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I have a numpy array,

a = np.zeros((5,2))

a = array([[0., 0.],
           [0., 0.],
           [0., 0.],
           [0., 0.],
           [0., 0.]])

Aim: Each value should have a probability of changing, p = 0.05, and the value it changes to is given by a sample from a normal distribution with mean = 1, st.dev = 0.2

So far, I have tried following This:

a[np.random.rand(*a.shape) < 0.05] = rng.normal(loc=1,scale=0.2)

This does change values randomly with p = 0.05 ,but all values are the same, which is not ideal.

So, how does one go about making sure that each sampled value is independent(without using for loop)?

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  • Just use np.random.seed(1), this will should be enough to get different values. Commented Aug 21, 2021 at 6:16
  • 2
    @Gray_Rhino That doesn't do anything to address his issue. Commented Aug 21, 2021 at 6:23
  • You have to use a for loop. The problem is you don't know how many numbers you'll need until the left side of the = is evaluated, and by then it's too late. Commented Aug 21, 2021 at 6:24
  • @Gray_Rhino, that would not work, I think the issue lies in the "rng.normal(loc=1,scale=0.2)". it gets a single output and sets it for all changing indices rather than sampling independently for every index Commented Aug 21, 2021 at 6:27
  • @TimRoberts for loop seems the only way then Commented Aug 21, 2021 at 6:28

1 Answer 1

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from scipy import stats

a = np.zeros((5,2))

a[:] = np.where(np.random.rand(*a.shape) < 0.05,
                stats.norm.rvs(loc=1,scale=0.2, size = a.shape),
                a)
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