Putting 2×2 matrices in ndarray without using "for"-CodePudding

Numpy

First, I define "x" and function "example" below:

import numpy as np
x=np.arange(1,4,1)
def example(d):
    return 2*d

In this simple case, if I put "x" into "example",

y=example(x)

I get "y", which has a same length as x and I can calculate quickly.

However, if I define example2 as below,

def example2(d):
    matrix=np.array([[d,0],[0,d])
    return matrix

and input "x" into example2,

y2=example2(x)

What I get is

[[array([1, 2, 3]) 0]
 [0 array([1, 2, 3])]]
<ipython-input-9-21f94eb7b4f4>:2: VisibleDeprecationWarning: Creating an ndarray from ragged nested sequences (which is a list-or-tuple of lists-or-tuples-or ndarrays with different lengths or shapes) is deprecated. If you meant to do this, you must specify 'dtype=object' when creating the ndarray
  matrix=np.array([[d,0],[0,d]])

There is a warning as well.

I really want to create a ndarray(or list) which contain a matrix in each term without using "for". I can do this if I use "append", but it takes much longer to finish calculations.

A1=np.array([[1,0],[0,1]])
A2=np.array([[2,0],[0,2]])
A3=np.array([[3,0],[0,3]])
y3=np.array([A1,A2,A3])

If I name each matrix A1,A2,A3, which are created in example2, what I want to do is to create y3 by just inputting "x" into a function I created.

Are there any solutions?

CodePudding user response：

not following but:


import numpy as np


x=np.arange(1,4,1)
def example(d):
    return 2*d


y=example(x)

print(y)

def example2(d):
    matrix=np.array([[d,0],[0,d]], dtype= object)
    return matrix



y2=example2(x)

print('\n___________________________')
print(y2)

gives:

[2 4 6]

___________________________
[[array([1, 2, 3]) 0]
 [0 array([1, 2, 3])]]

so no error, not sure it is was you were looking for, just post your desidered output to help the others to help you

CodePudding user response：

You could initial an array of the desired shape, and assign selected values:

In [93]: arr = np.zeros((3,2,2),int)
    ...: arr[:,0,0] = [1,2,3]
    ...: arr[:,1,1] = [4,5,6]
    ...: 
    ...: 
In [94]: arr
Out[94]: 
array([[[1, 0],
        [0, 4]],

       [[2, 0],
        [0, 5]],

       [[3, 0],
        [0, 6]]])

The same thing, but with one advanced-indexing assignment:

In [97]: arr = np.zeros((3,2,2),int)
    ...: arr[:,[0,1],[0,1]] = np.array([[1,2,3],[4,5,6]]).T 
In [98]: arr
Out[98]: 
array([[[1, 0],
        [0, 4]],

       [[2, 0],
        [0, 5]],

       [[3, 0],
        [0, 6]]])

This required several attempts, since I was getting broadcasting errors. Getting dimensions right when making assignments like this is tricky.