I want to rebuild the following logic with numpy broadcasting function such as np.where: From a 2d array check per row if the first element satisfies a condition. If the condition is true then return the first three elements as a row, else the last three elements.
A short MWE in form of a for-loop which I want to circumvent:
import numpy as np
array = np.array([
[1, 2, 3, 4],
[1, 2, 4, 2],
[2, 3, 4, 6]
])
new_array = np.zeros((array.shape[0], array.shape[1]-1))
for i, row in enumerate(array):
if row[0] == 1: new_array[i] = row[:3]
else: new_array[i] = row[-3:]
IIUC you want something like this:
condition = array[:,0]==1
new_array[condition,:] = array[condition,:3]
new_array[~condition,:] = array[~condition,-3:]
If you want to use np.where:
import numpy as np
array = np.array([
[1, 2, 3, 4],
[1, 2, 4, 2],
[2, 3, 4, 6]
])
cond = array[:, 0] == 1
np.where(cond[:, None], array[:,:3], array[:,-3:])
output:
array([[1, 2, 3],
[1, 2, 4],
[3, 4, 6]])
EDIT
slightly more concise version:
np.where(array[:, [0]] == 1, array[:,:3], array[:,-3:])