I want to create random variables in python and used the following below code
weights = np.random.random(10) but I want to create random variables such that one third of the weights should be zero. Is there any way possible? I have also tried below code but this is not what I want
weights = np.random.random(7)
weights.append(0, 0, 0)
With the clarification that you want the 0's to appear randomly, you can just use shuffle:
weights = np.random.random(7)
weights = np.append(weights,[0, 0, 0])
np.random.shuffle(weights)
weights[-3:] = [0,0,0]One simple way:
>>> import numpy as np
>>>
>>> a = np.clip(np.random.uniform(-0.5, 1, (100,)), 0, np.inf)
>>> a
array([0.39497669, 0.65003362, 0. , 0. , 0. ,
0.75545815, 0.30772786, 0.1805628 , 0. , 0. ,
0. , 0.82527704, 0. , 0.63983682, 0.89283051,
0.25173721, 0.18409163, 0.63631959, 0.59095185, 0. ,
0.85817311, 0. , 0.06769175, 0. , 0.67807471,
0.29805637, 0.03429861, 0.53077809, 0.32317273, 0.52346321,
0.22966515, 0.98175502, 0.54615167, 0. , 0.88853359,
0. , 0.70622272, 0.08106305, 0. , 0.8767082 ,
0.52920044, 0. , 0. , 0.29394736, 0.4097331 ,
0.77977164, 0.62860222, 0. , 0. , 0.14899124,
0.81880283, 0. , 0.1398242 , 0. , 0.50113732,
0. , 0.68872893, 0.15582668, 0. , 0.34789122,
0.18510949, 0.60281713, 0.21097922, 0.77419626, 0.29588479,
0.18890799, 0.9781896 , 0.96220508, 0.52201816, 0.71087763,
0. , 0.43540516, 0.99297503, 0. , 0.69248893,
0.05157044, 0. , 0.75131066, 0. , 0. ,
0.25627591, 0.53367521, 0.58151298, 0.85662171, 0.455367 ,
0. , 0. , 0.21293519, 0.52337335, 0. ,
0.68644488, 0. , 0. , 0.39695189, 0. ,
0.40860821, 0.84549468, 0. , 0.21247807, 0.59054669])
>>> np.count_nonzero(a)
67
It draws uniformly from [-0.5, 1] and then sets everything below zero to zero.
This will guarantee that approximately one third of your weights are 0:
weights = np.random.random(10)/np.random.choice([0,1],10,p=[0.3,0.7])
weights[np.isinf(weights)] = 0
# or
# weights[weights == np.inf] = 0
>>> weights
array([0. , 0.25715864, 0. , 0.80958258, 0.12880619,
0.48781856, 0.52278911, 0.76541417, 0.87736431, 0. ])
What it does is divides about 1/3 of your values by 0, giving you inf, then just replace the inf by 0
Alternatively, if you need it to be exactly 1/3 (or in your case, 3 out of 10), you can replace 1/3 of your weights with 0:
weights = np.random.random(10)
# Replace 3 with however many indices you want changed...
weights[np.random.choice(range(len(weights)),3,replace=False)] = 0
>>> weights
array([0. , 0.36839012, 0. , 0.51468295, 0.45694205,
0.23881473, 0.1223229 , 0.68440171, 0. , 0.15542469])
That selects 3 random indices from weights and replaces them with 0
size = 10
v = np.random.random(size)
v[np.random.randint(0, size, size // 3)] = 0
A little bit more optimized (because random number generation is not "cheap"):
v = np.zeros(size)
nnonzero = size - size // 3
idx = np.random.choice(size, nnonzero, replace=False)
v[idx] = np.random.random(nnonzero)
What about replacing the first third of items with 0 then shuffle it as following
weights = np.random.random(10)
weights[: weights.size / 3] = 0
np.random.shuffle(weights)
