Standard deviation from iterators

Question

I want this to be similar to the STL algorithms but I don't find it elegant nor concise at all:

#include <iostream> #include <iterator> #include <vector> #include <algorithm> using E = double; template <typename IT> E std_dev(IT begin, IT end){ auto N = std::distance(begin, end); E average = std::accumulate(begin, end, E()) / N; auto sum_term = [average](E init, E value)-> E{ return init + (value - average)*(value - average); }; E variance = std::accumulate(begin, end, E(), sum_term); return std::sqrt(variance * 1.0 / (N - 1)); } int main(){ std::vector<double> stuff {3.5, 3.4, 3.6, 3.9, 3.5, 3.5, 3.5, 3.5, 3.5}; std::cout << std_dev(stuff.begin(), stuff.end()) << "\n"; }

Please do not change the code in your questions after an answer has been posted, so as not to invalidate it. See What should I do when someone answers my question? — Phrancis
– Phrancis, Commented Mar 19, 2016 at 0:52

sehe · Accepted Answer · 2016-03-19 00:48:16Z

Firstly, make it correct.

N is integral, you could make it E so you don't accidentally do integer arithmetic.

N-1 is wrong.

Rename average to mean.

Don't hardcode E.

You get:

template <typename It, typename E = typename std::iterator_traits<It>::value_type> E std_dev(It begin, It end){ E N = std::distance(begin, end); E const mean = std::accumulate(begin, end, E()) / N; auto sum_term = [mean](E init, E value)-> E { return init + (value - mean)*(value - mean); }; E variance = std::accumulate(begin, end, E(), sum_term); return std::sqrt(variance / N); }

Slightly stylized, with comparison to Boost Accumulator:

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#include <iostream> #include <iterator> #include <vector> #include <algorithm> #include <boost/accumulators/accumulators.hpp> #include <boost/accumulators/statistics.hpp> template <typename It, typename E = typename std::iterator_traits<It>::value_type, typename R = typename std::common_type<double, E>::type> R std_dev_boost(It begin, It end){ namespace ba = boost::accumulators; ba::accumulator_set<R, ba::stats<ba::tag::variance> > accu; std::for_each(begin, end, std::ref(accu)); return std::sqrt(ba::variance(accu)); } template <typename It, typename E = typename std::iterator_traits<It>::value_type, typename R = typename std::common_type<double, E>::type> R std_dev(It b, It e) { R N = std::distance(b, e); R const mean = std::accumulate(b, e, R{}) / N; R variance = std::accumulate(b, e, R{}, [mean](R a, E v)-> R { return a + (v-mean)*(v-mean); }); return std::sqrt(variance / N); } int main(){ std::vector<int> stuff {35, 34, 36, 39, 35, 35, 35, 35, 35}; std::cout << std_dev_boost(stuff.begin(), stuff.end()) << "\n"; std::cout << std_dev (stuff.begin(), stuff.end()) << "\n"; }

Prints

1.34256 1.34256

@CaptainGiraffe Mmm. I was mistaken there. The N-1 seems to be the only mistake that made the outcome wrong :) — sehe
– sehe, Commented Mar 19, 2016 at 0:42
Ah. Well, see my update for how to derive a proper result type that works even if the input is an integer container. — sehe
– sehe, Commented Mar 19, 2016 at 0:49
The N-1 isn't wrong. It's just computing a different standard deviation. To be specific, if you're computing the standard deviation of "the universe", you use N. If you're computing the standard deviation of a sample, you use N-1. — Jerry Coffin
– Jerry Coffin, Commented Mar 19, 2016 at 1:51
@JerryCoffin Yeah, that was clear to me since Giraffe told me :) I was content contributing the style feedback — sehe
– sehe, Commented Mar 19, 2016 at 2:21

набиячлэвэлиь · Accepted Answer · 2016-03-19 00:37:58Z

A simple conciseness improvement would be to inline non-informative variables:

template <typename IT> E std_dev(IT begin, IT end) { auto N = std::distance(begin, end); E variance = std::accumulate(begin, end, E{}, [average=std::accumulate(begin, end, E{}) / N](E init, E value) -> E { return init + (value - average)*(value - average); }); return std::sqrt(variance * 1.0 / (N - 1)); }

average was moved into the lambda capture (C++14(?)) and the lambda itself was passed directly to std::accumulate.

@sehe No, I didn't remove them. I'm here for conciseness (as stated in the answer), not for correctness. — набиячлэвэлиь
– набиячлэвэлиь, Commented Mar 19, 2016 at 0:39
The main point of programming is to make it easier to read not more concise. — Loki Astari
– Loki Astari, Commented Mar 19, 2016 at 0:51
If you want it concise, use valarray. stackoverflow.com/a/1723071/179910 — Jerry Coffin
– Jerry Coffin, Commented Mar 19, 2016 at 1:17

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