Building NumPy array using values from another array

Question

Consider the following code:

import numpy as np 
index_info = np.matrix([[1, 1], [1, 2]])
value =  np.matrix([[0.5, 0.5]])
initial = np.zeros((3, 3))

How can I produce a matrix, final, which has the structure of initial with the elements specified by value at the locations specified by index_info WITHOUT a for loop? In this toy example, see below.

final =  np.matrix([[0, 0, 0], [0, 0.5, 0.5], [0, 0, 0]])

With a for loop, you can easily loop through all of the index's in index_info and value and use that to populate initial and form final. But is there a way to do so with vectorization (no for loop)?

Paul Panzer · Accepted Answer · 2018-03-03 20:17:04Z

2

Convert index_info to a tuple and use it to assign:

>>> initial[(*index_info,)]=value
>>> initial
array([[0. , 0. , 0. ],
       [0. , 0.5, 0.5],
       [0. , 0. , 0. ]])

Please note that use of the matrix class is discouraged. Use ndarray instead.

answered Mar 3, 2018 at 20:17

Paul Panzer

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Brad Solomon · Accepted Answer · 2018-03-03 20:17:19Z

1

You can do this with NumPy's array indexing:

>>> initial = np.zeros((3, 3))

>>> row = np.array([1, 1])
>>> col = np.array([1, 2])

>>> final = np.zeros_like(initial)
>>> final[row, col] = [0.5, 0.5]
>>> final
array([[0. , 0. , 0. ],
       [0. , 0.5, 0.5],
       [0. , 0. , 0. ]])

This is similar to @PaulPanzer's answer, where he is unpacking row and col from index_info all in one step. In other words:

row, col = (*index_info,)

answered Mar 3, 2018 at 20:17

Brad Solomon

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