Standard deviation is a statistic parameter that helps to estimate the dispersion of data series. It's usually calculated in two passes: first, you find a mean, and second, you calculate a square deviation of values from the mean:

```
double std_dev1(double a[], int n) {
if(n == 0)
return 0.0;
double sum = 0;
for(int i = 0; i < n; ++i)
sum += a[i];
double mean = sum / n;
double sq_diff_sum = 0;
for(int i = 0; i < n; ++i) {
double diff = a[i] - mean;
sq_diff_sum += diff * diff;
}
double variance = sq_diff_sum / n;
return sqrt(variance);
}
```

But you can do the same thing in one pass. Rewrite the formula in the following way:

```
double std_dev2(double a[], int n) {
if(n == 0)
return 0.0;
double sum = 0;
double sq_sum = 0;
for(int i = 0; i < n; ++i) {
sum += a[i];
sq_sum += a[i] * a[i];
}
double mean = sum / n;
double variance = sq_sum / n - mean * mean;
return sqrt(variance);
}
```

Unfortunately, the result will be inaccurate when the array contains large numbers (see the comments below).

This trick is mentioned in an old Soviet book about programmable calculators (here is the reference for Russian readers: Финк Л. Папа, мама, я и микрокалькулятор. — М.: Радио и связь, 1988).

## 21 comments

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Knuth actually has one pass algorithm like this

_M gets you mean, _C / (N-1) is variance

I've come across a problem using this algorithm where the variance calculation step outputs a negative zero (in OpenCL). Then the standard deviation step becomes the sqrt of a negative number, which results in NaN.

It might be wise to use the absolute value of the variance to avoid this.

ammendes, thank you so much

Thank you! Your derivation saved me thinking!

If the variance calculation is changed to this:

Wouldn't that be accurate, always, even for large numbers? this assumes there is no overflow, of course.

And maybe this (slightly different arrangement) might help prevent some overflows:

Never mind, it looks like those are equivalent. Sorry.