CODEWITHSUNDEEP
numpy
Stücke
Stücke

Reputation: 993

Array of 1-dimensional array and array row yields different results

I am having difficulty to understand why neither B or C equals A. How can I extract a specific row from f and calculate the same result as in A?

import numpy as np

L = np.array([
    [2.66667,1.33333],
    [0.8,1.6]
    ])

f = np.array([[0.5,0.333333]])

A = L*f.T

B = L*f[0,:]
C = L*f[0,:].T

print(20*'*')
print(A)
print(20*'*')
print(B)
print(20*'*')
print(C)
print(20*'*')

Output:

********************
[[1.333335  0.666665 ]
 [0.2666664 0.5333328]]
********************
[[1.333335   0.44444289]
 [0.4        0.5333328 ]]
********************
[[1.333335   0.44444289]
 [0.4        0.5333328 ]]
********************

Upvotes: 0

Views: 30

Answers (1)

hpaulj
hpaulj

Reputation: 231615

L = np.array([
    [2.66667,1.33333],
    [0.8,1.6]
    ])

f = np.array([[0.5,0.333333]])

L is (2,2), f is (1,2). f.T is (2,1)

A = L*f.T

This broadcasts the (2,2) with (2,1), replicating the the f.T columns to (2,2).

B = L*f[0,:]
C = L*f[0,:].T

f[0,:] is (2,) shape, 1d. f[0,:].T is the same because there's only one 1 axis to switch.

Here (2,2) times (2,) => (2,2) times (1,2), That's the same as L*f.

In A you are multiplying the columns of L by f:

In [235]: L[0,:]*f[0,:]
Out[235]: array([1.333335  , 0.44444289])

In B it's the rows of L:

In [236]: L[:,0]*f[0,:]
Out[236]: array([1.333335 , 0.2666664])

The mix and match of rows and columns could be clearer with L was (2,3) shaped.

edit

In [34]: np.array([L[:,i]*f[0] for i in range(2)]).T
Out[34]: 
array([[1.333335 , 0.666665 ],
       [0.2666664, 0.5333328]])
In [35]: np.array([L[i,:]*f[0] for i in range(2)])
Out[35]: 
array([[1.333335  , 0.44444289],
       [0.4       , 0.5333328 ]])

I got [235] and [236] switched.

Upvotes: 1

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