blocks|key|766616|text|我的答案与Josh+Greifer类似，但推广到样本协方差。样本方差只是样本协方差，但两个输入相同。这包括贝塞尔的相关性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|766617|++++template+<class+Iter>+typename+Iter::value_type+cov(const+Iter+&x,+const+Iter+&y)
++++{
++++++++double+sum_x+=+std::accumulate(std::begin(x),+std::end(x),+0.0);
++++++++double+sum_y+=+std::accumulate(std::begin(y),+std::end(y),+0.0);

++++++++double+mx+=++sum_x+/+x.size();
++++++++double+my+=++sum_y+/+y.size();

++++++++double+accum+=+0.0;

++++++++for+(auto+i+=+0;+i+<+x.size();+i%2B%2B)
++++++++{
++++++++++++accum+%2B=+(x.at(i)+-+mx)+*+(y.at(i)+-+my);
++++++++}

++++++++return+accum+/+(x.size()+-+1);
++++}|code-block|syntax|javascript|766618|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

My answer is similar as Josh Greifer but generalised to sample covariance. Sample variance is just sample covariance but with the two inputs identical. This includes Bessel's correlation.

<pre><code> template &lt;class Iter&gt; typename Iter::value_type cov(const Iter &amp;x, const Iter &amp;y)
 {
 double sum_x = std::accumulate(std::begin(x), std::end(x), 0.0);
 double sum_y = std::accumulate(std::begin(y), std::end(y), 0.0);

 double mx = sum_x / x.size();
 double my = sum_y / y.size();

 double accum = 0.0;

 for (auto i = 0; i &lt; x.size(); i++)
 {
 accum += (x.at(i) - mx) * (y.at(i) - my);
 }

 return accum / (x.size() - 1);
 }
</code></pre>

blocks|key|549964|text|创建您自己的容器：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|549965|template+<class+T>
class+statList+:+public+std::list<T>
{
++++public:
++++++++statList()+:+std::list<T>::list()+{}
++++++++~statList()+{}
++++++++T+mean()+{
+++++++++++return+accumulate(begin(),end(),0.0)/size();
++++++++}
++++++++T+stddev()+{
+++++++++++T+diff_sum+=+0;
+++++++++++T+m+=+mean();
+++++++++++for(iterator+it=+begin();+it+!=+end();+%2B%2Bit)
+++++++++++++++diff_sum+%2B=+((*it+-+m)*(*it+-m));
+++++++++++return+diff_sum/size();
++++++++}
};|code-block|syntax|javascript|549966|它确实有一些局限性，但当你知道自己在做什么的时候，它就会工作得很好。|549967|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Create your own container:

<pre><code>template &lt;class T&gt;
class statList : public std::list&lt;T&gt;
{
 public:
 statList() : std::list&lt;T&gt;::list() {}
 ~statList() {}
 T mean() {
 return accumulate(begin(),end(),0.0)/size();
 }
 T stddev() {
 T diff_sum = 0;
 T m = mean();
 for(iterator it= begin(); it != end(); ++it)
 diff_sum += ((*it - m)*(*it -m));
 return diff_sum/size();
 }
};
</code></pre>

It does have some limitations, but it works beautifully when you know what you are doing.

blocks|key|766732|text|比前面提到的版本快2倍-主要是因为transform()和inner_product()循环结合在一起。抱歉，我的快捷方式/typedefs/宏:+Flo+=+float。CR常量参考VFlo向量。在VS2010中测试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|766733|#define+fe(EL,+CONTAINER)+++for+each+(auto+EL+in+CONTAINER)++//VS2010
Flo+stdDev(VFlo+CR+crVec)+{
++++SZ++n+=+crVec.size();+++++++++++++++if+(n+<+2)+return+0.0f;
++++Flo+fSqSum+=+0.0f,+fSum+=+0.0f;
++++fe(f,+crVec)+fSqSum+%2B=+f+*+f;+++++++//+EDIT:+was+Cit(VFlo,+crVec)+{
++++fe(f,+crVec)+fSum+++%2B=+f;
++++Flo+fSumSq++++++=+fSum+*+fSum;
++++Flo+fSumSqDivN++=+fSumSq+/+n;
++++Flo+fSubSqSum+++=+fSqSum+-+fSumSqDivN;
++++Flo+fPreSqrt++++=+fSubSqSum+/+(n+-+1);
++++return+sqrt(fPreSqrt);
}|code-block|syntax|javascript|766734|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

2x faster than the versions before mentioned - mostly because transform() and inner_product() loops are joined. 
Sorry about my shortcut/typedefs/macro: Flo = float. CR const ref. VFlo - vector. Tested in VS2010

<pre><code>#define fe(EL, CONTAINER) for each (auto EL in CONTAINER) //VS2010
Flo stdDev(VFlo CR crVec) {
 SZ n = crVec.size(); if (n &lt; 2) return 0.0f;
 Flo fSqSum = 0.0f, fSum = 0.0f;
 fe(f, crVec) fSqSum += f * f; // EDIT: was Cit(VFlo, crVec) {
 fe(f, crVec) fSum += f;
 Flo fSumSq = fSum * fSum;
 Flo fSumSqDivN = fSumSq / n;
 Flo fSubSqSum = fSqSum - fSumSqDivN;
 Flo fPreSqrt = fSubSqSum / (n - 1);
 return sqrt(fPreSqrt);
}
</code></pre>

blocks|key|550051|text|对the+answer+by+musiphil进行了改进，您可以在没有临时向量diff的情况下编写标准差函数，只需使用一个带有C%2B%2B11+lambda功能的inner_product调用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|550052|double+stddev(std::vector<double>+const+&+func)
{
++++double+mean+=+std::accumulate(func.begin(),+func.end(),+0.0)+/+func.size();
++++double+sq_sum+=+std::inner_product(func.begin(),+func.end(),+func.begin(),+0.0,
++++++++[](double+const+&+x,+double+const+&+y)+{+return+x+%2B+y;+},
++++++++[mean](double+const+&+x,+double+const+&+y)+{+return+(x+-+mean)*(y+-+mean);+});
++++return+std::sqrt(sq_sum+/+func.size());
}|code-block|syntax|javascript|550053|我怀疑多次做减法比消耗额外的中间存储更便宜，而且我认为它更具可读性，但我还没有测试它的性能。|550054|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/7616783/671366^0|13|4|26|D|1|M|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Improving on <a href="https://stackoverflow.com/a/7616783/671366">the answer by musiphil</a>, you can write a standard deviation function without the temporary vector <code>diff</code>, just using a single <code>inner_product</code> call with the C++11 lambda capabilities:
<pre><code>double stddev(std::vector&lt;double&gt; const &amp; func)
{
 double mean = std::accumulate(func.begin(), func.end(), 0.0) / func.size();
 double sq_sum = std::inner_product(func.begin(), func.end(), func.begin(), 0.0,
 [](double const &amp; x, double const &amp; y) { return x + y; },
 [mean](double const &amp; x, double const &amp; y) { return (x - mean)*(y - mean); });
 return std::sqrt(sq_sum / func.size());
}
</code></pre>
I suspect doing the subtraction multiple times is cheaper than using up additional intermediate storage, and I think it is more readable, but I haven't tested the performance yet.

blocks|key|550099|text|以下优雅的递归解决方案似乎没有被提及，尽管它已经存在了很长一段时间。参考Knuth的计算机编程艺术，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|550100|mean_1+=+x_1,+variance_1+=+0;++++++++++++//initial+conditions;+edge+case;

//for+k+>=+2,+
mean_k+++++=+mean_k-1+%2B+(x_k+-+mean_k-1)+/+k;
variance_k+=+variance_k-1+%2B+(x_k+-+mean_k-1)+*+(x_k+-+mean_k);|code-block|syntax|javascript|550101|然后，对于n>=2值列表，标准差的估计值为：|offset|length|style|CODE|550102|stddev+=+std::sqrt(variance_n+/+(n-1)).+|550103|希望这能有所帮助！|550104|entityMap^0|0|0|5|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

It seems the following elegant recursive solution has not been mentioned, although it has been around for a long time. Referring to Knuth's Art of Computer Programming, 

<pre><code>mean_1 = x_1, variance_1 = 0; //initial conditions; edge case;

//for k &gt;= 2, 
mean_k = mean_k-1 + (x_k - mean_k-1) / k;
variance_k = variance_k-1 + (x_k - mean_k-1) * (x_k - mean_k);
</code></pre>

then for a list of <code>n&gt;=2</code> values, the estimate of the standard deviation is: 

<pre><code>stddev = std::sqrt(variance_n / (n-1)). 
</code></pre>

Hope this helps!

blocks|key|550155|text|为了以更好的精度计算样本均值，可以使用以下r步递归：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|550156|均值_k=1/k*(k-r)*mean_(k-r)%2B+sum_over_i_from_(n-r%2B1)_to_n(x_i)，|550157|其中，选择r是为了使求和分量彼此更接近。|550158|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

In order to calculate the sample mean with a better presicion the following r-step recursion can be used:
mean_k=1/k*[(k-r)*mean_(k-r) + sum_over_i_from_(n-r+1)_to_n(x_i)],
where r is chosen to make summation components closer to each other.

Is there a way to calculate mean and standard deviation for a vector containing samples using <a href="http://en.wikipedia.org/wiki/Boost_%28C%2B%2B_libraries%29" rel="noreferrer">Boost</a>? 

Or do I have to create an accumulator and feed the vector into it?

Calculate mean and standard deviation from a vector of samples in C++ using Boost

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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有没有办法使用计算包含样本的向量的均值和标准差？或者我必须创建一个累加器并将向量输入其中？

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