CODEWITHSUNDEEP
pythonarraysnumpyrandom
Cupitor
Cupitor

Reputation: 11637

Binary random array with a specific proportion of ones?

What is the efficient(probably vectorized with Matlab terminology) way to generate random number of zeros and ones with a specific proportion? Specially with Numpy?

As my case is special for 1/3, my code is:

import numpy as np 
a=np.mod(np.multiply(np.random.randomintegers(0,2,size)),3)

But is there any built-in function that could handle this more effeciently at least for the situation of K/N where K and N are natural numbers?

Upvotes: 93

Views: 86181

Answers (7)

Ali Chitsaz
Ali Chitsaz

Reputation: 1

You can generate a nd-array with random binary members (0 and 1) directly in one line through the following method. You can also use np.random.random() instead of np.random.uniform().

>>import numpy as np
>>np.array([[round(np.random.uniform()) for i in range(3)] for j in  range(3)])
array([[1, 0, 0],
       [1, 1, 1],
       [0, 1, 0]])
>>

Upvotes: 0

Galactic Ketchup
Galactic Ketchup

Reputation: 504

Simple one-liner: you can avoid using lists of integers and probability distributions, which are unintuitive and overkill for this problem in my opinion, by simply working with bools first and then casting to int if necessary (though leaving it as a bool array should work in most cases).

>>> import numpy as np
>>> np.random.random(9) < 1/3.
array([False,  True,  True,  True,  True, False, False, False, False])   
>>> (np.random.random(9) < 1/3.).astype(int)
array([0, 0, 0, 0, 0, 1, 0, 0, 1])

Upvotes: 1

joelostblom
joelostblom

Reputation: 48879

Another way of getting the exact number of ones and zeroes is to sample indices without replacement using np.random.choice:

arr_len = 30
num_ones = 8

arr = np.zeros(arr_len, dtype=int)
idx = np.random.choice(range(arr_len), num_ones, replace=False)
arr[idx] = 1

Out:

arr

array([0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1,
       0, 0, 0, 0, 0, 1, 0, 0])

Upvotes: 2

Jaime
Jaime

Reputation: 67427

Yet another approach, using np.random.choice:

>>> np.random.choice([0, 1], size=(10,), p=[1./3, 2./3])
array([0, 1, 1, 1, 1, 0, 0, 0, 0, 0])

Upvotes: 130

Abhijit
Abhijit

Reputation: 63707

If I understand your problem correctly, you might get some help with numpy.random.shuffle

>>> def rand_bin_array(K, N):
    arr = np.zeros(N)
    arr[:K]  = 1
    np.random.shuffle(arr)
    return arr

>>> rand_bin_array(5,15)
array([ 0.,  1.,  0.,  1.,  1.,  1.,  0.,  0.,  0.,  1.,  0.,  0.,  0.,
        0.,  0.])

Upvotes: 23

mdml
mdml

Reputation: 22882

A simple way to do this would be to first generate an ndarray with the proportion of zeros and ones you want:

>>> import numpy as np
>>> N = 100
>>> K = 30 # K zeros, N-K ones
>>> arr = np.array([0] * K + [1] * (N-K))
>>> arr
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1])

Then you can just shuffle the array, making the distribution random:

>>> np.random.shuffle(arr)
>>> arr
array([1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0,
       1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1,
       1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1,
       0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1,
       1, 1, 1, 0, 1, 1, 1, 1])

Note that this approach will give you the exact proportion of zeros/ones you request, unlike say the binomial approach. If you don't need the exact proportion, then the binomial approach will work just fine.

Upvotes: 52

Warren Weckesser
Warren Weckesser

Reputation: 114791

You can use numpy.random.binomial. E.g. suppose frac is the proportion of ones:

In [50]: frac = 0.15

In [51]: sample = np.random.binomial(1, frac, size=10000)

In [52]: sample.sum()
Out[52]: 1567

Upvotes: 21

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