'Standard deviation javascript
I am trying to get the standard deviation of a user input string. I have as follows, but it returns the wrong value for SD. The calculation should go as follows: Sum values/number values = mean Square (sum each value-mean) Sum squares/number values.
Assistance appreciated (and explanation if possible):
function sum() {
var val = document.getElementById('userInput').value;
var temp = val.split(" ");
var total = 0;
var v;
var mean = total / temp.length;
var total1 = 0;
var v1;
var temp23;
var square;
for (var i = 0; i < temp.length; i++) {
v = parseFloat(temp[i]);
total += v;
}
mean = total / temp.length;
for (var i = 0; i < temp.length; i++) {
v1 = parseFloat(Math.pow(temp[i] - mean), 2);
total1 += v1;
}
temp23 = total1 / temp.length;
square = Math.sqrt(temp23);
document.write(total + '<br />');
document.write(mean + '<br />');
document.write(square);
}<html>
<head>
</head>
<body>
<form id="input">
<textarea id="userInput" rows=20 cols=20></textarea>
<input id="Run" type=Button value="run" onClick="sum()" />
</form>
</body>
</html>Solution 1:[1]
I think the (main) problem is on this line:
v1 = parseFloat(Math.pow(temp[i]-mean),2);
Should be:
v1 = Math.pow(parseFloat(temp[i])-mean),2);
Your code is trying to use the string in temp[i] as a number and subtract mean from it, and then square it, and then parse the resulting value. Need to parseFloat before using it in a calculation. Also you've got the ,2 outside the closing parenenthesis for the Math.pow call so the squaring won't work either.
Would be helpful to use more meaningful variable names too, I mean, e.g., you have a variable called "square" that holds the result of a square-root operation.
P.S. You need to add some error checking in case the user enters non-numeric data. Check that the result of parseFloat() is not NaN. I'd be inclined to do an initial loop through the array parsing and checking for valid numbers, storing the parsed numbers in a second array (or writing them back to the first array), and if any are invalid give the user an error message at that point and stop. Then in your actual calculations you don't have to worry about parsing as you go (or, in your case, parsing again in the second loop).
Solution 2:[2]
Shorthand method for getting standard deviation from an array if you don't like lots of code:
function getStandardDeviation (array) {
const n = array.length
const mean = array.reduce((a, b) => a + b) / n
return Math.sqrt(array.map(x => Math.pow(x - mean, 2)).reduce((a, b) => a + b) / n)
}
Solution 3:[3]
For anyone looking for a more generic solution, here's a standard deviation function added to the Array#. The function expects to be called on an array of numbers.
Array.prototype.stanDeviate = function(){
var i,j,total = 0, mean = 0, diffSqredArr = [];
for(i=0;i<this.length;i+=1){
total+=this[i];
}
mean = total/this.length;
for(j=0;j<this.length;j+=1){
diffSqredArr.push(Math.pow((this[j]-mean),2));
}
return (Math.sqrt(diffSqredArr.reduce(function(firstEl, nextEl){
return firstEl + nextEl;
})/this.length));
};
Solution 4:[4]
function getStandardDeviation(numbersArr) {
// CALCULATE AVERAGE
var total = 0;
for(var key in numbersArr)
total += numbersArr[key];
var meanVal = total / numbersArr.length;
// CALCULATE AVERAGE
// CALCULATE STANDARD DEVIATION
var SDprep = 0;
for(var key in numbersArr)
SDprep += Math.pow((parseFloat(numbersArr[key]) - meanVal),2);
var SDresult = Math.sqrt(SDprep/(numbersArr.length-1));
// CALCULATE STANDARD DEVIATION
return SDresult;
}
var numbersArr = [10, 11, 12, 13, 14];
alert(getStandardDeviation(numbersArr));Solution 5:[5]
This ES6 implementation matches Excel's built in STDEV.P and STDEV.S (.S is the default when STDEV is called). Passing in the true flag for usePopulation here will match Excel's STDEV.P
const standardDeviation = (arr, usePopulation = false) => {
const mean = arr.reduce((acc, val) => acc + val, 0) / arr.length;
return Math.sqrt(
arr.reduce((acc, val) => acc.concat((val - mean) ** 2), []).reduce((acc, val) => acc + val, 0) /
(arr.length - (usePopulation ? 0 : 1))
);
};
console.log('STDEV.S =>',
standardDeviation([
10, 2, 38, 23, 38, 23, 21
])
);
console.log('STDEV.P =>',
standardDeviation([
10, 2, 38, 23, 38, 23, 21
], true)
);Solution 6:[6]
Standard Deviation is quite simple.
- Calcuate the mean
- Square the subtraction of each number against the mean. Average these new numbers again to get the variance.
- Finally square root the variance to get the standard deviation.
function standardDeviation(numArray) {
const mean = numArray.reduce((s, n) => s + n) / numArray.length;
const variance = numArray.reduce((s, n) => s + (n - mean) ** 2, 0) / (numArray.length - 1);
return Math.sqrt(variance);
}
if you have a array of strings, remember to convert array of floats first array.map(s => parseFloat(s))
Solution 7:[7]
You can use mathjs library
math.std(array)
Solution 8:[8]
Quick implementation of the standard deviation function:
const sd = numbers => {
const mean = numbers.reduce((acc, n) => acc + n) / numbers.length;
return Math.sqrt(
numbers.reduce((acc, n) => (n - mean) ** 2) / numbers.length
);
};
Corrected SD version:
const correctedSd = numbers => {
const mean = numbers.reduce((acc, n) => acc + n) / numbers.length;
return Math.sqrt(
numbers.reduce((acc, n) => (n - mean) ** 2) / (numbers.length - 1)
);
};
Solution 9:[9]
This function works and produces the same result as numjs
const st = (numbers) => {
const mean = numbers.reduce((acc, item) => acc + item) / numbers.length;
return Math.sqrt(numbers.reduce((acc, item) => acc + Math.pow((parseFloat(item) -mean), 2)))
}Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | zingi |
| Solution 3 | cssimsek |
| Solution 4 | Scott |
| Solution 5 | Vinnie James |
| Solution 6 | |
| Solution 7 | tal weissler |
| Solution 8 | |
| Solution 9 | Diana Lisovenko |