convert np.where array to list

Question

I try to get indices of an array using np.where and wish to concatenate the lists in such a manner that it gives me a 1D list. Is it possible?

l = np.array([10,20,14,10,23,5,10,1,2,3,10,5,6,5,10])
y= np.where(l==10)
p=np.where(l==5)

If I print y and p, they give me

(array([ 0,  3,  6, 10, 14]),)

(array([ 5, 11, 13]),)

Upon appending it results in a list of tuples. However the output I want is this:

[0,3,6,10,14,5,11,13]

Answer 1

Since there are many other solutions, I'll show you another way.

You can use np.isin to test for good values in your array:

goovalues = {5, 10}
np.where(np.isin(l, goodvalues))[0].tolist() #  [0, 3, 6, 10, 14, 5, 11, 13]

Answer 2

You could concatinate both arrays and then convert the result to a list:

result = np.concatenate((y[0], p[0])).tolist()

Answer 3

You can access the lists with y[0] and p[0], then append the results. Just add the line:

r = np.append(y[0], p[0])

and r will be a np.array with the values you asked. Use list(r) if you want it as a list.

Answer 4

A way to do this with concatenate:

import numpy as np

l = np.array([10,20,14,10,23,5,10,1,2,3,10,5,6,5,10])
y = np.where(l==10)[0]
p = np.where(l==5)[0]
k = np.concatenate((y, p))

print(k) # [ 0  3  6 10 14  5 11 13]

Answer 5

An other addition to the existing ones in one line.

l = np.array([10,20,14,10,23,5,10,1,2,3,10,5,6,5,10])

y = np.where((l == 10) | (l == 5))[0]

Numpy works with operators such as & (and), | (or), and ~ (not). The where function returns a tuple in case you pass a boolean array and hence the index 0.

Hope this helps.

Answer 6

Try this

y = [items for items in y[0]]
p = [items for items in p[0]]

then

new_list = y + p