用另一个数组中的值替换 numpy 数组中的所有 -1

Question

I have two numpy arrays, eg:

x = np.array([4, -1, 1, -1, -1, 2])
y = np.array([1, 2, 3, 4, 5, 6])

I would like to replace all -1 in x with numbers in y , that are not in x . First -1 with first number in y that is not in x ( 3 ), second -1 with second number in y that is not in x ( 5 ), ... So final x should be:

[4 3 1 5 6 2]

I created this function:

import numpy as np
import time

start = time.time()

def fill(x, y):
    x_i = 0
    y_i = 0

    while x_i < len(x) and y_i < len(y):
        if x[x_i] != -1: # If current value in x is not -1.
            x_i += 1
            continue

        if y[y_i] in x: # If current value in y is already in x.
            y_i += 1
            continue

        x[x_i] = y[y_i] # Replace current -1 in x by current value in y.


for i in range(10000):
    x = np.array([4, -1, 1, -1, -1, 2])
    y = np.array([1, 2, 3, 4, 5, 6])
    fill(x, y)

end = time.time()
print(end - start) # 0.296

It's working, but I need run this function many times (eg milion times), so I would like to optimize it. Is there any way?

Answer 1

You could do:

import numpy as np

x = np.array([4, -1, 1, -1, -1, 2])
y = np.array([1, 2, 3, 4, 5, 6])

# create set of choices
sx = set(x)
choices = np.array([yi for yi in y if yi not in sx])

# assign new values
x[x == -1] = choices[:(x == -1).sum()]

print(x)

Output

[4 3 1 5 6 2]

Answer 2

y_not_in_x = np.setdiff1d(y, x)
x_eq_neg1 = x == -1
n_neg1s = np.sum(x_eq_neg1)
x[x_eq_neg1] = y_not_in_x[:n_neg1s]