Issue
How can I do the indexing of some arrays used as indices? I have the following six 2D arrays like this-
array([[2, 0],
[3, 0],
[3, 1],
[5, 0],
[5, 1],
[5, 2]])
I want to use these arrays as indices and put the value 10 in the corresponding indices of a new empty matrix. The output should look like this-
array([[ 0, 0, 0],
[ 0, 0, 0],
[10, 0, 0],
[10, 10, 0],
[ 0, 0, 0],
[10, 10, 10]])
So far I have tried this-
from numpy import*
a = array([[2,0],[3,0],[3,1],[5,0],[5,1],[5,2]])
b = zeros((6,3),dtype ='int32')
b[a] = 10
But this gives me the wrong output.
Solution
In [1]: import numpy as np
In [2]: a = np.array([[2,0],[3,0],[3,1],[5,0],[5,1],[5,2]])
In [3]: b = np.zeros((6,3), dtype='int32')
In [4]: b[a[:,0], a[:,1]] = 10
In [5]: b
Out[5]:
array([[ 0, 0, 0],
[ 0, 0, 0],
[10, 0, 0],
[10, 10, 0],
[ 0, 0, 0],
[10, 10, 10]])
Why it works:
If you index b
with two numpy arrays in an assignment,
b[x, y] = z
then think of NumPy as moving simultaneously over each element of x
and each element of y
and each element of z
(let's call them xval
, yval
and zval
), and assigning to b[xval, yval] the value zval
. When z
is a constant, "moving over z
just returns the same value each time.
That's what we want, with x
being the first column of a
and y
being the second column of a
. Thus, choose x = a[:, 0]
, and y = a[:, 1]
.
b[a[:,0], a[:,1]] = 10
Why b[a] = 10
does not work
When you write b[a]
, think of NumPy as creating a new array by moving over each element of a
, (let's call each one idx
) and placing in the new array the value of b[idx]
at the location of idx
in a
.
idx
is a value in a
. So it is an int32. b
is of shape (6,3), so b[idx]
is a row of b
of shape (3,). For example, when idx
is
In [37]: a[1,1]
Out[37]: 0
b[a[1,1]]
is
In [38]: b[a[1,1]]
Out[38]: array([0, 0, 0])
So
In [33]: b[a].shape
Out[33]: (6, 2, 3)
So let's repeat: NumPy is creating a new array by moving over each element of a
and placing in the new array the value of b[idx]
at the location of idx
in a
. As idx
moves over a
, an array of shape (6,2) would be created. But since b[idx]
is itself of shape (3,), at each location in the (6,2)-shaped array, a (3,)-shaped value is being placed. The result is an array of shape (6,2,3).
Now, when you make an assignment like
b[a] = 10
a temporary array of shape (6,2,3) with values b[a]
is created, then the assignment is performed. Since 10 is a constant, this assignment places the value 10 at each location in the (6,2,3)-shaped array.
Then the values from the temporary array are reassigned back to b
.
See reference to docs. Thus the values in the (6,2,3)-shaped array are copied back to the (6,3)-shaped b
array. Values overwrite each other. But the main point is you do not obtain the assignments you desire.
Answered By - unutbu
0 comments:
Post a Comment
Note: Only a member of this blog may post a comment.