Map range
From Rosetta Code
You are encouraged to solve this task according to the task description, using any language you may know.
The task is to write a function/subroutine/... that takes two ranges and a real number, and returns the mapping of the real number from the first to the second range. Use this function to map values from the range [0, 10] to the range [-1, 0].
Extra credit: Show additional idiomatic ways of performing the mapping, using tools available to the language.
[edit] ACL2
(defun mapping (a1 a2 b1 b2 s)
(+ b1 (/ (* (- s a1)
(- b2 b1))
(- a2 a1))))
[edit] Ada
with Ada.Text_IO;
procedure Map is
type First_Range is new Float range 0.0 .. 10.0;
type Second_Range is new Float range -1.0 .. 0.0;
function Translate (Value : First_Range) return Second_Range is
B1 : Float := Float (Second_Range'First);
B2 : Float := Float (Second_Range'Last);
A1 : Float := Float (First_Range'First);
A2 : Float := Float (First_Range'Last);
Result : Float;
begin
Result := B1 + (Float (Value) - A1) * (B2 - B1) / (A2 - A1);
return Second_Range (Result);
end;
function Translate (Value : Second_Range) return First_Range is
B1 : Float := Float (First_Range'First);
B2 : Float := Float (First_Range'Last);
A1 : Float := Float (Second_Range'First);
A2 : Float := Float (Second_Range'Last);
Result : Float;
begin
Result := B1 + (Float (Value) - A1) * (B2 - B1) / (A2 - A1);
return First_Range (Result);
end;
Test_Value : First_Range := First_Range'First;
begin
loop
Ada.Text_IO.Put_Line (First_Range'Image (Test_Value) & " maps to: "
& Second_Range'Image (Translate (Test_Value)));
exit when Test_Value = First_Range'Last;
Test_Value := Test_Value + 1.0;
end loop;
end Map;
Output:
0.00000E+00 maps to: -1.00000E+00 1.00000E+00 maps to: -9.00000E-01 2.00000E+00 maps to: -8.00000E-01 3.00000E+00 maps to: -7.00000E-01 4.00000E+00 maps to: -6.00000E-01 5.00000E+00 maps to: -5.00000E-01 6.00000E+00 maps to: -4.00000E-01 7.00000E+00 maps to: -3.00000E-01 8.00000E+00 maps to: -2.00000E-01 9.00000E+00 maps to: -1.00000E-01 1.00000E+01 maps to: 0.00000E+00
[edit] AutoHotkey
mapRange(a1, a2, b1, b2, s)
{
return b1 + (s-a1)*(b2-b1)/(a2-a1)
}
out := "Mapping [0,10] to [-1,0] at intervals of 1:`n"
Loop 11
out .= "f(" A_Index-1 ") = " mapRange(0,10,-1,0,A_Index-1) "`n"
MsgBox % out
[edit] Axiom
Axiom provides a Segment domain for intervals. The following uses a closure for a mapRange function over fields, which provides for some generality.
)abbrev package TESTP TestPackageUse:
TestPackage(R:Field) : with
mapRange: (Segment(R), Segment(R)) -> (R->R)
== add
mapRange(fromRange, toRange) ==
(a1,a2,b1,b2) := (lo fromRange,hi fromRange,lo toRange,hi toRange)
(x:R):R +-> b1+(x-a1)*(b2-b1)/(a2-a1)
f := mapRange(1..10,a..b)Output:
[(xi,f xi) for xi in 1..10]
b + 8a 2b + 7a b + 2a 4b + 5a 5b + 4a
[(1,a), (2,------), (3,-------), (4,------), (5,-------), (6,-------),
9 9 3 9 9
2b + a 7b + 2a 8b + a
(7,------), (8,-------), (9,------), (10,b)]
3 9 9
Type: List(Tuple(Fraction(Polynomial(Integer))))
[edit] bc
/* map s from [a, b] to [c, d] */Output:
define m(a, b, c, d, s) {
return (c + (s - a) * (d - c) / (b - a))
}
scale = 6 /* division to 6 decimal places */
"[0, 10] => [-1, 0]
"
for (i = 0; i <= 10; i += 2) {
/*
* If your bc(1) has a print statement, you can try
* print i, " => ", m(0, 10, -1, 0, i), "\n"
*/
i; " => "; m(0, 10, -1, 0, i)
}
quit
[0, 10] => [-1, 0] 0 => -1.000000 2 => -.800000 4 => -.600000 6 => -.400000 8 => -.200000 10 => 0.000000
[edit] Bracmat
( ( mapRange
= a1,a2,b1,b2,s
. !arg:(?a1,?a2.?b1,?b2.?s)
& !b1+(!s+-1*!a1)*(!b2+-1*!b1)*(!a2+-1*!a1)^-1
)
& out$"Mapping [0,10] to [-1,0] at intervals of 1:"
& 0:?n
& whl
' ( !n:~>10
& out$("f(" !n ") = " flt$(mapRange$(0,10.-1,0.!n),2))
& 1+!n:?n
)
);
Output:
Mapping [0,10] to [-1,0] at intervals of 1: f( 0 ) = -1,00*10E0 f( 1 ) = -9,00*10E-1 f( 2 ) = -8,00*10E-1 f( 3 ) = -7,00*10E-1 f( 4 ) = -6,00*10E-1 f( 5 ) = -5,00*10E-1 f( 6 ) = -4,00*10E-1 f( 7 ) = -3,00*10E-1 f( 8 ) = -2,00*10E-1 f( 9 ) = -1,00*10E-1 f( 10 ) = 0
[edit] C
#include <stdio.h>
double mapRange(double a1,double a2,double b1,double b2,double s)
{
return b1 + (s-a1)*(b2-b1)/(a2-a1);
}
int main()
{
int i;
puts("Mapping [0,10] to [-1,0] at intervals of 1:");
for(i=0;i<=10;i++)
{
printf("f(%d) = %g\n",i,mapRange(0,10,-1,0,i));
}
return 0;
}
The output is :
Mapping [0,10] to [-1,0] at intervals of 1:
f(0) = -1
f(1) = -0.9
f(2) = -0.8
f(3) = -0.7
f(4) = -0.6
f(5) = -0.5
f(6) = -0.4
f(7) = -0.3
f(8) = -0.2
f(9) = -0.1
f(10) = 0
[edit] C++
This example defines a template function to handle the mapping, using two std::pair objects to define the source and destination ranges. It returns the provided value mapped into the target range.
It's not written efficiently; certainly, there can be fewer explicit temporary variables. The use of the template offers a choice in types for precision and accuracy considerations, though one area for improvement might be to allow a different type for intermediate calculations.
#include <iostream>
#include <utility>
template<typename tVal>
tVal map_value(std::pair<tVal,tVal> a, std::pair<tVal, tVal> b, tVal inVal)
{
tVal inValNorm = inVal - a.first;
tVal aUpperNorm = a.second - a.first;
tVal normPosition = inValNorm / aUpperNorm;
tVal bUpperNorm = b.second - b.first;
tVal bValNorm = normPosition * bUpperNorm;
tVal outVal = b.first + bValNorm;
return outVal;
}
int main()
{
std::pair<float,float> a(0,10), b(-1,0);
for(float value = 0.0; 10.0 >= value; ++value)
std::cout << "map_value(" << value << ") = " << map_value(a, b, value) << std::endl;
return 0;
}
Output:
map_value(0) = -1 map_value(1) = -0.9 map_value(2) = -0.8 map_value(3) = -0.7 map_value(4) = -0.6 map_value(5) = -0.5 map_value(6) = -0.4 map_value(7) = -0.3 map_value(8) = -0.2 map_value(9) = -0.1 map_value(10) = 0
[edit] Clojure
(defn maprange [[a1 a2] [b1 b2] s]
(+ b1 (/ (* (- s a1) (- b2 b1)) (- a2 a1))))
> (doseq [s (range 11)]
(printf "%2s maps to %s\n" s (maprange [0 10] [-1 0] s)))
0 maps to -1
1 maps to -9/10
2 maps to -4/5
3 maps to -7/10
4 maps to -3/5
5 maps to -1/2
6 maps to -2/5
7 maps to -3/10
8 maps to -1/5
9 maps to -1/10
10 maps to 0
[edit] D
import std.stdio;
double mapRange(in double[] a, in double[] b, in double s)
pure nothrow {
return b[0] + ((s - a[0]) * (b[1] - b[0]) / (a[1] - a[0]));
}
void main() {
const r1 = [0.0, 10.0];
const r2 = [-1.0, 0.0];
foreach (s; 0 .. 11)
writefln("%2d maps to %5.2f", s, mapRange(r1, r2, s));
}
- Output:
0 maps to -1.00 1 maps to -0.90 2 maps to -0.80 3 maps to -0.70 4 maps to -0.60 5 maps to -0.50 6 maps to -0.40 7 maps to -0.30 8 maps to -0.20 9 maps to -0.10 10 maps to 0.00
[edit] Erlang
-module(map_range).
-export([map_value/3]).
map_value({A1,A2},{B1,B2},S) ->
B1 + (S - A1) * (B2 - B1) / (A2 - A1).
[edit] Euphoria
function map_range(sequence a, sequence b, atom s)
return b[1]+(s-a[1])*(b[2]-b[1])/(a[2]-a[1])
end function
for i = 0 to 10 do
printf(1, "%2g maps to %4g\n", {i, map_range({0,10},{-1,0},i)})
end for
Output:
0 maps to -1 1 maps to -0.9 2 maps to -0.8 3 maps to -0.7 4 maps to -0.6 5 maps to -0.5 6 maps to -0.4 7 maps to -0.3 8 maps to -0.2 9 maps to -0.1 10 maps to 0
[edit] Factor
USE: locals
:: map-range ( a1 a2 b1 b2 x -- y )
x a1 - b2 b1 - * a2 a1 - / b1 + ;
Or:
USING: locals infix ;
:: map-range ( a1 a2 b1 b2 x -- y )
[infix
b1 + (x - a1) * (b2 - b1) / (a2 - a1)
infix] ;
Test run:
10 iota [| x | 0 10 -1 0 x map-range ] map . ! { -1 -9/10 -4/5 -7/10 -3/5 -1/2 -2/5 -3/10 -1/5 -1/10 }
[edit] Fantom
class FRange
{
const Float low
const Float high
// in constructing a range, ensure the low value is smaller than high
new make (Float low, Float high)
{
this.low = ( low <= high ? low : high )
this.high = ( low <= high ? high : low )
}
// return range as a string
override Str toStr () { "[$low,$high]" }
// return a point in given range interpolated into this range
Float remap (Float point, FRange given)
{
this.low + (point - given.low) * (this.high - this.low) / (given.high - given.low)
}
}
class Main
{
public static Void main ()
{
range1 := FRange (0f, 10f)
range2 := FRange (-1f, 0f)
11.times |Int n|
{
m := range2.remap (n.toFloat, range1)
echo ("Value $n in ${range1} maps to $m in ${range2}")
}
}
}
Output:
Value 0 in [0.0,10.0] maps to -1.0 in [-1.0,0.0] Value 1 in [0.0,10.0] maps to -0.9 in [-1.0,0.0] Value 2 in [0.0,10.0] maps to -0.8 in [-1.0,0.0] Value 3 in [0.0,10.0] maps to -0.7 in [-1.0,0.0] Value 4 in [0.0,10.0] maps to -0.6 in [-1.0,0.0] Value 5 in [0.0,10.0] maps to -0.5 in [-1.0,0.0] Value 6 in [0.0,10.0] maps to -0.4 in [-1.0,0.0] Value 7 in [0.0,10.0] maps to -0.30000000000000004 in [-1.0,0.0] Value 8 in [0.0,10.0] maps to -0.19999999999999996 in [-1.0,0.0] Value 9 in [0.0,10.0] maps to -0.09999999999999998 in [-1.0,0.0] Value 10 in [0.0,10.0] maps to 0.0 in [-1.0,0.0]
[edit] Forth
\ linear interpolation
: lerp ( b2 b1 a2 a1 s -- t )
fover f-
frot frot f- f/
frot frot fswap fover f- frot f*
f+ ;
: test 11 0 do 0e -1e 10e 0e i s>f lerp f. loop ;
There is less stack shuffling if you use origin and range instead of endpoints for intervals. (o = a1, r = a2-a1)
: lerp ( o2 r2 r1 o1 s -- t ) fswap f- fswap f/ f* f+ ;
: test 11 0 do -1e 1e 10e 0e i s>f lerp f. loop ;
[edit] Fortran
program Map
implicit none
real :: t
integer :: i
do i = 0, 10
t = Maprange((/0.0, 10.0/), (/-1.0, 0.0/), real(i))
write(*,*) i, " maps to ", t
end do
contains
function Maprange(a, b, s)
real :: Maprange
real, intent(in) :: a(2), b(2), s
Maprange = (s-a(1)) * (b(2)-b(1)) / (a(2)-a(1)) + b(1)
end function Maprange
end program Map
[edit] Go
Basic task
package main
import "fmt"
type rangeBounds struct {
b1, b2 float64
}
func mapRange(x, y rangeBounds, n float64) float64 {
return y.b1 + (n - x.b1) * (y.b2 - y.b1) / (x.b2 - x.b1)
}
func main() {
r1 := rangeBounds{0, 10}
r2 := rangeBounds{-1, 0}
for n := float64(0); n <= 10; n += 2 {
fmt.Println(n, "maps to", mapRange(r1, r2, n))
}
}
Output:
0 maps to -1 2 maps to -0.8 4 maps to -0.6 6 maps to -0.4 8 maps to -0.19999999999999996 10 maps to 0
Extra credit
First, a function literal replaces the mapping function specified by the basic task. This allows a simpler parameter signature and also allows things to be precomputed for efficiency. newMapRange checks the direction of the first range and if it is decreasing, reverses both ranges. This simplifies an out-of-range check in the function literal. Also, the slope and intercept of the linear function are computed. This allows the range mapping to use the slope intercept formula which is computationally more efficient that the two point formula.
Second, ", ok" is a Go idiom. It takes advantage of Go's multiple return values and multiple assignment to return a success/failure disposition. In the case of this task, the result t is undefined if the input s is out of range.
package main
import "fmt"
type rangeBounds struct {
b1, b2 float64
}
func newRangeMap(xr, yr rangeBounds) func(float64) (float64, bool) {
// normalize direction of ranges so that out-of-range test works
if xr.b1 > xr.b2 {
xr.b1, xr.b2 = xr.b2, xr.b1
yr.b1, yr.b2 = yr.b2, yr.b1
}
// compute slope, intercept
m := (yr.b2 - yr.b1) / (xr.b2 - xr.b1)
b := yr.b1 - m*xr.b1
// return function literal
return func(x float64) (y float64, ok bool) {
if x < xr.b1 || x > xr.b2 {
return 0, false // out of range
}
return m*x + b, true
}
}
func main() {
rm := newRangeMap(rangeBounds{0, 10}, rangeBounds{-1, 0})
for s := float64(-2); s <= 12; s += 2 {
t, ok := rm(s)
if ok {
fmt.Printf("s: %5.2f t: %5.2f\n", s, t)
} else {
fmt.Printf("s: %5.2f out of range\n", s)
}
}
}
Output:
s: -2.00 out of range s: 0.00 t: -1.00 s: 2.00 t: -0.80 s: 4.00 t: -0.60 s: 6.00 t: -0.40 s: 8.00 t: -0.20 s: 10.00 t: 0.00 s: 12.00 out of range
[edit] Groovy
def mapRange(a1, a2, b1, b2, s) {
b1 + ((s - a1) * (b2 - b1)) / (a2 - a1)
}
(0..10).each { s ->
println(s + " in [0, 10] maps to " + mapRange(0, 10, -1, 0, s) + " in [-1, 0].")
}
Output:
0 in [0, 10] maps to -1 in [-1, 0]. 1 in [0, 10] maps to -0.9 in [-1, 0]. 2 in [0, 10] maps to -0.8 in [-1, 0]. 3 in [0, 10] maps to -0.7 in [-1, 0]. 4 in [0, 10] maps to -0.6 in [-1, 0]. 5 in [0, 10] maps to -0.5 in [-1, 0]. 6 in [0, 10] maps to -0.4 in [-1, 0]. 7 in [0, 10] maps to -0.3 in [-1, 0]. 8 in [0, 10] maps to -0.2 in [-1, 0]. 9 in [0, 10] maps to -0.1 in [-1, 0]. 10 in [0, 10] maps to 0 in [-1, 0].
[edit] Haskell
Rather than handling only floating point numbers, the mapping function takes any number implementing the Fractional typeclass, which in our example also includes exact Rational numbers.
import Data.Ratio
import Text.Printf
-- Map a value from the range [a1,a2] to the range [b1,b2]. We don't check
-- for empty ranges.
mapRange :: (Fractional a) => (a, a) -> (a, a) -> a -> a
mapRange (a1,a2) (b1,b2) s = b1+(s-a1)*(b2-b1)/(a2-a1)
main = do
-- Perform the mapping over floating point numbers.
putStrLn "---------- Floating point ----------"
mapM_ (\n -> prtD n . mapRange (0,10) (-1,0) $ fromIntegral n) [0..10]
-- Perform the same mapping over exact rationals.
putStrLn "---------- Rationals ----------"
mapM_ (\n -> prtR n . mapRange (0,10) (-1,0) $ n%1) [0..10]
where prtD :: PrintfType r => Integer -> Double -> r
prtD n x = printf "%2d -> %6.3f\n" n x
prtR :: PrintfType r => Integer -> Rational -> r
prtR n x = printf "%2d -> %s\n" n (show x)
Output:
---------- Floating point ---------- 0 -> -1.000 1 -> -0.900 2 -> -0.800 3 -> -0.700 4 -> -0.600 5 -> -0.500 6 -> -0.400 7 -> -0.300 8 -> -0.200 9 -> -0.100 10 -> 0.000 ---------- Rationals ---------- 0 -> (-1) % 1 1 -> (-9) % 10 2 -> (-4) % 5 3 -> (-7) % 10 4 -> (-3) % 5 5 -> (-1) % 2 6 -> (-2) % 5 7 -> (-3) % 10 8 -> (-1) % 5 9 -> (-1) % 10 10 -> 0 % 1
[edit] Icon and Unicon
record Range(a, b)
# note, we force 'n' to be real, which means recalculation will
# be using real numbers, not integers
procedure remap (range1, range2, n : real)
if n < range2.a | n > range2.b then fail # n out of given range
return range1.a + (n - range2.a) * (range1.b - range1.a) / (range2.b - range2.a)
end
procedure range_string (range)
return "[" || range.a || ", " || range.b || "]"
end
procedure main ()
range1 := Range (0, 10)
range2 := Range (-1, 0)
# if i is out of range1, then 'remap' fails, so only valid changes are written
every i := -2 to 12 do {
if m := remap (range2, range1, i)
then write ("Value " || i || " in " || range_string (range1) ||
" maps to " || m || " in " || range_string (range2))
}
end
Icon does not permit the type declaration, as Unicon does. For Icon, replace 'remap' with:
procedure remap (range1, range2, n)
n *:= 1.0
if n < range2.a | n > range2.b then fail # n out of given range
return range1.a + (n - range2.a) * (range1.b - range1.a) / (range2.b - range2.a)
end
Output:
Value 0 in [0, 10] maps to -1.0 in [-1, 0] Value 1 in [0, 10] maps to -0.9 in [-1, 0] Value 2 in [0, 10] maps to -0.8 in [-1, 0] Value 3 in [0, 10] maps to -0.7 in [-1, 0] Value 4 in [0, 10] maps to -0.6 in [-1, 0] Value 5 in [0, 10] maps to -0.5 in [-1, 0] Value 6 in [0, 10] maps to -0.4 in [-1, 0] Value 7 in [0, 10] maps to -0.3 in [-1, 0] Value 8 in [0, 10] maps to -0.2 in [-1, 0] Value 9 in [0, 10] maps to -0.1 in [-1, 0] Value 10 in [0, 10] maps to 0.0 in [-1, 0]
[edit] J
maprange=:2 :0
'a1 a2'=.m
'b1 b2'=.n
b1+((y-a1)*b2-b1)%a2-a1
)
NB. this version defers all calculations to runtime, but mirrors exactly the task formulation
Or
maprange=:2 :0
'a1 a2'=.m
'b1 b2'=.n
b1 + ((b2-b1)%a2-a1) * -&a1
)
NB. this version precomputes the scaling ratio
Example use:
2 4 maprange 5 11 (2.718282 3 3.141592)
7.15485 8 8.42478
or
adjust=:2 4 maprange 5 11 NB. save the derived function as a named entity
adjust 2.718282 3 3.141592
7.15485 8 8.42478
Required example:
0 10 maprange _1 0 i.11
_1 _0.9 _0.8 _0.7 _0.6 _0.5 _0.4 _0.3 _0.2 _0.1 0
[edit] Java
public class Range {
public static void main(String[] args){
for(float s = 0;s <= 10; s++){
System.out.println(s + " in [0, 10] maps to "+
mapRange(0, 10, -1, 0, s)+" in [-1, 0].");
}
}
public static double mapRange(double a1, double a2, double b1, double b2, double s){
return b1 + ((s - a1)*(b2 - b1))/(a2 - a1);
}
}
Output:
0.0 in [0, 10] maps to -1.0 in [-1, 0]. 1.0 in [0, 10] maps to -0.9 in [-1, 0]. 2.0 in [0, 10] maps to -0.8 in [-1, 0]. 3.0 in [0, 10] maps to -0.7 in [-1, 0]. 4.0 in [0, 10] maps to -0.6 in [-1, 0]. 5.0 in [0, 10] maps to -0.5 in [-1, 0]. 6.0 in [0, 10] maps to -0.4 in [-1, 0]. 7.0 in [0, 10] maps to -0.30000000000000004 in [-1, 0]. 8.0 in [0, 10] maps to -0.19999999999999996 in [-1, 0]. 9.0 in [0, 10] maps to -0.09999999999999998 in [-1, 0]. 10.0 in [0, 10] maps to 0.0 in [-1, 0].
The differences in 7, 8, and 9 come from double math. Similar issues show even when using float types.
[edit] JavaScript
mapRange = function(t, from, to) {
return to[0]+(t-from[0])*(to[1]-to[0])/(from[1]-from[0]);
}
[edit] K
f:{[a1;a2;b1;b2;s] b1+(s-a1)*(b2-b1)%(a2-a1)}
+(a; f[0;10;-1;0]'a:!11)
((0;-1.0)
(1;-0.9)
(2;-0.8)
(3;-0.7)
(4;-0.6)
(5;-0.5)
(6;-0.4)
(7;-0.3)
(8;-0.2)
(9;-0.1)
(10;0.0))
[edit] Logo
to interpolate :s :a1 :a2 :b1 :b2
output (:s-:a1) / (:a2-:a1) * (:b2-:b1) + :b1
end
for [i 0 10] [print interpolate :i 0 10 -1 0]
[edit] Lua
function map_range( a1, a2, b1, b2, s )
return b1 + (s-a1)*(b2-b1)/(a2-a1)
end
for i = 0, 10 do
print( string.format( "f(%d) = %f", i, map_range( 0, 10, -1, 0, i ) ) )
end
[edit] Mathematica
Such a function is already built in
Rescale[#,{0,10},{-1,0}]&/@Range[0,10]
Output: {-1., -0.9, -0.8, -0.7, -0.6, -0.5, -0.4, -0.3, -0.2, -0.1, 0.}
[edit] Maxima
maprange(a, b, c, d) := buildq([e: ratsimp(('x - a)*(d - c)/(b - a) + c)],
lambda([x], e))$
f: maprange(0, 10, -1, 0);
[edit] Nemerle
using System;
using System.Console;
module Maprange
{
Maprange(a : double * double, b : double * double, s : double) : double
{
def (a1, a2) = a; def (b1, b2) = b;
b1 + (((s - a1) * (b2 - b1))/(a2 - a1))
}
Main() : void
{
foreach (i in [0 .. 10])
WriteLine("{0, 2:f0} maps to {1:f1}", i, Maprange((0.0, 10.0), (-1.0, 0.0), i));
}
}
[edit] Objeck
bundle Default {
class Range {
function : MapRange(a1:Float, a2:Float, b1:Float, b2:Float, s:Float) ~ Float {
return b1 + (s-a1)*(b2-b1)/(a2-a1);
}
function : Main(args : String[]) ~ Nil {
"Mapping [0,10] to [-1,0] at intervals of 1:"->PrintLine();
for(i := 0.0; i <= 10.0; i += 1;) {
IO.Console->Print("f(")->Print(i->As(Int))->Print(") = ")->PrintLine(MapRange(0.0, 10.0, -1.0, 0.0, i));
};
}
}
}
Output:
Mapping [0,10] to [-1,0] at intervals of 1: f(0) = -1 f(1) = -0.9 f(2) = -0.8 f(3) = -0.7 f(4) = -0.6 f(5) = -0.5 f(6) = -0.4 f(7) = -0.3 f(8) = -0.2 f(9) = -0.1 f(10) = 0
[edit] OCaml
let map_range (a1, a2) (b1, b2) s =
b1 +. ((s -. a1) *. (b2 -. b1) /. (a2 -. a1))
let () =
print_endline "Mapping [0,10] to [-1,0] at intervals of 1:";
for i = 0 to 10 do
Printf.printf "f(%d) = %g\n" i (map_range (0.0, 10.0) (-1.0, 0.0) (float i))
done
Output:
Mapping [0,10] to [-1,0] at intervals of 1: f(0) = -1 f(1) = -0.9 f(2) = -0.8 f(3) = -0.7 f(4) = -0.6 f(5) = -0.5 f(6) = -0.4 f(7) = -0.3 f(8) = -0.2 f(9) = -0.1 f(10) = 0
If range mapping is used in a heavy computational task we can reduce the number of calculations made using partial application and currying:
let map_range (a1, a2) (b1, b2) =
let v = (b2 -. b1) /. (a2 -. a1) in
function s ->
b1 +. ((s -. a1) *. v)
let () =
print_endline "Mapping [0,10] to [-1,0] at intervals of 1:";
let p = (map_range (0.0, 10.0) (-1.0, 0.0)) in
for i = 0 to 10 do
Printf.printf "f(%d) = %g\n" i (p (float i))
done
[edit] PARI/GP
Usage (e.g.): map([1,10],[0,5],8.)
map(r1,r2,x)=r2[1]+(x-r1[1])*(r2[2]-r2[1])/(r1[2]-r1[1])
[edit] Pascal
Program Map(output);
function MapRange(fromRange, toRange: array of real; value: real): real;
begin
MapRange := (value-fromRange[0]) * (toRange[1]-toRange[0]) / (fromRange[1]-fromRange[0]) + toRange[0];
end;
var
i: integer;
begin
for i := 0 to 10 do
writeln (i, ' maps to: ', MapRange([0.0, 10.0], [-1.0, 0.0], i):4:2);
end.
Output:
:> ./MapRange 0 maps to: -1.00 1 maps to: -0.90 2 maps to: -0.80 3 maps to: -0.70 4 maps to: -0.60 5 maps to: -0.50 6 maps to: -0.40 7 maps to: -0.30 8 maps to: -0.20 9 maps to: -0.10 10 maps to: 0.00
[edit] Perl
#!/usr/bin/perl -w
use strict ;
sub mapValue {
my ( $range1 , $range2 , $number ) = @_ ;
return ( $range2->[ 0 ] +
(( $number - $range1->[ 0 ] ) * ( $range2->[ 1 ] - $range2->[ 0 ] ) ) / ( $range1->[ -1 ]
- $range1->[ 0 ] ) ) ;
}
my @numbers = 0..10 ;
my @interval = ( -1 , 0 ) ;
print "The mapped value for $_ is " . mapValue( \@numbers , \@interval , $_ ) . " !\n" foreach @numbers ;
Output:
The mapped value for 0 is -1 ! The mapped value for 1 is -0.9 ! The mapped value for 2 is -0.8 ! The mapped value for 3 is -0.7 ! The mapped value for 4 is -0.6 ! The mapped value for 5 is -0.5 ! The mapped value for 6 is -0.4 ! The mapped value for 7 is -0.3 ! The mapped value for 8 is -0.2 ! The mapped value for 9 is -0.1 ! The mapped value for 10 is 0 !
[edit] Perl 6
use v6;
# Author: P. Seebauer
sub the_function(Range $a, Range $b, $s ) {
my ($a1, $a2, $b1, $b2) = ($a, $b)».bounds;
return $b1 + (($s-$a1) * ($b2-$b1) / ($a2-$a1));
}
for ^11 -> $x {say "$x maps to {the_function(0..10,-1..0, $x)}"}
%perl6 map_range.p6 0 maps to -1 1 maps to -0.9 2 maps to -0.8 3 maps to -0.7 4 maps to -0.6 5 maps to -0.5 6 maps to -0.4 7 maps to -0.3 8 maps to -0.2 9 maps to -0.1 10 maps to 0
A more idiomatic way would be to return a closure that does the mapping without have to supply the ranges every time:
sub getmapper(Range $a, Range $b) {
my ($a1, $a2, $b1, $b2) = ($a, $b)».bounds;
return -> $s { $b1 + (($s-$a1) * ($b2-$b1) / ($a2-$a1)) }
}
my &mapper = getmapper(0 .. 10, -1 .. 0);
for ^11 -> $x {say "$x maps to &mapper($x)"}
[edit] PicoLisp
(scl 1)
(de mapRange (Val A1 A2 B1 B2)
(+ B1 (*/ (- Val A1) (- B2 B1) (- A2 A1))) )
(for Val (range 0 10.0 1.0)
(prinl
(format (mapRange Val 0 10.0 -1.0 0) *Scl) ) )
Output:
-1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0.0
[edit] PL/I
map: procedure options (main); /* 24/11/2011 */
declare (a1, a2, b1, b2) float;
declare d fixed decimal (3,1);
do d = 0 to 10 by 0.9, 10;
put skip edit ( d, ' maps to ', map(0, 10, -1, 0, d) ) (f(5,1), a, f(10,6));
end;
map: procedure (a1, a2, b1, b2, s) returns (float);
declare (a1, a2, b1, b2, s) float;
return (b1 + (s - a1)*(b2 - b1) / (a2 - a1) );
end map;
end map;
OUTPUT:
0.0 maps to -1.000000 0.9 maps to -0.910000 1.8 maps to -0.820000 2.7 maps to -0.730000 3.6 maps to -0.640000 4.5 maps to -0.550000 5.4 maps to -0.460000 6.3 maps to -0.370000 7.2 maps to -0.280000 8.1 maps to -0.190000 9.0 maps to -0.100000 9.9 maps to -0.010000 10.0 maps to 0.000000
[edit] PureBasic
Structure RR
a.f
b.f
EndStructure
Procedure.f MapRange(*a.RR, *b.RR, s)
Protected.f a1, a2, b1, b2
a1=*a\a: a2=*a\b
b1=*b\a: b2=*b\b
ProcedureReturn b1 + ((s - a1) * (b2 - b1) / (a2 - a1))
EndProcedure
;- Test the function
If OpenConsole()
Define.RR Range1, Range2
Range1\a=0: Range1\b=10
Range2\a=-1:Range2\b=0
;
For i=0 To 10
PrintN(RSet(Str(i),2)+" maps to "+StrF(MapRange(@Range1, @Range2, i),1))
Next
EndIf
0 maps to -1.0 1 maps to -0.9 2 maps to -0.8 3 maps to -0.7 4 maps to -0.6 5 maps to -0.5 6 maps to -0.4 7 maps to -0.3 8 maps to -0.2 9 maps to -0.1 10 maps to 0.0
[edit] Python
>>> def maprange( a, b, s):
(a1, a2), (b1, b2) = a, b
return b1 + ((s - a1) * (b2 - b1) / (a2 - a1))
>>> for s in range(11):
print("%2g maps to %g" % (s, maprange( (0, 10), (-1, 0), s)))
0 maps to -1
1 maps to -0.9
2 maps to -0.8
3 maps to -0.7
4 maps to -0.6
5 maps to -0.5
6 maps to -0.4
7 maps to -0.3
8 maps to -0.2
9 maps to -0.1
10 maps to 0
[edit] REXX
(The different versions don't differ idiomaticly but just in style.)
All program versions could be made more robust by checking if the high and low ends of rangeA aren't equal.
[edit] version 1
/*REXX program maps numbers from one range to another range. */
rangeA='0 10'
rangeB='-1 0'
do j=0 to 10
say right(j,3) ' maps to ' mapRange(rangeA, rangeB, j)
end
exit
/*─────────────────────────────────────MapRANGE subroutine──────────────*/
mapRange: procedure; arg a1 a2,b1 b2,x; return b1+(x-a1)*(b2-b1)/(a2-a1)
Output:
0 maps to -1 1 maps to -0.9 2 maps to -0.8 3 maps to -0.7 4 maps to -0.6 5 maps to -0.5 6 maps to -0.4 7 maps to -0.3 8 maps to -0.2 9 maps to -0.1 10 maps to 0
[edit] version 2
/*REXX program maps numbers from one range to another range. */
do j=0 to 10
say right(j,3) ' maps to ' mapRange(0 10, -1 0, j)
end
exit
/*─────────────────────────────────────MapRANGE subroutine──────────────*/
mapRange: procedure; arg a1 a2,b1 b2,x; return b1+(x-a1)*(b2-b1)/(a2-a1)
[edit] version 3
/*REXX program maps numbers from one range to another range. */
rangeA='0 10'; parse var rangeA a1 a2
rangeB='-1 0'; parse var rangeB b1 b2
do j=0 for 11
say right(j,3) ' maps to ' b1+(x-a1)*(b2-b1)/(a2-a1)
end
[edit] Ruby
def map_range(a, b, s)
af, al, bf, bl = a.first, a.last, b.first, b.last
bf + (s - af).*(bl - bf).quo(al - af)
end
11.times do |s|
s *= 1.0 # force floating point
print s, " maps to ", map_range((0..10), (-1..0), s), "\n"
end
Numeric#quo acts like Numeric#/ but gives more precise result when both operands are Integers.
Output using floating-point arithmetic:
0 maps to -1.0 1 maps to -0.9 2 maps to -0.8 3 maps to -0.7 4 maps to -0.6 5 maps to -0.5 6 maps to -0.4 7 maps to -0.30000000000000004 8 maps to -0.19999999999999996 9 maps to -0.09999999999999998 10 maps to 0.0
To use rational arithmetic, delete s *= 1.0 and either require 'rational', or use Ruby 1.9 (which has Rational in the core library).
Output using rational arithmetic:
0 maps to -1/1 1 maps to -9/10 2 maps to -4/5 3 maps to -7/10 4 maps to -3/5 5 maps to -1/2 6 maps to -2/5 7 maps to -3/10 8 maps to -1/5 9 maps to -1/10 10 maps to 0/1
[edit] Scala
def mapRange(a1:Double, a2:Double, b1:Double, b2:Double, x:Double):Double=b1+(x-a1)*(b2-b1)/(a2-a1)
for(i <- 0 to 10)
println("%2d in [0, 10] maps to %5.2f in [-1, 0]".format(i, mapRange(0,10, -1,0, i)))
Output:
0 in [0, 10] maps to -1,00 in [-1, 0] 1 in [0, 10] maps to -0,90 in [-1, 0] 2 in [0, 10] maps to -0,80 in [-1, 0] 3 in [0, 10] maps to -0,70 in [-1, 0] 4 in [0, 10] maps to -0,60 in [-1, 0] 5 in [0, 10] maps to -0,50 in [-1, 0] 6 in [0, 10] maps to -0,40 in [-1, 0] 7 in [0, 10] maps to -0,30 in [-1, 0] 8 in [0, 10] maps to -0,20 in [-1, 0] 9 in [0, 10] maps to -0,10 in [-1, 0] 10 in [0, 10] maps to 0,00 in [-1, 0]
[edit] Seed7
$ include "seed7_05.s7i";
include "float.s7i";
const func float: mapRange (in float: a1, in float: a2, in float: b1, in float: b2, ref float: s) is
return b1 + (s-a1)*(b2-b1)/(a2-a1);
const proc: main is func
local
var integer: number is 0;
begin
writeln("Mapping [0,10] to [-1,0] at intervals of 1:");
for number range 0 to 10 do
writeln("f(" <& number <& ") = " <& mapRange(0.0, 10.0, -1.0, 0.0, flt(number)) digits 1);
end for;
end func;
Output:
Mapping [0,10] to [-1,0] at intervals of 1: f(0) = -1.0 f(1) = -0.9 f(2) = -0.8 f(3) = -0.7 f(4) = -0.6 f(5) = -0.5 f(6) = -0.4 f(7) = -0.3 f(8) = -0.2 f(9) = -0.1 f(10) = 0.0
[edit] Tcl
package require Tcl 8.5
proc rangemap {rangeA rangeB value} {
lassign $rangeA a1 a2
lassign $rangeB b1 b2
expr {$b1 + ($value - $a1)*double($b2 - $b1)/($a2 - $a1)}
}
Demonstration (using a curried alias to bind the ranges mapped from and to):
interp alias {} demomap {} rangemap {0 10} {-1 0}
for {set i 0} {$i <= 10} {incr i} {
puts [format "%2d -> %5.2f" $i [demomap $i]]
}
Output:
0 -> -1.00 1 -> -0.90 2 -> -0.80 3 -> -0.70 4 -> -0.60 5 -> -0.50 6 -> -0.40 7 -> -0.30 8 -> -0.20 9 -> -0.10 10 -> 0.00
[edit] Ursala
The function f is defined using pattern matching and substitution, taking a pair of pairs of interval endpoints and a number as parameters, and returning a number.
#import flo
f((("a1","a2"),("b1","b2")),"s") = plus("b1",div(minus("s","a1"),minus("a2","a1")))
#cast %eL
test = f* ((0.,10.),(-1.,0.))-* ari11/0. 10.
output:
< -1.000000e+00, -9.000000e-01, -8.000000e-01, -7.000000e-01, -6.000000e-01, -5.000000e-01, -4.000000e-01, -3.000000e-01, -2.000000e-01, -1.000000e-01, 0.000000e+00>
A more idiomatic way is to define f as a second order function
f(("a1","a2"),("b1","b2")) "s" = ...
with the same right hand side as above, so that it takes a pair of intervals and returns a function mapping numbers in one interval to numbers in the other.
An even more idiomatic way is to use the standard library function plin, which takes an arbitrarily long list of interval endpoints and returns a piecewise linear interpolation function.
