blocks|key|3202152|text|从你想要的评论来看：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3202153|H+=+magic(5);+%25Suppose+this+is+your+matrix
LimiteInferiore+=+23;+%25And+suppose+this+is+your+treshold

Media_b=+NaN(size(H,1),1);+%25Or+perhaps+zeros,+whatever+you+like+to+show+by+default

idx+=+any(H+>+treshold,2);
Media_b(idx)+=+mean(H(H>treshold),2);|code-block|syntax|javascript|3202154|如果它存在，它将给出行平均值，而NaN则给出其他值。我相信您不需要NaN平均值，因为NaN值不会计算为>treshold。|offset|length|style|CODE|3202155|entityMap^0|0|0|1F|9|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Judging from the comments you want this:

<pre><code>H = magic(5); %Suppose this is your matrix
LimiteInferiore = 23; %And suppose this is your treshold

Media_b= NaN(size(H,1),1); %Or perhaps zeros, whatever you like to show by default

idx = any(H &gt; treshold,2);
Media_b(idx) = mean(H(H&gt;treshold),2);
</code></pre>

It will give you the rowmean if it exists, and NaN otherwise. I believe you won't need NaN mean as NaN values won't evaluate as <code>&gt;treshold</code>.

blocks|key|2060589|text|对于矩阵H和阈值T，可以使用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2060590|M+=+nanmean(H%2B0./(H>T),2);|code-block|syntax|javascript|2060591|就你而言：|2060592|Media_b+=+nanmean(W_tot_migl%2B0./(W_tot_migl>LimiteInferiore),2);|2060593|或者自己计算：|2060594|M+=+sum(H.*(H>T),2)./sum(H>T,2);|2060595|就你的情况而言：|2060596|Media_b+=+sum(W_tot_migl.*(W_tot_migl>LimiteInferiore),2)./sum(W_tot_migl>LimiteInferiore,2);|2060597|请注意，如果行中没有大于阈值的值，则这两个方法都返回NaN。|2060598|但是，通常您可以使用accumarray来完成这一任务：|2060599|H2+=+H>T;
[I,~]+=+find(H2);
M+=+accumarray(I,+H(H2),+[size(H,1)+1],+@mean);|2060600|如果一行中没有大于阈值的元素，则此方法返回零。|2060601|entityMap^0|4|1|8|1|0|0|0|0|0|0|0|0|Q|3|0|A|A|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1C|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1G|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|1L|8|@$9|1M|A|1N|B|C]]|D|@]|E|$]]|$1|10|3|11|5|H|7|1O|8|@]|D|@]|E|$I|J]]|$1|12|3|13|5|6|7|1P|8|@]|D|@]|E|$]]|$1|14|3|-4|5|6|7|1Q|8|@]|D|@]|E|$]]]|15|$]]

For matrix <code>H</code> and threshold <code>T</code> you can use:

<pre><code>M = nanmean(H+0./(H&gt;T),2);
</code></pre>

in your case:

<pre><code>Media_b = nanmean(W_tot_migl+0./(W_tot_migl&gt;LimiteInferiore),2);
</code></pre>

OR calculate it yourself:

<pre><code>M = sum(H.*(H&gt;T),2)./sum(H&gt;T,2);
</code></pre>

For your case:

<pre><code>Media_b = sum(W_tot_migl.*(W_tot_migl&gt;LimiteInferiore),2)./sum(W_tot_migl&gt;LimiteInferiore,2);
</code></pre>

Note that both methods return <code>NaN</code> if there are no values greater than the threshold in a row.

However, in general you can use <code>accumarray</code> to accomplish this:

<pre><code>H2 = H&gt;T;
[I,~] = find(H2);
M = accumarray(I, H(H2), [size(H,1) 1], @mean);
</code></pre>

This method returns zero if there are no elements larger than the threshold in a row.

I need to compute the mean of a matrix but only for those values that are greater than a specified number. Right now I come with this idea:

<pre><code>Media = mean(W_tot_migl,2);

H = W_tot_migl;
H(H&lt;LimiteInferiore) = nan;
Media_b = nanmean(H,2);
</code></pre>

Is there any way to avoid to create another matrix <code>H = W_tot_migl</code>?

Matrix mean of values >x

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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我需要计算矩阵的平均值，但只计算那些大于指定数的值。现在我带着这样的想法：Media = mean(W_tot_migl,2);H = W_tot_migl;H(H<LimiteInferiore) = nan;Media_b = nanmean(H,2);有没有办法避免创建另一个矩阵H = W_tot_migl？

问矩阵均值>x
EN

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问矩阵均值>xEN