m else m - 1
[List.head x2, median x2.[..lowerEnd], median x2, median x2.[m..], last x2]
[<EntryPoint>]
let main _ =
let x1 = [
[15.0; 6.0; 42.0; 41.0; 7.0; 36.0; 49.0; 40.0; 39.0; 47.0; 43.0];
[36.0; 40.0; 7.0; 39.0; 41.0; 15.0];
[
0.14082834; 0.09748790; 1.73131507; 0.87636009; -1.95059594;
0.73438555; -0.03035726; 1.46675970; -0.74621349; -0.72588772;
0.63905160; 0.61501527; -0.98983780; -1.00447874; -0.62759469;
0.66206163; 1.04312009; -0.10305385; 0.75775634; 0.32566578
]
]
for a in x1 do
let y = fivenum a
Console.WriteLine("{0}", y);
0 // return an integer exit code
{{out}}
[(6, 25.5, 40, 42.5, 49)]
[(7, 15, 37.5, 40, 41)]
[(-1.95059594, -0.676741205, 0.23324706, 0.746070945, 1.73131507)]
Factor
USING: combinators combinators.smart kernel math
math.statistics prettyprint sequences sorting ;
IN: rosetta-code.five-number
<PRIVATE
: bisect ( seq -- lower upper )
dup length even? [ halves ]
[ dup midpoint@ 1 + [ head ] [ tail* ] 2bi ] if ;
: (fivenum) ( seq -- summary )
natural-sort {
[ infimum ]
[ bisect drop median ]
[ median ]
[ bisect nip median ]
[ supremum ]
} cleave>array ;
PRIVATE>
ERROR: fivenum-empty data ;
ERROR: fivenum-nan data ;
: fivenum ( seq -- summary )
{
{ [ dup empty? ] [ fivenum-empty ] }
{ [ dup [ fp-nan? ] any? ] [ fivenum-nan ] }
[ (fivenum) ]
} cond ;
: fivenum-demo ( -- )
{ 15 6 42 41 7 36 49 40 39 47 43 }
{ 36 40 7 39 41 15 }
{ 0.14082834 0.09748790 1.73131507 0.87636009
-1.95059594 0.73438555 -0.03035726 1.46675970
-0.74621349 -0.72588772 0.63905160 0.61501527
-0.98983780 -1.00447874 -0.62759469 0.66206163
1.04312009 -0.10305385 0.75775634 0.32566578 }
[ fivenum . ] tri@ ;
MAIN: fivenum-demo
{{out}}
{ 6 25+1/2 40 42+1/2 49 }
{ 7 15 37+1/2 40 41 }
{ -1.95059594 -0.676741205 0.23324706 0.746070945 1.73131507 }
Go
{{trans|Perl}}
package main
import (
"fmt"
"math"
"sort"
)
func fivenum(a []float64) (n5 [5]float64) {
sort.Float64s(a)
n := float64(len(a))
n4 := float64((len(a)+3)/2) / 2
d := []float64{1, n4, (n + 1) / 2, n + 1 - n4, n}
for e, de := range d {
floor := int(de - 1)
ceil := int(math.Ceil(de - 1))
n5[e] = .5 * (a[floor] + a[ceil])
}
return
}
var (
x1 = []float64{36, 40, 7, 39, 41, 15}
x2 = []float64{15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43}
x3 = []float64{
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578,
}
)
func main() {
fmt.Println(fivenum(x1))
fmt.Println(fivenum(x2))
fmt.Println(fivenum(x3))
}
{{out}}
[7 15 37.5 40 41]
[6 25.5 40 42.5 49]
[-1.95059594 -0.676741205 0.23324706 0.746070945 1.73131507]
'''Alternate:'''
This solution is aimed at handling larger data sets more efficiently. It replaces the O(n log n) sort with O(n) quickselect. It also does not attempt to reproduce the R result exactly, to average values to get a median of an even number of data values, or otherwise estimate quantiles. The quickselect here leaves the input partitioned around the selected value, which allows another small optimization: The first quickselect call partitions the full input around the median. The second call, to get the first quartile, thus only has to process the partition up to the median. The third call, to get the minimum, only has to process the partition up to the first quartile. The 3rd quartile and maximum are obtained similarly.
package main
import (
"fmt"
"math/rand"
)
func fivenum(a []float64) (n [5]float64) {
last := len(a) - 1
m := last / 2
n[2] = qsel(a, m)
q1 := len(a) / 4
n[1] = qsel(a[:m], q1)
n[0] = qsel(a[:q1], 0)
a = a[m:]
q3 := last - m - q1
n[3] = qsel(a, q3)
a = a[q3:]
n[4] = qsel(a, len(a)-1)
return
}
func qsel(a []float64, k int) float64 {
for len(a) > 1 {
px := rand.Intn(len(a))
pv := a[px]
last := len(a) - 1
a[px], a[last] = a[last], pv
px = 0
for i, v := range a[:last] {
if v < pv {
a[px], a[i] = v, a[px]
px++
}
}
a[px], a[last] = pv, a[px]
if px == k {
return pv
}
if k < px {
a = a[:px]
} else {
a = a[px+1:]
k -= px + 1
}
}
return a[0]
}
var (
x1 = []float64{36, 40, 7, 39, 41, 15}
x2 = []float64{15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43}
x3 = []float64{
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578,
}
)
func main() {
fmt.Println(fivenum(x1))
fmt.Println(fivenum(x2))
fmt.Println(fivenum(x3))
}
{{out}}
[7 15 36 40 41]
[6 15 40 43 49]
[-1.95059594 -0.62759469 0.14082834 0.73438555 1.73131507]
J
'''Solution'''
midpts=: (1 + #) <:@(] , -:@[ , -) -:@<.@-:@(3 + #) NB. mid points of y
quartiles=: -:@(+/)@((<. ,: >.)@midpts { /:~@]) NB. quartiles of y
fivenum=: <./ , quartiles , >./ NB. fivenum summary of y
'''Example Usage'''
test1=: 15 6 42 41 7 36 49 40 39 47 43
test2=: 36 40 7 39 41 15
test3=: , 0 ". ];._2 noun define
0.14082834 0.09748790 1.73131507 0.87636009 -1.95059594
0.73438555 -0.03035726 1.46675970 -0.74621349 -0.72588772
0.63905160 0.61501527 -0.98983780 -1.00447874 -0.62759469
0.66206163 1.04312009 -0.10305385 0.75775634 0.32566578
)
fivenum &> test1;test2;test3
6 25.5 40 42.5 49
7 15 37.5 40 41
_1.9506 _0.676741 0.233247 0.746071 1.73132
Java
{{trans|Kotlin}}
import java.util.Arrays;
public class Fivenum {
static double median(double[] x, int start, int endInclusive) {
int size = endInclusive - start + 1;
if (size <= 0) throw new IllegalArgumentException("Array slice cannot be empty");
int m = start + size / 2;
return (size % 2 == 1) ? x[m] : (x[m - 1] + x[m]) / 2.0;
}
static double[] fivenum(double[] x) {
for (Double d : x) {
if (d.isNaN())
throw new IllegalArgumentException("Unable to deal with arrays containing NaN");
}
double[] result = new double[5];
Arrays.sort(x);
result[0] = x[0];
result[2] = median(x, 0, x.length - 1);
result[4] = x[x.length - 1];
int m = x.length / 2;
int lowerEnd = (x.length % 2 == 1) ? m : m - 1;
result[1] = median(x, 0, lowerEnd);
result[3] = median(x, m, x.length - 1);
return result;
}
public static void main(String[] args) {
double xl[][] = {
{15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0},
{36.0, 40.0, 7.0, 39.0, 41.0, 15.0},
{
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
}
};
for (double[] x : xl) System.out.printf("%s\n\n", Arrays.toString(fivenum(x)));
}
}
{{out}}
[6.0, 25.5, 40.0, 42.5, 49.0]
[7.0, 15.0, 37.5, 40.0, 41.0]
[-1.95059594, -0.676741205, 0.23324706, 0.746070945, 1.73131507]
Julia
{{works with|Julia|0.6}}
function mediansorted(x::AbstractVector{T}, i::Integer, l::Integer)::T where T
len = l - i + 1
len > zero(len) || throw(ArgumentError("Array slice cannot be empty."))
mid = i + len ÷ 2
return isodd(len) ? x[mid] : (x[mid-1] + x[mid]) / 2
end
function fivenum(x::AbstractVector{T}) where T<:AbstractFloat
r = Vector{T}(5)
xs = sort(x)
mid::Int = length(xs) ÷ 2
lowerend::Int = isodd(length(xs)) ? mid : mid - 1
r[1] = xs[1]
r[2] = mediansorted(xs, 1, lowerend)
r[3] = mediansorted(xs, 1, endof(xs))
r[4] = mediansorted(xs, mid, endof(xs))
r[end] = xs[end]
return r
end
for v in ([15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0],
[36.0, 40.0, 7.0, 39.0, 41.0, 15.0],
[0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578])
println("# ", v, "\n -> ", fivenum(v))
end
{{out}}
# [15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0]
-> [6.0, 15.0, 40.0, 42.0, 49.0]
# [36.0, 40.0, 7.0, 39.0, 41.0, 15.0]
-> [7.0, 11.0, 37.5, 39.5, 41.0]
# [0.140828, 0.0974879, 1.73132, 0.87636, -1.9506, 0.734386, -0.0303573, 1.46676, -0.746213, -0.725888, 0.639052, 0.615015, -0.989838, -1.00448, -0.627595,0.662062, 1.04312, -0.103054, 0.757756, 0.325666]
-> [-1.9506, -0.725888, 0.233247, 0.734386, 1.73132]
Kotlin
The following uses Tukey's method for calculating the lower and upper quartiles (or 'hinges') which is what the R function, fivenum, appears to use.
As arrays containing NaNs and nulls cannot really be dealt with in a sensible fashion in Kotlin, they've been excluded altogether.
// version 1.2.21
fun median(x: DoubleArray, start: Int, endInclusive: Int): Double {
val size = endInclusive - start + 1
require (size > 0) { "Array slice cannot be empty" }
val m = start + size / 2
return if (size % 2 == 1) x[m] else (x[m - 1] + x[m]) / 2.0
}
fun fivenum(x: DoubleArray): DoubleArray {
require(x.none { it.isNaN() }) { "Unable to deal with arrays containing NaN" }
val result = DoubleArray(5)
x.sort()
result[0] = x[0]
result[2] = median(x, 0, x.size - 1)
result[4] = x[x.lastIndex]
val m = x.size / 2
var lowerEnd = if (x.size % 2 == 1) m else m - 1
result[1] = median(x, 0, lowerEnd)
result[3] = median(x, m, x.size - 1)
return result
}
fun main(args: Array<String>) {
var xl = listOf(
doubleArrayOf(15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0),
doubleArrayOf(36.0, 40.0, 7.0, 39.0, 41.0, 15.0),
doubleArrayOf(
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
)
)
xl.forEach { println("${fivenum(it).asList()}\n") }
}
{{out}}
[6.0, 25.5, 40.0, 42.5, 49.0]
[7.0, 15.0, 37.5, 40.0, 41.0]
[-1.95059594, -0.676741205, 0.23324706, 0.746070945, 1.73131507]
Lua
function slice(tbl, low, high)
local copy = {}
for i=low or 1, high or #tbl do
copy[#copy+1] = tbl[i]
end
return copy
end
-- assumes that tbl is sorted
function median(tbl)
m = math.floor(#tbl / 2) + 1
if #tbl % 2 == 1 then
return tbl[m]
end
return (tbl[m-1] + tbl[m]) / 2
end
function fivenum(tbl)
table.sort(tbl)
r0 = tbl[1]
r2 = median(tbl)
r4 = tbl[#tbl]
m = math.floor(#tbl / 2)
if #tbl % 2 == 1 then
low = m
else
low = m - 1
end
r1 = median(slice(tbl, nil, low+1))
r3 = median(slice(tbl, low+2, nil))
return r0, r1, r2, r3, r4
end
x1 = {
{15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0},
{36.0, 40.0, 7.0, 39.0, 41.0, 15.0},
{
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
}
}
for i,x in ipairs(x1) do
print(fivenum(x))
end
{{out}}
6 25.5 40 43 49
7 15 37.5 40 41
-1.95059594 -0.676741205 0.23324706 0.746070945 1.73131507
=={{header|Modula-2}}==
MODULE Fivenum;
FROM FormatString IMPORT FormatString;
FROM LongStr IMPORT RealToStr;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE WriteLongReal(v : LONGREAL);
VAR buf : ARRAY[0..63] OF CHAR;
BEGIN
RealToStr(v, buf);
WriteString(buf)
END WriteLongReal;
PROCEDURE WriteArray(arr : ARRAY OF LONGREAL);
VAR i : CARDINAL;
BEGIN
WriteString("[");
FOR i:=0 TO HIGH(arr) DO
WriteLongReal(arr[i]);
WriteString(", ")
END;
WriteString("]")
END WriteArray;
(* Assumes that the input is sorted *)
PROCEDURE Median(x : ARRAY OF LONGREAL; beg,end : CARDINAL) : LONGREAL;
VAR m,cnt : CARDINAL;
BEGIN
cnt := end - beg + 1;
m := cnt / 2;
IF cnt MOD 2 = 1 THEN
RETURN x[beg + m]
END;
RETURN (x[beg + m - 1] + x[beg + m]) / 2.0
END Median;
TYPE Summary = ARRAY[0..4] OF LONGREAL;
PROCEDURE Fivenum(input : ARRAY OF LONGREAL) : Summary;
PROCEDURE Sort();
VAR
i,j : CARDINAL;
t : LONGREAL;
BEGIN
FOR i:=0 TO HIGH(input) DO
FOR j:=0 TO HIGH(input) DO
IF (i#j) AND (input[i] < input[j]) THEN
t := input[i];
input[i] := input[j];
input[j] := t
END
END
END
END Sort;
VAR
result : Summary;
size,m,low : CARDINAL;
BEGIN
size := HIGH(input);
Sort();
result[0] := input[0];
result[2] := Median(input,0,size);
result[4] := input[size];
m := size / 2;
IF (size MOD 2 = 1) THEN
low := m
ELSE
low := m - 1
END;
result[1] := Median(input, 0, m);
result[3] := Median(input, m+1, size);
RETURN result;
END Fivenum;
TYPE
A6 = ARRAY[0..5] OF LONGREAL;
A11 = ARRAY[0..10] OF LONGREAL;
A20 = ARRAY[0..19] OF LONGREAL;
VAR
a6 : A6;
a11 : A11;
a20 : A20;
BEGIN
a11 := A11{15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0};
WriteArray(Fivenum(a11));
WriteLn;
WriteLn;
a6 := A6{36.0, 40.0, 7.0, 39.0, 41.0, 15.0};
WriteArray(Fivenum(a6));
WriteLn;
WriteLn;
a20 := A20{
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
};
WriteArray(Fivenum(a20));
WriteLn;
ReadChar
END Fivenum.
{{out}}
[6.000000000000000, 25.499999999999900, 40.000000000000000, 42.499999999999900, 49.000000000000000, ]
[7.000000000000000, 15.000000000000000, 35.500000000000000, 40.000000000000000, 40.499999999999900, ]
[-1.950594000000000, -0.676741205000000, 0.233247060000000, 0.746070945000000, 1.731315070000000, ]
Perl
use POSIX qw(ceil floor);
sub fivenum {
my(@array) = @_;
my $n = scalar @array;
die "No values were entered into fivenum!" if $n == 0;
my @x = sort {$a <=> $b} @array;
my $n4 = floor(($n+3)/2)/2;
my @d = (1, $n4, ($n +1)/2, $n+1-$n4, $n);
my @sum_array;
for my $e (0..4) {
my $floor = floor($d[$e]-1);
my $ceil = ceil($d[$e]-1);
push @sum_array, (0.5 * ($x[$floor] + $x[$ceil]));
}
return @sum_array;
}
my @x = (15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43);
my @tukey = fivenum(\@x);
say join (',', @tukey);
#----------
@x = (36, 40, 7, 39, 41, 15),
@tukey = fivenum(\@x);
say join (',', @tukey);
#----------
@x = (0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578);
@tukey = fivenum(\@x);
say join (',', @tukey);
{{out}}
6,25.5,40,42.5,49
7,15,37.5,40,41
-1.95059594,-0.676741205,0.23324706,0.746070945,1.73131507
Perl 6
{{trans|Perl}}
sub fourths ( Int $end ) {
my $end_22 = $end div 2 / 2;
return 0, $end_22, $end/2, $end - $end_22, $end;
}
sub fivenum ( @nums ) {
my @x = @nums.sort(+*)
or die 'Input must have at least one element';
my @d = fourths(@x.end);
return ( @x[@d».floor] Z+ @x[@d».ceiling] ) »/» 2;
}
say .&fivenum for [15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43],
[36, 40, 7, 39, 41, 15], [
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578,
];
{{out}}
(6 25.5 40 42.5 49)
(7 15 37.5 40 41)
(-1.95059594 -0.676741205 0.23324706 0.746070945 1.73131507)
Phix
function median(sequence tbl, integer lo, hi)
integer l = hi-lo+1
integer m = lo+floor(l/2)
if remainder(l,2)=1 then
return tbl[m]
end if
return (tbl[m-1]+tbl[m])/2
end function
function fivenum(sequence tbl)
tbl = sort(tbl)
integer l = length(tbl),
m = floor(l/2)+remainder(l,2)
atom r1 = tbl[1],
r2 = median(tbl,1,m),
r3 = median(tbl,1,l),
r4 = median(tbl,m+1,l),
r5 = tbl[l]
return {r1, r2, r3, r4, r5}
end function
constant x1 = {15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43},
x2 = {36, 40, 7, 39, 41, 15},
x3 = {0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578}
?fivenum(x1)
?fivenum(x2)
?fivenum(x3)
{{out}}
{6,25.5,40,43,49}
{7,15,37.5,40,41}
{-1.95059594,-0.676741205,0.23324706,0.746070945,1.73131507}
Python
Python: Standard commands
{{trans|Perl}} '''Work with: Python 2'''
'''Work with: Python 3'''
from __future__ import division
import math
import sys
def fivenum(array):
n = len(array)
if n == 0:
print("you entered an empty array.")
sys.exit()
x = sorted(array)
n4 = math.floor((n+3.0)/2.0)/2.0
d = [1, n4, (n+1)/2, n+1-n4, n]
sum_array = []
for e in range(5):
floor = int(math.floor(d[e] - 1))
ceil = int(math.ceil(d[e] - 1))
sum_array.append(0.5 * (x[floor] + x[ceil]))
return sum_array
x = [0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555, -0.03035726, 1.46675970,
-0.74621349, -0.72588772, 0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469, 0.66206163,
1.04312009, -0.10305385, 0.75775634, 0.32566578]
y = fivenum(x)
print(y)
{{out}}
[-1.95059594, -0.676741205, 0.23324706, 0.746070945, 1.73131507]
Python: Pandas library
There are many ways to compute this kind of summary statistics (see [[wp:Percentile#Definitions]]). The Python Pandas library supports a range.
[https://pandas.pydata.org/ Pandas] is a well known Python library. Its [https://pandas.pydata.org/pandas-docs/version/0.17.0/generated/pandas.DataFrame.describe.html Dataframe.describe] method produces summary stats from data.
(Though these 25% and 75% values do '''not''' correspond to the Fivenum Tukey quartile values specified in this task)
import pandas as pd
pd.DataFrame([1, 2, 3, 4, 5, 6]).describe()
{{out}}
0
count 6.000000
mean 3.500000
std 1.870829
min 1.000000
25% 2.250000
50% 3.500000
75% 4.750000
max 6.000000
To get the fivenum values asked for, the [https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.quantile.html pandas.DataFrame.quantile] function can be used:
import pandas as pd
pd.DataFrame([1, 2, 3, 4, 5, 6]).quantile([.0, .25, .50, .75, 1.00], interpolation='nearest')
{{out}}
0
0.00 1
0.25 2
0.50 3
0.75 5
1.00 6
The interpolation value supports more of the differing ways of calculation in use.
===Python: Functional – without imports=== '''Works with: Python 3'''
# fiveNums :: [Float] -> (Float, Float, Float, Float, Float)
def fiveNums(xs):
def median(xs):
lng = len(xs)
m = lng // 2
return xs[m] if (
0 != lng % 2
) else (xs[m - 1] + xs[m]) / 2
ys = sorted(xs)
lng = len(ys)
m = lng // 2
return (
ys[0],
median(ys[0:(m + (lng % 2))]),
median(ys),
median(ys[m:]),
ys[-1]
) if 0 < lng else None
# TEST --------------------------------------------------------------------
for xs in [[15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43],
[36, 40, 7, 39, 41, 15],
[
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578
]]:
print(
fiveNums(xs)
)
{{Out}}
(6, 25.5, 40, 42.5, 49)
(7, 15, 37.5, 40, 41)
(-1.95059594, -0.676741205, 0.23324706, 0.746070945, 1.73131507)
R
The '''fivenum''' function is built-in, see [https://stat.ethz.ch/R-manual/R-devel/library/stats/html/fivenum.html R manual].
x <- c(0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578)
fivenum(x)
'''Output'''
[1] -1.9505959 -0.6767412 0.2332471 0.7460709 1.7313151
Racket
Racket's =quantile= functions use a different method to Tukey; so a new implementation was made.
#lang racket/base
(require math/private/statistics/quickselect)
;; racket's quantile uses "Method 1" of https://en.wikipedia.org/wiki/Quartile
;; Tukey (fivenum) uses "Method 2", so we will need a specialist median
(define (fivenum! data-v)
(define (tukey-median start end)
(define-values (n/2 parity) (quotient/remainder (- end start) 2))
(define mid (+ start n/2))
(if (zero? parity)
(/ (+ (data-kth-value! (+ mid (sub1 parity))) (data-kth-value! mid)) 2)
(data-kth-value! mid)))
(define n-data (let ((l (vector-length data-v)))
(if (zero? l)
(raise-argument-error 'data-v "nonempty (Vectorof Real)" data-v)
l)))
(define (data-kth-value! n) (kth-value! data-v n <))
(define subset-size (let-values (((n/2 parity) (quotient/remainder n-data 2))) (+ n/2 parity)))
(vector (data-kth-value! 0)
(tukey-median 0 subset-size)
(tukey-median 0 n-data)
(tukey-median (- n-data subset-size) n-data)
(data-kth-value! (sub1 n-data))))
(define (fivenum data-seq)
(fivenum! (if (and (vector? data-seq) (not (immutable? data-seq)))
data-seq
(for/vector ((datum data-seq)) datum))))
(module+ test
(require rackunit
racket/vector)
(check-equal? #(14 14 14 14 14) (fivenum #(14)) "Minimal case")
(check-equal? #(8 11 14 17 20) (fivenum #(8 14 20)) "3-value case")
(check-equal? #(8 11 15 18 20) (fivenum #(8 14 16 20)) "4-value case")
(define x1-seq #(36 40 7 39 41 15))
(define x1-v (vector-copy x1-seq))
(check-equal? x1-seq x1-v "before fivenum! sequence and vector were not `equal?`")
(check-equal? #(7 15 #e37.5 40 41) (fivenum! x1-v) "Test against Go results x1")
(check-not-equal? x1-seq x1-v "fivenum! did not mutate mutable input vectors")
(check-equal? #(6 #e25.5 40 #e42.5 49) (fivenum #(15 6 42 41 7 36 49 40 39 47 43)) "Test against Go results x2")
(check-equal? #(-1.95059594 -0.676741205 0.23324706 0.746070945 1.73131507)
(fivenum (vector 0.14082834 0.09748790 1.73131507 0.87636009 -1.95059594 0.73438555
-0.03035726 1.46675970 -0.74621349 -0.72588772 0.63905160 0.61501527
-0.98983780 -1.00447874 -0.62759469 0.66206163 1.04312009 -0.10305385
0.75775634 0.32566578))
"Test against Go results x3"))
This program passes its tests silently.
REXX
Programming note: this REXX program uses a unity─based array.
/*REXX program computes the five─number summary (LO─value, p25, medium, p75, HI─value).*/
parse arg x
if x='' then x= 15 6 42 41 7 36 49 40 39 47 43 /*Not specified? Then use the defaults*/
say 'input numbers: ' space(x) /*display the original list of numbers.*/
call 5num /*invoke the five─number function. */
say ' five─numbers: ' result /*display " " " results. */
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
bSort: procedure expose @.; parse arg n; m=n-1 /*N: the number of @ array elements.*/
do m=m for m by -1 until ok; ok= 1 /*keep sorting the @ array 'til done.*/
do j=1 for m; k= j + 1; if @.j<=@.k then iterate /*In order? Good.*/
parse value @.j @.k 0 with @.k @.j ok /*swap 2 elements; flag as ¬done.*/
end /*j*/
end /*m*/; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
med: arg s,e; $=e-s+1; m=s+$%2; if $//2 then return @.m; _=m-1; return (@._+@.m)/2
/*──────────────────────────────────────────────────────────────────────────────────────*/
5num: #= words(x); if #==0 then return '***error*** array is empty.'
parse var x . 1 LO . 1 HI . /*assume values for LO and HI (for now)*/
q2= # % 2; er= '***error*** element'
do j=1 for #; @.j= word(x, j)
if \datatype(@.j, 'N') then return er j "isn't numeric: " @.j
LO= min(LO, @.j); HI= max(HI, @.j)
end /*j*/ /* [↑] traipse thru array, find min,max*/
call bSort # /*use a bubble sort (easiest to code). */
if #//2 then p25= q2 /*calculate the second quartile number. */
else p25= q2 - 1 /* " " " " " */
return LO med(1, p25) med(1, #) med(q2, #) HI /*return list of 5 numbers.*/
{{out|output|text= when using the default input of: 15 6 42 41 7 36 49 40 39 47 43 }}
input numbers: 15 6 42 41 7 36 49 40 39 47 43
five─numbers: 6 15 40 42 49
{{out|output|text= when using the (internal) default inputs of: 36 40 7 39 41 15 }}
input numbers: 36 40 7 39 41 15
five─numbers: 7 11 37.5 39.5 41
Ruby
{{trans|Perl}}
def fivenum(array)
sorted_arr = array.sort
n = array.size
n4 = (((n + 3).to_f / 2.to_f) / 2.to_f).floor
d = Array.[](1, n4, ((n.to_f + 1) / 2).to_i, n + 1 - n4, n)
sum_array = []
(0..4).each do |e| # each loops have local scope, for loops don't
index_floor = (d[e] - 1).floor
index_ceil = (d[e] - 1).ceil
sum_array.push(0.5 * (sorted_arr[index_floor] + sorted_arr[index_ceil]))
end
sum_array
end
test_array = [15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43]
tukey_array = fivenum(test_array)
p tukey_array
test_array = [36, 40, 7, 39, 41, 15]
tukey_array = fivenum(test_array)
p tukey_array
test_array = [0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578]
tukey_array = fivenum(test_array)
p tukey_array
{{out}}
[6.0, 15.0, 40.0, 43.0, 49.0]
[7.0, 15.0, 36.0, 40.0, 41.0]
[-1.95059594, -0.72588772, 0.14082834, 0.75775634, 1.73131507]
SAS
/* build a dataset */
data test;
do i=1 to 10000;
x=rannor(12345);
output;
end;
keep x;
run;
/* compute the five numbers */
proc means data=test min p25 median p75 max;
var x;
run;
'''Output'''
| Analysis Variable : x | ||||
| Minimum | 25th Pctl | Median | 75th Pctl | Maximum |
| -4.0692299 | -0.6533022 | 0.0066299 | 0.6768043 | 4.1328026 |
Scala
Array based solution
import java.util
object Fivenum extends App {
val xl = Array(
Array(15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0),
Array(36.0, 40.0, 7.0, 39.0, 41.0, 15.0),
Array(0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527, -0.98983780,
-1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578)
)
for (x <- xl) println(f"${util.Arrays.toString(fivenum(x))}%s\n\n")
def fivenum(x: Array[Double]): Array[Double] = {
require(x.forall(!_.isNaN), "Unable to deal with arrays containing NaN")
def median(x: Array[Double], start: Int, endInclusive: Int): Double = {
val size = endInclusive - start + 1
require(size > 0, "Array slice cannot be empty")
val m = start + size / 2
if (size % 2 == 1) x(m) else (x(m - 1) + x(m)) / 2.0
}
val result = new Array[Double](5)
util.Arrays.sort(x)
result(0) = x(0)
result(2) = median(x, 0, x.length - 1)
result(4) = x(x.length - 1)
val m = x.length / 2
val lowerEnd = if (x.length % 2 == 1) m else m - 1
result(1) = median(x, 0, lowerEnd)
result(3) = median(x, m, x.length - 1)
result
}
}
{{Out}}See it running in your browser by [https://scalafiddle.io/sf/8s0OdOO/2 ScalaFiddle (JavaScript, non JVM)] or by [https://scastie.scala-lang.org/Ady3dSnoRRKNhCaZYIVbig Scastie (JVM)].
Sidef
{{trans|Perl 6}}
func fourths(e) {
var t = ((e>>1) / 2)
[0, t, e/2, e - t, e]
}
func fivenum(nums) {
var x = nums.sort
var d = fourths(x.end)
([x[d.map{.floor}]] ~Z+ [x[d.map{.ceil}]]) »/» 2
}
var nums = [
[15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43],
[36, 40, 7, 39, 41, 15], [
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594,
0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772,
0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469,
0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578,
]]
nums.each { say fivenum(_).join(', ') }
{{out}}
6, 25.5, 40, 42.5, 49
7, 15, 37.5, 40, 41
-1.95059594, -0.676741205, 0.23324706, 0.746070945, 1.73131507
Stata
First build a dataset:
clear
set seed 17760704
qui set obs 10000
gen x=rnormal()
The '''[https://www.stata.com/help.cgi?summarize summarize]''' command produces all the required statistics, and more:
qui sum x, detail
di r(min),r(p25),r(p50),r(p75),r(max)
'''Output'''
-3.6345866 -.66536 .0026834 .68398139 3.7997103
It's also possible to use the '''[https://www.stata.com/help.cgi?tabstat tabstat]''' command
tabstat x, s(mi q ma)
'''Output'''
variable | min p25 p50 p75 max
-------------+--------------------------------------------------
x | -3.634587 -.66536 .0026834 .6839814 3.79971
----------------------------------------------------------------
Another example:
clear
mat a=0.14082834\0.09748790\1.73131507\0.87636009\-1.95059594\ ///
0.73438555\-0.03035726\1.46675970\-0.74621349\-0.72588772\ ///
0.63905160\0.61501527\-0.98983780\-1.00447874\-0.62759469\ ///
0.66206163\1.04312009\-0.10305385\0.75775634\0.32566578
svmat a
tabstat a1, s(mi q ma)
'''Output'''
variable | min p25 p50 p75 max
-------------+--------------------------------------------------
a1 | -1.950596 -.6767412 .2332471 .746071 1.731315
----------------------------------------------------------------
VBA
Uses [[Sorting_algorithms/Quicksort#VBA|Quicksort]]. {{trans|Phix}}
Option Base 1
Private Function median(tbl As Variant, lo As Integer, hi As Integer)
Dim l As Integer: l = hi - lo + 1
Dim m As Integer: m = lo + WorksheetFunction.Floor_Precise(l / 2)
If l Mod 2 = 1 Then
median = tbl(m)
Else
median = (tbl(m - 1) + tbl(m)) / 2
End if
End Function
Private Function fivenum(tbl As Variant) As Variant
Sort tbl, UBound(tbl)
Dim l As Integer: l = UBound(tbl)
Dim m As Integer: m = WorksheetFunction.Floor_Precise(l / 2) + l Mod 2
Dim r(5) As String
r(1) = CStr(tbl(1))
r(2) = CStr(median(tbl, 1, m))
r(3) = CStr(median(tbl, 1, l))
r(4) = CStr(median(tbl, m + 1, l))
r(5) = CStr(tbl(l))
fivenum = r
End Function
Public Sub main()
Dim x1 As Variant, x2 As Variant, x3 As Variant
x1 = [{15, 6, 42, 41, 7, 36, 49, 40, 39, 47, 43}]
x2 = [{36, 40, 7, 39, 41, 15}]
x3 = [{0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555, -0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385, 0.75775634, 0.32566578}]
Debug.Print Join(fivenum(x1), " | ")
Debug.Print Join(fivenum(x2), " | ")
Debug.Print Join(fivenum(x3), " | ")
End Sub
{{out}}
6 | 25,5 | 40 | 43 | 49
7 | 15 | 37,5 | 40 | 41
-1,95059594 | -0,676741205 | 0,23324706 | 0,746070945 | 1,73131507
Visual Basic .NET
{{trans|C#}}
Imports System.Runtime.CompilerServices
Imports System.Text
Module Module1
<Extension()>
Function AsString(Of T)(c As ICollection(Of T), Optional format As String = "{0}") As String
Dim sb As New StringBuilder("[")
Dim it = c.GetEnumerator()
If it.MoveNext() Then
sb.AppendFormat(format, it.Current)
End If
While it.MoveNext()
sb.Append(", ")
sb.AppendFormat(format, it.Current)
End While
Return sb.Append("]").ToString()
End Function
Function Median(x As Double(), start As Integer, endInclusive As Integer) As Double
Dim size = endInclusive - start + 1
If size <= 0 Then
Throw New ArgumentException("Array slice cannot be empty")
End If
Dim m = start + size \ 2
Return If(size Mod 2 = 1, x(m), (x(m - 1) + x(m)) / 2.0)
End Function
Function Fivenum(x As Double()) As Double()
For Each d In x
If Double.IsNaN(d) Then
Throw New ArgumentException("Unable to deal with arrays containing NaN")
End If
Next
Array.Sort(x)
Dim result(4) As Double
result(0) = x.First()
result(2) = Median(x, 0, x.Length - 1)
result(4) = x.Last()
Dim m = x.Length \ 2
Dim lowerEnd = If(x.Length Mod 2 = 1, m, m - 1)
result(1) = Median(x, 0, lowerEnd)
result(3) = Median(x, m, x.Length - 1)
Return result
End Function
Sub Main()
Dim x1 = {
New Double() {15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0},
New Double() {36.0, 40.0, 7.0, 39.0, 41.0, 15.0},
New Double() {
0.14082834, 0.0974879, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.4667597, -0.74621349, -0.72588772, 0.6390516, 0.61501527,
-0.9898378, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578
}
}
For Each x In x1
Dim result = Fivenum(x)
Console.WriteLine(result.AsString("{0:F8}"))
Next
End Sub
End Module
{{out}}
[6.00000000, 25.50000000, 40.00000000, 42.50000000, 49.00000000]
[7.00000000, 15.00000000, 37.50000000, 40.00000000, 41.00000000]
[-1.95059594, -0.67674121, 0.23324706, 0.74607095, 1.73131507]
zkl
Uses GNU GSL library.
var [const] GSL=Import("zklGSL"); // libGSL (GNU Scientific Library)
fcn fiveNum(v){ // V is a GSL Vector, --> min, 1st qu, median, 3rd qu, max
v.sort();
return(v.min(),v.quantile(0.25),v.median(),v.quantile(0.75),v.max())
}
fiveNum(GSL.VectorFromData(
15.0, 6.0, 42.0, 41.0, 7.0, 36.0, 49.0, 40.0, 39.0, 47.0, 43.0)).println();
println(fiveNum(GSL.VectorFromData(36.0, 40.0, 7.0, 39.0, 41.0, 15.0)));
v:=GSL.VectorFromData(
0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555,
-0.03035726, 1.46675970, -0.74621349, -0.72588772, 0.63905160, 0.61501527,
-0.98983780, -1.00447874, -0.62759469, 0.66206163, 1.04312009, -0.10305385,
0.75775634, 0.32566578);
println(fiveNum(v));
{{out}}
L(6,25.5,40,42.5,49)
L(7,20.25,37.5,39.75,41)
L(-1.9506,-0.652168,0.233247,0.740228,1.73132)