Generating multiple new arrays using numpy

Question

Say I have the following two different states:

state1 = [0, 1, 0, 1, 1] state2 = [1, 1, 0, 0, 1] 

And I want to generate n number of new states by only changing the values at where these states differ; and I want to do so randomly. So I did the following:

state1 = np.array(state1) differ = np.where(state1 != state2)[0] # --> array([0, 3], dtype=int64) rand = [random.choices([1, 0], k=len(differ)) for _ in range(4)] states = [state1.copy() for _ in range(4)] for i in range(len(states)): states[i][differ] = rand[i] # Will replace the values at locations where they are mismatched only 

Out:

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

This does the job but I was wondering if there is a better way (to avoid 2 for loops), but this was the best I could come up with. Any ideas on how to do the same thing but better/cleaner?

Answer 1

You could use np.random.choice() to generate the states, while using a==b and a!=b as masks.

def new_states(a,b,n): return [ a*(a==b) + np.random.choice(2,len(a))*(a!=b) for _ in range(n) ] state1 = np.array([0, 1, 0, 1, 1]) state2 = np.array([1, 1, 0, 0, 1]) print(new_states(state1,state2,4))