blocks|key|3184540|text|有很多方法可以做到这一点。一种可能性是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3184541|tmp=round(rand(size(vector1)))
res+=+tmp.*vector1+%2B+(1-tmp).*vector2|code-block|syntax|javascript|3184542|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|$]]

There are many ways you could do this. One possibility is:

<pre><code>tmp=round(rand(size(vector1)))
res = tmp.*vector1 + (1-tmp).*vector2
</code></pre>

blocks|key|1030313|text|要获得一个混合样本，您可以使用以下代码片段的想法(不是最优的，但可能足够清楚)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1030314|a++++++++=+[1,+3,+4,+7];
b++++++++=+[2,+5,+6,+8];
selector+=+randn(size(a));
sample+++=+a.*(selector>0)+%2B+b.*(selector<=0);|code-block|syntax|javascript|1030315|对于n示例，将上面的代码放在for循环中：|offset|length|style|CODE|1030316|for+k=1:n
++++%25+Sample+code+(without+initial+"samplee"+assignments)
++++%25+Here+do+stuff+with+the+sample
end;|1030317|entityMap^0|0|0|2|1|E|3|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

To get one mixed sample, you may use the idea of the following code snippet (not the optimal one, but maybe clear enough):

<pre><code>a = [1, 3, 4, 7];
b = [2, 5, 6, 8];
selector = randn(size(a));
sample = a.*(selector&gt;0) + b.*(selector&lt;=0);
</code></pre>

For <code>n</code> samples put the above code in a <code>for</code> loop:

<pre><code>for k=1:n
 % Sample code (without initial "samplee" assignments)
 % Here do stuff with the sample
end;
</code></pre>

blocks|key|3184638|text|更一般的情况是，如果X是一个矩阵，并且对于每一行，您希望从随机选择的列中抽取样本，则可以使用循环执行此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3184639|y+=+zeros(size(X,1),1);
for+ii+=+1:size(X,1)
++++y(ii)+=+X(ii,ceil(rand*size(X,2)));
end|code-block|syntax|javascript|3184640|您可以通过sub2ind使用巧妙的索引来避免循环：|3184641|idx_n+=+ceil(rand(size(X,1),1)*size(X,2));
idx+++=+sub2ind(size(X),(1:size(X,1))',idx_n);
y+=+X(idx);|3184642|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

More generally, if X is a matrix and for each row you want to take a sample from a column chosen at random, you can do this with a loop:

<pre><code>y = zeros(size(X,1),1);
for ii = 1:size(X,1)
 y(ii) = X(ii,ceil(rand*size(X,2)));
end
</code></pre>

You can avoid the loop using clever indexing via sub2ind:

<pre><code>idx_n = ceil(rand(size(X,1),1)*size(X,2));
idx = sub2ind(size(X),(1:size(X,1))',idx_n);
y = X(idx);
</code></pre>

blocks|key|4652458|text|如果我理解你的问题，你选择了两个随机数。首先，决定是选择向量1还是向量2；接下来，从所选向量中选择一个元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4652459|下面的代码利用了vector1和vector2长度相同的事实：|4652460|N+=+1000;
sampleMatrix+=+[vector1+vector2];
M+=+numel(sampleMatrix);
randIndex+=+ceil(rand(1,N)*M);+%25+N+random+numbers+from+1+to+M
randomNumbers+=+sampleMatrix(randIndex);+%25+sample+N+times+from+the+matrix|code-block|syntax|javascript|4652461|然后，您可以使用来显示结果，例如|4652462|figure;+hist(randomNumbers);+%25+draw+a+histogram+of+numbers+drawn|4652463|当vector1和vector2具有不同的元素时，就会遇到问题。如果将它们连接起来，最终将更频繁地从较长的向量中挑选元素。一种解决方法是从两个数组中创建随机采样，然后在它们之间进行选择：|4652464|M1+=+numel(vector1);
M2+=+numel(vector2);
r1+=+ceil(rand(1,N)*M1);
r2+=+ceil(rand(1,N)*M2);
randMat+=+[vector1(r1(:))+vector2(r2(:))];+%25+two+columns,+now+pick+one+or+the+other
randPick+=+ceil(rand(1,N)*2);
randomNumbers+=+[randMat(randPick==1,+1);+randMat(randPick==2,+2)];|4652465|在重新阅读时，您可能只想为向量的所有元素选择"element+1+from+1+or+2"，然后选择"element+2+from+1+or+2"，依此类推。在这种情况下，请这么做|4652466|N=numel(vector1);
randPick+=+ceil(rand(1,N)*2);
randMat=[vector1(:)+vector2(:)];
randomNumbers+=+[randMat(randPick==1,+1);+randMat(randPick==2,+2)];|4652467|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

If I understand your question, you are choosing two random numbers. First you decide whether to select vector 1 or vector 2; next you pick an element from the chosen vector.

The following code takes advantage of the fact that vector1 and vector2 are the same length:

<pre><code>N = 1000;
sampleMatrix = [vector1 vector2];
M = numel(sampleMatrix);
randIndex = ceil(rand(1,N)*M); % N random numbers from 1 to M
randomNumbers = sampleMatrix(randIndex); % sample N times from the matrix
</code></pre>

You can then display the result with, for instance

<pre><code>figure; hist(randomNumbers); % draw a histogram of numbers drawn
</code></pre>

When vector1 and vector2 have different elements, you run into a problem. If you concatenate them, you will end up picking elements from the longer vector more often. One way around this is to create random samplings from both arrays, then choose between them:

<pre><code>M1 = numel(vector1);
M2 = numel(vector2);
r1 = ceil(rand(1,N)*M1);
r2 = ceil(rand(1,N)*M2);
randMat = [vector1(r1(:)) vector2(r2(:))]; % two columns, now pick one or the other
randPick = ceil(rand(1,N)*2);
randomNumbers = [randMat(randPick==1, 1); randMat(randPick==2, 2)];
</code></pre>

On re-reading, maybe you just want to pick "element 1 from either 1 or 2", then "element 2 from either 1 or 2", etc for all the elements of the vector. In that case, do

<pre><code>N=numel(vector1);
randPick = ceil(rand(1,N)*2);
randMat=[vector1(:) vector2(:)];
randomNumbers = [randMat(randPick==1, 1); randMat(randPick==2, 2)];
</code></pre>

blocks|key|3184676|text|这个问题可以使用函数datasample来解决。将两个向量合并为一个并应用函数。与其他答案中的手工版本相比，我更喜欢这种方法。它给了你更多的灵活性来选择你真正想要的东西，同时又是一行程序。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3184677|entityMap|0|LINK|mutability|MUTABLE|url|https://de.mathworks.com/help/stats/datasample.html^0|A|A|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

This problem can be solved using the function <a href="https://de.mathworks.com/help/stats/datasample.html" rel="nofollow noreferrer">datasample</a>.
Combine both vectors into one and apply the function. I like this approach more than the handcrafted versions in the other answers. It gives you much more flexibility in choosing what you actually want, while being a one-liner.

How does one create a vector that is composed of a random sampling of two other vectors?

For example 

Vector 1 [1, 3, 4, 7], Vector 2 [2, 5, 6, 8] 
Random Vector [random draw from vector 1 or 2 (value 1 or 2), random draw from vector 1 or 2 (value 3 or 5)... etc]

Finally, how can one ask matlab to repeat this process n times to draw a distribution of results?

Thank you,

Creating a vector with random sampling of two vectors in matlab

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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如何创建一个由两个其他向量的随机采样组成的向量？例如矢量1 1，3，4，7，矢量2 2，5，6，8随机矢量从矢量1或2(值1或2)随机抽取，从矢量1或2(值3或5)随机抽取...等最后，如何让matlab重复此过程n次以绘制结果分布？谢谢,

问在matlab中创建具有两个向量的随机采样的向量
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