Create a row Matrix

Question

Konstantinos 2015-1-12

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评论： John D'Errico 2015-1-13

I want to create a row matrix of 20 elements (1x20), which consists of 1s and 0s. 1s are generated with rate:r and 0s are generated with rate:(1-r), where r could be any random number between 0 and 1 ( 0<r<1 ).

Any help could be useful. Thanks in advance!

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Answer 1

Star Strider 2015-1-12

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编辑：Star Strider 2015-1-12

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This is one option:

r = 0.5;
idx = randperm(20, ceil(r*20));
rv = zeros(1,20);
rv(idx) = 1;

The logic is relatively straightforward: ‘idx’ uses rendperm to generate a matrix of index positions, then generates ‘rv’ and uses ‘idx’ to determine the r*20 values that are set to 1.

With r = 0.5, this yields:

rv =
    Columns 1 through 17
       0     1     0     0     1     0     0     1     1     1     1     0     1     0     0     1     1
    Columns 18 through 20
       1     0     0

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Konstantinos 2015-1-12

Thanks a lot guys for your time. I really appreciate your help!

Star Strider 2015-1-12

Our pleasure!

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Answer 2

John D'Errico 2015-1-12

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编辑：John D'Errico 2015-1-12

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Well, since a test returns a 0 or 1 (i.e., true or false) what is the probability that each element of a set of random samples, taken by the call rand(1,20), is greater than r? What is the probability that they are less than r?

x = (rand(1,20) >= r);

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John D'Errico 2015-1-13

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Sigh. I think you do not understand basic probability.

There is NO reason that a fair coin tossed 20 times will always land heads up exactly 10 of them. Similarly, with r = 0.37 for a binomial sample, the probability of receiving exactly 8 ones and 12 zeros is not even the MOST probable event! I'll show this as a simulation, but it is trivial to compute the exact probabilities as I do below for a couple of cases.

hist(sum(rand(1e7,20)<0.37,2),0:20)
grid on

As this simulation shows with a sample size of 1e7 sets of 20 samples, 7 ones is actually more probable than 8.

We can compute the actual probability easily enough. Thus we get 7 ones with probability:

nchoosek(20,7)*0.37^7*(1-0.37)^13
ans =
         0.181239550087757

8 ones is indeed less probable.

nchoosek(20,8)*0.37^8*(1-0.37)^12
ans =
         0.172968697603594

If you prefer simulation to back up the mathematics...

R = sum(rand(1e7,20)<0.37,2);
sum(R == 7)
ans =
     1812235
sum(R == 8)
ans =
     1729849

The point is, IF you insist that for a probability of getting a 1 there to be 0.37, AND that the actual number of ones will be 8, then you really do not have a random sample with probability r=0.37.

I think it really is time for you to do some reading. This is probability 101.

Star Strider 2015-1-13

@John — That seems to be the crux of the issue. I voted for your answer because in Konstantinos’ original Question, it was not possible to determine what was desired. By my reading, both your Answer and mine are equally valid.

John D'Errico 2015-1-13

Star - Oh, I don't dispute that you may arguably have a valid answer of the question. Questions are frequently slightly ambiguous, so they can often be read many ways. My patience today is clearly stressed by probabilistically meaningless statements like this:

"if r = 0.37, then the number of ones I want is: N1 = ceil(r*20)=8 1s and No=(1 - N1)=12 0s"

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