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I basically have a number, say 100.
I can increase it by 10 percent every time. So, :
1st run it would become 110,
2nd 121,
3rd 133 and so on.
I can have count of how much the value was increased. But how to expotentialy decrease the amount knowing the number of times it has been increased back to 100?
Is there a way to do it with simple math op like ** instead of looping the current value amount of times it has been altered by 10 percents?
I know I can just store it in additionl column like base_number=100 or something when I need the basic value, but I would like to know if its possible to do by one-liner calculations?
So your basic question is, how do you invert and find x_0 given a known n and:
x_n = x_0 * 1.1^n
Looks like we can simply divide through by 1.1^n
x_n/(1.1^n) = x_0
So you can either calculate 1.1^n with pow(1.1, n) and divide x_n (your "increased" value) by that, or just loop and reduce like you increased:
//1.
$original = $increased/pow(1.1, n);
//2.
$original = $increased;
for ($i = 0; $i < n; $i++) {
$original = $original / 1.1;
}
So in your example, let's say our $increased is known to be 133, and n=3. Then using the first method:
$original = 133 / (1.1^3) = 133 / 1.33 = 100
Let's make a simple example and try to find a formula :
100 * 1.10 = 110;
110 * 1.10 = 121;
121 * 1.10 = 133.1;
So right now we have :
basic_number (will be bn) * increase_value (will be iv) = basic_number2;
bn2 * iv = bn3;
bn3 * iv = bn4;
We can write it too :
bn * iv = bn2;
bn * iv * iv = bn3;
bn * iv * iv * iv = bn4;
And so we have the beginning of a formula :
bn * iv^3 = bn4;
Now what you will have as data according to your post is :
X : the number of increase
bnX : the basic number increase X time
iv : the increase value
And you want to find bn according to those value :
bn * iv^X = bnX;
bn = bnX / iv^X;
bn = bnX / (iv * iv * iv ... * iv); // X time
So with PHP it could look like this :
$X = /* the number of incease */;
$bnX = /* the basic number increase X time */;
$iv = /* the increase value */;
for($i = 0; $i < $X; $i++) {
$bnX = $bnX / $iv;
}
This way you will if you echo $bnX; at the end of the loop, you should get your basic number !
You can try to make a php formula to do it and use it every time :
// 1/ Using a loop
function getBasicNumber($bnX, $X, $iv) {
for($i = 0; $i < $X; $i++) {
$bnX = $bnX / $iv;
}
return $bnX;
}
EDIT
// 2/ Using pow
function getBasicNumber($bnX, $X, $iv) {
return $bnX / pow($X, $iv);
}
// 3/ Using '**'
function getBasicNumber($bnX, $X, $iv) {
return $bnX / $X**$iv;
}
This way you just have to do :
echo getBasicNumber(133.1, 3, 1.10); // 100 here for example
Hope it's clear? But still, it's more a maths problem than a programming one
I would like to validate a number to be between 10 and 20, or 30 and 40 or equal to 99.
I could imagine the code to be something similar to this, but it does not work:
In Entity\NameOfFile.php:
/*
* #Assert\Range(
* min=10,
* max=20
* )
*
* #Assert\Range(
* min=30,
* max=40
* )
*
* #Assert\IdenticalTo(value = 99)
*/
private $myVariable;
Or maybe something similar to this:
/*
* #Assert\Choice({
* Range({min = 10, max = 20}),
* Range({min = 10, max = 20}),
* 99
* )}
*/
private $myVariable;
I also added min and max messages.
In the first option, apparently only the first Assert is taken into consideration and the second one is ignored. The second option, does not work at all.
Ps: Please, a solution without Regex
EDIT:
Following the advice of M Khalid Juanid, the code looks like this:
/**
* #Assert\Expression(
* "this.getmyVariable() in 10..20 or this.getmyVariable() in 30..40 or this.getmyVariable() == 99",
* message="Your error message", groups{"groupA"}
* )
*private $myVariable;
*/
***
if($appendForm->isValid(array($groupA))
{
***
}
It works fine, but only when the validation is not assigned to groupA. How can, the validation be assigned to a group, in this case?
You can use #Assert\Expression for your mulitple criteria
/**
* #Assert\Expression(
* "this.checkRange()",
* message="Your error message"
* )
*/
class YourEntityName
{
public function checkRange(){
if(
($this->yourPorperty >= 10 && $this->yourPorperty <= 20)
|| ($this->yourPorperty >= 30 && $this->yourPorperty <= 40)
|| ($this->yourPorperty == 90)
){
return true;
}
return false;
}
}
As per docs The expression option is the expression that must return true in order for validation to pass
Even more simpler according to the documentation
/**
* #Assert\Expression(
* "this.getYourPorperty() in 10..20 or this.getYourPorperty() in 30..40 or this.getYourPorperty() == 90",
* message="Your error message"
* )
*/
You can use a custom validation constraint. To create one you can refer to the official documentation: http://symfony.com/doc/3.4/validation/custom_constraint.html
You will have to put all your constraints in the validate function of the Validator class.
Since Symfony 5.1, there is a new validator that can check if at least one of several constraints is met.
AtLeastOneOf
/**
* #Assert\AtLeastOneOf({
* #Assert\Range(
* min=10,
* max=20
* ),
* #Assert\Range(
* min=30,
* max=40
* ),
* #Assert\IdenticalTo(value = 99)
* })
*/
private $myVariable;
I have encrypted 2 numbers using paillier cryptosystem. encrypted value of the numbers are bigInteger when I want to divide them the value is the decimal number
for example: first encrypted value of a 9 is 12446486760457687016046
and encrypted value of 3 is 98647617673416817617. the result of division is likely to be decimal. and final result is 0 in this case because paillier get bigInteger as parameter. how can I divide them?
public class Paillier {
/**
* p and q are two large primes.
* lambda = lcm(p-1, q-1) = (p-1)*(q-1)/gcd(p-1, q-1).
*/
private BigInteger p, q, lambda;
/**
* n = p*q, where p and q are two large primes.
*/
public BigInteger n;
/**
* nsquare = n*n
*/
public BigInteger nsquare;
/**
* a random integer in Z*_{n^2} where gcd (L(g^lambda mod n^2), n) = 1.
*/
private BigInteger g;
/**
* number of bits of modulus
*/
private int bitLength;
/**
* Constructs an instance of the Paillier cryptosystem.
* #param bitLengthVal number of bits of modulus
* #param certainty The probability that the new BigInteger represents a prime number will exceed (1 - 2^(-certainty)). The execution time of this constructor is proportional to the value of this parameter.
*/
public Paillier(int bitLengthVal, int certainty) {
KeyGeneration(bitLengthVal, certainty);
}
/**
* Constructs an instance of the Paillier cryptosystem with 512 bits of modulus and at least 1-2^(-64) certainty of primes generation.
*/
public Paillier() {
KeyGeneration(512, 64);
}
/**
* Sets up the public key and private key.
* #param bitLengthVal number of bits of modulus.
* #param certainty The probability that the new BigInteger represents a prime number will exceed (1 - 2^(-certainty)). The execution time of this constructor is proportional to the value of this parameter.
*/
public void KeyGeneration(int bitLengthVal, int certainty) {
bitLength = bitLengthVal;
/*Constructs two randomly generated positive BigIntegers that are probably prime, with the specified bitLength and certainty.*/
p = new BigInteger(bitLength / 2, certainty, new Random());
q = new BigInteger(bitLength / 2, certainty, new Random());
n = p.multiply(q);
nsquare = n.multiply(n);
g = new BigInteger("2");
lambda = p.subtract(BigInteger.ONE).multiply(q.subtract(BigInteger.ONE)).divide(
p.subtract(BigInteger.ONE).gcd(q.subtract(BigInteger.ONE)));
/* check whether g is good.*/
if (g.modPow(lambda, nsquare).subtract(BigInteger.ONE).divide(n).gcd(n).intValue() != 1) {
System.out.println("g is not good. Choose g again.");
System.exit(1);
}
}
/**
* Encrypts plaintext m. ciphertext c = g^m * r^n mod n^2. This function explicitly requires random input r to help with encryption.
* #param m plaintext as a BigInteger
* #param r random plaintext to help with encryption
* #return ciphertext as a BigInteger
*/
public BigInteger Encryption(BigInteger m, BigInteger r) {
return g.modPow(m, nsquare).multiply(r.modPow(n, nsquare)).mod(nsquare);
}
/**
* Encrypts plaintext m. ciphertext c = g^m * r^n mod n^2. This function automatically generates random input r (to help with encryption).
* #param m plaintext as a BigInteger
* #return ciphertext as a BigInteger
*/
public BigInteger Encryption(BigInteger m) {
BigInteger r = new BigInteger(bitLength, new Random());
return g.modPow(m, nsquare).multiply(r.modPow(n, nsquare)).mod(nsquare);
}
/**
* Decrypts ciphertext c. plaintext m = L(c^lambda mod n^2) * u mod n, where u = (L(g^lambda mod n^2))^(-1) mod n.
* #param c ciphertext as a BigInteger
* #return plaintext as a BigInteger
*/
public BigInteger Decryption(BigInteger c) {
BigInteger u = g.modPow(lambda, nsquare).subtract(BigInteger.ONE).divide(n).modInverse(n);
return c.modPow(lambda, nsquare).subtract(BigInteger.ONE).divide(n).multiply(u).mod(n);
}
/**
* main function
* #param str intput string
*/
public static void main(String[] str) {
/* instantiating an object of Paillier cryptosystem*/
Paillier paillier = new Paillier();
BigInteger o1 = (paillier.Encryption(new BigInteger("9")));
BigInteger o2 = (paillier.Encryption(new BigInteger("3")));
BigInteger od = o2.divide(o1);
System.out.println(paillier.Decryption(od));
As I explained before, in cryptographic applications, it is common to use the multiplicative inverse instead of division.
In grade school, I learned to divide 9 by 3: 9 ÷ 3 = 3. A bit later I learned that multiplying by the reciprocal of the divisor would do the same thing: 9 × ⅓ = 3. For rational numbers, ⅓ is the multiplicative inverse of 3: 3 × ⅓ = 1
In modular arithmetic, the multiplicative inverse is analogous to a reciprocal. Suppose I'm working with numbers modulo 256. I want to "divide" by 3. As above, I can do it using the multiplicative inverse of the "divisor". 3 × 171 mod 256 = 1, so 171 is the multiplicative inverse of 3. But wait, 9 × 171 = 1539. Shouldn't it be 3? Oh, wait, we forgot that we are working modulo 256: 1539 mod 256 = 3.
In your example, you have two numbers that could be used as a modulus, n or nsquare. I believe that if you study a bit, you'll discover which to use when performing homomorphic arithmetic with Paillier. Then you can use the modInverse() function to perform your "subtraction".
Paillier paillier = new Paillier();
BigInteger o1 = (paillier.Encryption(new BigInteger("9")));
BigInteger o2 = (paillier.Encryption(new BigInteger("3")));
BigInteger od = o1.multiply(o2.modInverse(???));
System.out.println(paillier.Decryption(od));
As a silly toy example, suppose
x=4.5
w=c(1,2,4,6,7)
I wonder if there is a simple R function that finds the index of the closest match to x in w. So if foo is that function, foo(w,x) would return 3. The function match is the right idea, but seems to apply only for exact matches.
Solutions here (e.g. which.min(abs(w - x)), which(abs(w-x)==min(abs(w-x))), etc.) are all O(n) instead of log(n) (I'm assuming that w is already sorted).
R>findInterval(4.5, c(1,2,4,5,6))
[1] 3
will do that with price-is-right matching (closest without going over).
You can use data.table to do a binary search:
dt = data.table(w, val = w) # you'll see why val is needed in a sec
setattr(dt, "sorted", "w") # let data.table know that w is sorted
Note that if the column w isn't already sorted, then you'll have to use setkey(dt, w) instead of setattr(.).
# binary search and "roll" to the nearest neighbour
dt[J(x), roll = "nearest"]
# w val
#1: 4.5 4
In the final expression the val column will have the you're looking for.
# or to get the index as Josh points out
# (and then you don't need the val column):
dt[J(x), .I, roll = "nearest", by = .EACHI]
# w .I
#1: 4.5 3
# or to get the index alone
dt[J(x), roll = "nearest", which = TRUE]
#[1] 3
See match.closest() from the MALDIquant package:
> library(MALDIquant)
> match.closest(x, w)
[1] 3
x = 4.5
w = c(1,2,4,6,7)
closestLoc = which(min(abs(w-x)))
closestVal = w[which(min(abs(w-x)))]
# On my phone- please pardon typos
If your vector is lengthy, try a 2-step approach:
x = 4.5
w = c(1,2,4,6,7)
sdev = sapply(w,function(v,x) abs(v-x), x = x)
closestLoc = which(min(sdev))
for maddeningly long vectors (millions of rows!, warning- this will actually be slower for data which is not very, very, very large.)
require(doMC)
registerDoMC()
closestLoc = which(min(foreach(i = w) %dopar% {
abs(i-x)
}))
This example is just to give you a basic idea of leveraging parallel processing when you have huge data. Note, I do not recommend you use it for simple & fast functions like abs().
To do this on character vectors, Martin Morgan suggested this function on R-help:
bsearch7 <-
function(val, tab, L=1L, H=length(tab))
{
b <- cbind(L=rep(L, length(val)), H=rep(H, length(val)))
i0 <- seq_along(val)
repeat {
updt <- M <- b[i0,"L"] + (b[i0,"H"] - b[i0,"L"]) %/% 2L
tabM <- tab[M]
val0 <- val[i0]
i <- tabM < val0
updt[i] <- M[i] + 1L
i <- tabM > val0
updt[i] <- M[i] - 1L
b[i0 + i * length(val)] <- updt
i0 <- which(b[i0, "H"] >= b[i0, "L"])
if (!length(i0)) break;
}
b[,"L"] - 1L
}
NearestValueSearch = function(x, w){
## A simple binary search algo
## Assume the w vector is sorted so we can use binary search
left = 1
right = length(w)
while(right - left > 1){
middle = floor((left + right) / 2)
if(x < w[middle]){
right = middle
}
else{
left = middle
}
}
if(abs(x - w[right]) < abs(x - w[left])){
return(right)
}
else{
return(left)
}
}
x = 4.5
w = c(1,2,4,6,7)
NearestValueSearch(x, w) # return 3
Based on #neal-fultz answer, here is a simple function that uses findInterval():
get_closest_index <- function(x, vec){
# vec must be sorted
iv <- findInterval(x, vec)
dist_left <- x - vec[ifelse(iv == 0, NA, iv)]
dist_right <- vec[iv + 1] - x
ifelse(! is.na(dist_left) & (is.na(dist_right) | dist_left < dist_right), iv, iv + 1)
}
values <- c(-15, -0.01, 3.1, 6, 10, 100)
grid <- c(-2, -0.1, 0.1, 3, 7)
get_closest_index(values, grid)
#> [1] 1 2 4 5 5 5
Created on 2020-05-29 by the reprex package (v0.3.0)
You can always implement custom binary search algorithm to find the closest value. Alternately, you can leverage standard implementation of libc bsearch(). You can use other binary search implementations as well, but it does not change the fact that you have to implement the comparing function carefully to find the closest element in array. The issue with standard binary search implementation is that it is meant for exact comparison. That means your improvised comparing function needs to do some kind of exactification to figure out if an element in array is close-enough. To achieve it, the comparing function needs to have awareness of other elements in the array, especially following aspects:
position of the current element (one which is being compared with the
key).
the distance with key and how it compares with neighbors (previous
or next element).
To provide this extra knowledge in comparing function, the key needs to be packaged with additional information (not just the key value). Once the comparing function have awareness on these aspects, it can figure out if the element itself is closest. When it knows that it is the closest, it returns "match".
The the following C code finds the closest value.
#include <stdio.h>
#include <stdlib.h>
struct key {
int key_val;
int *array_head;
int array_size;
};
int compar(const void *k, const void *e) {
struct key *key = (struct key*)k;
int *elem = (int*)e;
int *arr_first = key->array_head;
int *arr_last = key->array_head + key->array_size -1;
int kv = key->key_val;
int dist_left;
int dist_right;
if (kv == *elem) {
/* easy case: if both same, got to be closest */
return 0;
} else if (key->array_size == 1) {
/* easy case: only element got to be closest */
return 0;
} else if (elem == arr_first) {
/* element is the first in array */
if (kv < *elem) {
/* if keyval is less the first element then
* first elem is closest.
*/
return 0;
} else {
/* check distance between first and 2nd elem.
* if distance with first elem is smaller, it is closest.
*/
dist_left = kv - *elem;
dist_right = *(elem+1) - kv;
return (dist_left <= dist_right) ? 0:1;
}
} else if (elem == arr_last) {
/* element is the last in array */
if (kv > *elem) {
/* if keyval is larger than the last element then
* last elem is closest.
*/
return 0;
} else {
/* check distance between last and last-but-one.
* if distance with last elem is smaller, it is closest.
*/
dist_left = kv - *(elem-1);
dist_right = *elem - kv;
return (dist_right <= dist_left) ? 0:-1;
}
}
/* condition for remaining cases (other cases are handled already):
* - elem is neither first or last in the array
* - array has atleast three elements.
*/
if (kv < *elem) {
/* keyval is smaller than elem */
if (kv <= *(elem -1)) {
/* keyval is smaller than previous (of "elem") too.
* hence, elem cannot be closest.
*/
return -1;
} else {
/* check distance between elem and elem-prev.
* if distance with elem is smaller, it is closest.
*/
dist_left = kv - *(elem -1);
dist_right = *elem - kv;
return (dist_right <= dist_left) ? 0:-1;
}
}
/* remaining case: (keyval > *elem) */
if (kv >= *(elem+1)) {
/* keyval is larger than next (of "elem") too.
* hence, elem cannot be closest.
*/
return 1;
}
/* check distance between elem and elem-next.
* if distance with elem is smaller, it is closest.
*/
dist_right = *(elem+1) - kv;
dist_left = kv - *elem;
return (dist_left <= dist_right) ? 0:1;
}
int main(int argc, char **argv) {
int arr[] = {10, 20, 30, 40, 50, 60, 70};
int *found;
struct key k;
if (argc < 2) {
return 1;
}
k.key_val = atoi(argv[1]);
k.array_head = arr;
k.array_size = sizeof(arr)/sizeof(int);
found = (int*)bsearch(&k, arr, sizeof(arr)/sizeof(int), sizeof(int),
compar);
if(found) {
printf("found closest: %d\n", *found);
} else {
printf("closest not found. absurd! \n");
}
return 0;
}
Needless to say that bsearch() in above example should never fail (unless the array size is zero).
If you implement your own custom binary search, essentially you have to embed same comparing logic in the main body of binary search code (instead of having this logic in comparing function in above example).
So I am trying to figure out how to take a range of numbers and scale the values down to fit a range. The reason for wanting to do this is that I am trying to draw ellipses in a java swing jpanel. I want the height and width of each ellipse to be in a range of say 1-30. I have methods that find the minimum and maximum values from my data set, but I won't have the min and max until runtime. Is there an easy way to do this?
Let's say you want to scale a range [min,max] to [a,b]. You're looking for a (continuous) function that satisfies
f(min) = a
f(max) = b
In your case, a would be 1 and b would be 30, but let's start with something simpler and try to map [min,max] into the range [0,1].
Putting min into a function and getting out 0 could be accomplished with
f(x) = x - min ===> f(min) = min - min = 0
So that's almost what we want. But putting in max would give us max - min when we actually want 1. So we'll have to scale it:
x - min max - min
f(x) = --------- ===> f(min) = 0; f(max) = --------- = 1
max - min max - min
which is what we want. So we need to do a translation and a scaling. Now if instead we want to get arbitrary values of a and b, we need something a little more complicated:
(b-a)(x - min)
f(x) = -------------- + a
max - min
You can verify that putting in min for x now gives a, and putting in max gives b.
You might also notice that (b-a)/(max-min) is a scaling factor between the size of the new range and the size of the original range. So really we are first translating x by -min, scaling it to the correct factor, and then translating it back up to the new minimum value of a.
Here's some JavaScript for copy-paste ease (this is irritate's answer):
function scaleBetween(unscaledNum, minAllowed, maxAllowed, min, max) {
return (maxAllowed - minAllowed) * (unscaledNum - min) / (max - min) + minAllowed;
}
Applied like so, scaling the range 10-50 to a range between 0-100.
var unscaledNums = [10, 13, 25, 28, 43, 50];
var maxRange = Math.max.apply(Math, unscaledNums);
var minRange = Math.min.apply(Math, unscaledNums);
for (var i = 0; i < unscaledNums.length; i++) {
var unscaled = unscaledNums[i];
var scaled = scaleBetween(unscaled, 0, 100, minRange, maxRange);
console.log(scaled.toFixed(2));
}
0.00, 18.37, 48.98, 55.10, 85.71, 100.00
Edit:
I know I answered this a long time ago, but here's a cleaner function that I use now:
Array.prototype.scaleBetween = function(scaledMin, scaledMax) {
var max = Math.max.apply(Math, this);
var min = Math.min.apply(Math, this);
return this.map(num => (scaledMax-scaledMin)*(num-min)/(max-min)+scaledMin);
}
Applied like so:
[-4, 0, 5, 6, 9].scaleBetween(0, 100);
[0, 30.76923076923077, 69.23076923076923, 76.92307692307692, 100]
For convenience, here is Irritate's algorithm in a Java form. Add error checking, exception handling and tweak as necessary.
public class Algorithms {
public static double scale(final double valueIn, final double baseMin, final double baseMax, final double limitMin, final double limitMax) {
return ((limitMax - limitMin) * (valueIn - baseMin) / (baseMax - baseMin)) + limitMin;
}
}
Tester:
final double baseMin = 0.0;
final double baseMax = 360.0;
final double limitMin = 90.0;
final double limitMax = 270.0;
double valueIn = 0;
System.out.println(Algorithms.scale(valueIn, baseMin, baseMax, limitMin, limitMax));
valueIn = 360;
System.out.println(Algorithms.scale(valueIn, baseMin, baseMax, limitMin, limitMax));
valueIn = 180;
System.out.println(Algorithms.scale(valueIn, baseMin, baseMax, limitMin, limitMax));
90.0
270.0
180.0
Here's how I understand it:
What percent does x lie in a range
Let's assume you have a range from 0 to 100. Given an arbitrary number from that range, what "percent" from that range does it lie in? This should be pretty simple, 0 would be 0%, 50 would be 50% and 100 would be 100%.
Now, what if your range was 20 to 100? We cannot apply the same logic as above (divide by 100) because:
20 / 100
doesn't give us 0 (20 should be 0% now). This should be simple to fix, we just need to make the numerator 0 for the case of 20. We can do that by subtracting:
(20 - 20) / 100
However, this doesn't work for 100 anymore because:
(100 - 20) / 100
doesn't give us 100%. Again, we can fix this by subtracting from the denominator as well:
(100 - 20) / (100 - 20)
A more generalized equation for finding out what % x lies in a range would be:
(x - MIN) / (MAX - MIN)
Scale range to another range
Now that we know what percent a number lies in a range, we can apply it to map the number to another range. Let's go through an example.
old range = [200, 1000]
new range = [10, 20]
If we have a number in the old range, what would the number be in the new range? Let's say the number is 400. First, figure out what percent 400 is within the old range. We can apply our equation above.
(400 - 200) / (1000 - 200) = 0.25
So, 400 lies in 25% of the old range. We just need to figure out what number is 25% of the new range. Think about what 50% of [0, 20] is. It would be 10 right? How did you arrive at that answer? Well, we can just do:
20 * 0.5 = 10
But, what about from [10, 20]? We need to shift everything by 10 now. eg:
((20 - 10) * 0.5) + 10
a more generalized formula would be:
((MAX - MIN) * PERCENT) + MIN
To the original example of what 25% of [10, 20] is:
((20 - 10) * 0.25) + 10 = 12.5
So, 400 in the range [200, 1000] would map to 12.5 in the range [10, 20]
TLDR
To map x from old range to new range:
OLD PERCENT = (x - OLD MIN) / (OLD MAX - OLD MIN)
NEW X = ((NEW MAX - NEW MIN) * OLD PERCENT) + NEW MIN
I came across this solution but this does not really fit my need. So I digged a bit in the d3 source code. I personally would recommend to do it like d3.scale does.
So here you scale the domain to the range. The advantage is that you can flip signs to your target range. This is useful since the y axis on a computer screen goes top down so large values have a small y.
public class Rescale {
private final double range0,range1,domain0,domain1;
public Rescale(double domain0, double domain1, double range0, double range1) {
this.range0 = range0;
this.range1 = range1;
this.domain0 = domain0;
this.domain1 = domain1;
}
private double interpolate(double x) {
return range0 * (1 - x) + range1 * x;
}
private double uninterpolate(double x) {
double b = (domain1 - domain0) != 0 ? domain1 - domain0 : 1 / domain1;
return (x - domain0) / b;
}
public double rescale(double x) {
return interpolate(uninterpolate(x));
}
}
And here is the test where you can see what I mean
public class RescaleTest {
#Test
public void testRescale() {
Rescale r;
r = new Rescale(5,7,0,1);
Assert.assertTrue(r.rescale(5) == 0);
Assert.assertTrue(r.rescale(6) == 0.5);
Assert.assertTrue(r.rescale(7) == 1);
r = new Rescale(5,7,1,0);
Assert.assertTrue(r.rescale(5) == 1);
Assert.assertTrue(r.rescale(6) == 0.5);
Assert.assertTrue(r.rescale(7) == 0);
r = new Rescale(-3,3,0,1);
Assert.assertTrue(r.rescale(-3) == 0);
Assert.assertTrue(r.rescale(0) == 0.5);
Assert.assertTrue(r.rescale(3) == 1);
r = new Rescale(-3,3,-1,1);
Assert.assertTrue(r.rescale(-3) == -1);
Assert.assertTrue(r.rescale(0) == 0);
Assert.assertTrue(r.rescale(3) == 1);
}
}
I sometimes find a variation of this useful.
Wrapping the scale function in a class so that I do not need to pass around the min/max values if scaling the same ranges in several places
Adding two small checks that ensures that the result value stays within the expected range.
Example in JavaScript:
class Scaler {
constructor(inMin, inMax, outMin, outMax) {
this.inMin = inMin;
this.inMax = inMax;
this.outMin = outMin;
this.outMax = outMax;
}
scale(value) {
const result = (value - this.inMin) * (this.outMax - this.outMin) / (this.inMax - this.inMin) + this.outMin;
if (result < this.outMin) {
return this.outMin;
} else if (result > this.outMax) {
return this.outMax;
}
return result;
}
}
This example along with a function based version comes from the page https://writingjavascript.com/scaling-values-between-two-ranges
Based on Charles Clayton's response, I included some JSDoc, ES6 tweaks, and incorporated suggestions from the comments in the original response.
/**
* Returns a scaled number within its source bounds to the desired target bounds.
* #param {number} n - Unscaled number
* #param {number} tMin - Minimum (target) bound to scale to
* #param {number} tMax - Maximum (target) bound to scale to
* #param {number} sMin - Minimum (source) bound to scale from
* #param {number} sMax - Maximum (source) bound to scale from
* #returns {number} The scaled number within the target bounds.
*/
const scaleBetween = (n, tMin, tMax, sMin, sMax) => {
return (tMax - tMin) * (n - sMin) / (sMax - sMin) + tMin;
}
if (Array.prototype.scaleBetween === undefined) {
/**
* Returns a scaled array of numbers fit to the desired target bounds.
* #param {number} tMin - Minimum (target) bound to scale to
* #param {number} tMax - Maximum (target) bound to scale to
* #returns {number} The scaled array.
*/
Array.prototype.scaleBetween = function(tMin, tMax) {
if (arguments.length === 1 || tMax === undefined) {
tMax = tMin; tMin = 0;
}
let sMax = Math.max(...this), sMin = Math.min(...this);
if (sMax - sMin == 0) return this.map(num => (tMin + tMax) / 2);
return this.map(num => (tMax - tMin) * (num - sMin) / (sMax - sMin) + tMin);
}
}
// ================================================================
// Usage
// ================================================================
let nums = [10, 13, 25, 28, 43, 50], tMin = 0, tMax = 100,
sMin = Math.min(...nums), sMax = Math.max(...nums);
// Result: [ 0.0, 7.50, 37.50, 45.00, 82.50, 100.00 ]
console.log(nums.map(n => scaleBetween(n, tMin, tMax, sMin, sMax).toFixed(2)).join(', '));
// Result: [ 0, 30.769, 69.231, 76.923, 100 ]
console.log([-4, 0, 5, 6, 9].scaleBetween(0, 100).join(', '));
// Result: [ 50, 50, 50 ]
console.log([1, 1, 1].scaleBetween(0, 100).join(', '));
.as-console-wrapper { top: 0; max-height: 100% !important; }
I've taken Irritate's answer and refactored it so as to minimize the computational steps for subsequent computations by factoring it into the fewest constants. The motivation is to allow a scaler to be trained on one set of data, and then be run on new data (for an ML algo). In effect, it's much like SciKit's preprocessing MinMaxScaler for Python in usage.
Thus, x' = (b-a)(x-min)/(max-min) + a (where b!=a) becomes x' = x(b-a)/(max-min) + min(-b+a)/(max-min) + a which can be reduced to two constants in the form x' = x*Part1 + Part2.
Here's a C# implementation with two constructors: one to train, and one to reload a trained instance (e.g., to support persistence).
public class MinMaxColumnSpec
{
/// <summary>
/// To reduce repetitive computations, the min-max formula has been refactored so that the portions that remain constant are just computed once.
/// This transforms the forumula from
/// x' = (b-a)(x-min)/(max-min) + a
/// into x' = x(b-a)/(max-min) + min(-b+a)/(max-min) + a
/// which can be further factored into
/// x' = x*Part1 + Part2
/// </summary>
public readonly double Part1, Part2;
/// <summary>
/// Use this ctor to train a new scaler.
/// </summary>
public MinMaxColumnSpec(double[] columnValues, int newMin = 0, int newMax = 1)
{
if (newMax <= newMin)
throw new ArgumentOutOfRangeException("newMax", "newMax must be greater than newMin");
var oldMax = columnValues.Max();
var oldMin = columnValues.Min();
Part1 = (newMax - newMin) / (oldMax - oldMin);
Part2 = newMin + (oldMin * (newMin - newMax) / (oldMax - oldMin));
}
/// <summary>
/// Use this ctor for previously-trained scalers with known constants.
/// </summary>
public MinMaxColumnSpec(double part1, double part2)
{
Part1 = part1;
Part2 = part2;
}
public double Scale(double x) => (x * Part1) + Part2;
}