Created
April 16, 2012 17:22
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Calculate difference in percentage between 2 hex colors
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function color_meter(cwith, ccolor) { | |
if (!cwith && !ccolor) return; | |
var _cwith = (cwith.charAt(0)=="#") ? cwith.substring(1,7) : cwith; | |
var _ccolor = (ccolor.charAt(0)=="#") ? ccolor.substring(1,7) : ccolor; | |
var _r = parseInt(_cwith.substring(0,2), 16); | |
var _g = parseInt(_cwith.substring(2,4), 16); | |
var _b = parseInt(_cwith.substring(4,6), 16); | |
var __r = parseInt(_ccolor.substring(0,2), 16); | |
var __g = parseInt(_ccolor.substring(2,4), 16); | |
var __b = parseInt(_ccolor.substring(4,6), 16); | |
var p1 = (_r / 255) * 100; | |
var p2 = (_g / 255) * 100; | |
var p3 = (_b / 255) * 100; | |
var perc1 = Math.round((p1 + p2 + p3) / 3); | |
var p1 = (__r / 255) * 100; | |
var p2 = (__g / 255) * 100; | |
var p3 = (__b / 255) * 100; | |
var perc2 = Math.round((p1 + p2 + p3) / 3); | |
return Math.abs(perc1 - perc2); | |
} |
Let's say that we found the percent difference between two blues to be 20 percent. Now we have a green color and we want it to be exactly the same percentage of difference (20 percent). Is there a way to rewrite the code/formula to find this?
Sort of like: x - y = 20%,
z +/- 20% = ?
It's been a while since math class, but isn't is redundant to multiply both the numerator and the denominator by 100?
Also I'm curious if this method is documented somewhere as a standard method?
I don't see it listed on the Wikipedia page for Color Difference on the topic The easiest method shown there is the Euclidean colour difference formula:
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Might as well live demo it, huh?
https://plnkr.co/edit/ERaf37?p=preview