I want to generate a random array of size N which only contains 0 and 1, I want my array to have some ratio between 0 and 1. For example, 90% of the array be 1 and the remaining 10% be 0 (I want this 90% to be random along with the whole array).
right now I have:
randomLabel = np.random.randint(2, size=numbers) But I can't control the ratio between 0 and 1.
If you want an exact 1:9 ratio:
nums = numpy.ones(1000) nums[:100] = 0 numpy.random.shuffle(nums) If you want independent 10% probabilities:
nums = numpy.random.choice([0, 1], size=1000, p=[.1, .9]) or
nums = (numpy.random.rand(1000) > 0.1).astype(int) You could use a binomial distribution:
np.random.binomial(n=1, p=0.9, size=[1000]) Without using numpy, you could do as follows:
import random percent = 90 nums = percent * [1] + (100 - percent) * [0] random.shuffle(nums) Its difficult to get an exact count but you can get approximate answer by assuming that random.random returns a uniform distribution. This is strictly not the case, but is only approximately true. If you have a truly uniform distribution then it is possible. You can try something like the following:
In [33]: p = random.random(10000) In [34]: p[p <= 0.1] = 0 In [35]: p[p > 0] = 1 In [36]: sum(p == 0) Out[36]: 997 In [37]: sum(p == 1) Out[37]: 9003 Hope this helps ...
I recently needed to do something similar as part of a compression experiment. Posting the code to generate [n] random bytes with a given ration of zeros to ones for the next person who googles...
import random from bitstring import BitArray def get_by_fives(): for i in range(60, 100, 5): percent_0 = i percent_1 = 100-i set_count = 0 ba = BitArray(8388608) # 1MB for j in range(0, len(ba)): die_roll = random.randint(0, 100) set = die_roll > percent_0 if set: ba.set(True, j) set_count +=1 file_name = "{}_percent_bits_set".format(percent_1, percent_0) print("writing {} with set_count {} to disk".format(file_name, set_count)) with open(file_name, 'wb') as f: ba.tofile(f) if __name__ == "__main__": get_by_fives() 