I want to generate random binary numbers (0 or 1) having N length size. The tricky part is that, It should be able to control the probability of having either more 1 or 0. For example, I want total 100 random numbers with 0 having probability of 40% and 1 having probability of 60%. Please help.
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4C++ or python? Those are vastly different languages. – Lukas-T Aug 04 '21 at 08:47
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For Python, see https://stackoverflow.com/questions/4265988/generate-random-numbers-with-a-given-numerical-distribution – Sefan Aug 04 '21 at 08:48
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3Step 1: work out how to generate a 0 or 1 with the stated probabilities. Step 2: do this N times. Step 3: do *that* 100 times. – Botje Aug 04 '21 at 08:49
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1`(np.random.rand(100)>0.4).astype(int)` – Julien Aug 04 '21 at 08:50
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For C++, [`std::bernoulli_distribution`](https://en.cppreference.com/w/cpp/numeric/random/bernoulli_distribution) – Yksisarvinen Aug 04 '21 at 08:50
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C++ or python any one is good. – Liril Silvi Aug 04 '21 at 09:12
3 Answers
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A general solution for controlling this distribution is as follows:
First generate a uniform random number between 0-100 (or 0-1000 for more control, i.e if you need 60.1% chance for a number)
Then if the number is below or equal to 60, assign 1, now you have a 60% chance to assign 1.
I hope this helps, I think you will figure it out.
GustavD
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I did not get 60.1% chance. It still has around ±10-20% of deviation. Any solution? – Liril Silvi Aug 05 '21 at 05:50
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Well, if you flip a coin 10 times most of the time you won't get exactly 5 heads and 5 tails. Maybe you asked your original question wrong. If you need exactly a certain number of 1's and 0's, I think you want to create an array with exactly the number of elements you want, and then insert those elements at random positions in a new array. – GustavD Aug 05 '21 at 08:51
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I made an array of N numbers with 30% 1s and 70% 0s. e.g. [1 1 1 0 0 0 0 0 0 0]. and then used random shuffle function to randomly shuffle the positions of each element. and it produced the results :[0 0 1 0 1 0 0 0 0 0]. Which looks good to me. Thanks. – Liril Silvi Aug 06 '21 at 13:05
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You can always store the count of 0s and 1s. Here since you need a total of 100 random numbers with 0 having the probability of 40% and 1 having the probability of 60% so lets initialize count_0=40, count_1=60, total_count=100. Now you can generate a random number between 0 and 100.
Let us assume the first randomly generated number happens to be 35. Since 35 is less than 40 so first result happens to be 0. Decrement count_0. Decrement total_count. Now, for the next random no pick a randomly generated number between 0 and 99(total_count). This time if the randomly generated number is less than or equal to count_0 ( which equals 39 ) you set the result to be 0 else 1.
Follow this process for 100 iterations to generate 100 random 0s and 1s. This approach uses no extra space and O(n) time complexity where n is the number of 0s and 1s to be generated.
Srishti Ahuja
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In Python:
prob = 0.6 #p = prob of having 1
n_samples = np.random.choice([0,1], size=N, p=[prob, 1-prob])
