how does multiplication differ for NumPy Matrix vs Array classes?

Question:

The numpy docs recommend using array instead of matrix for working with matrices. However, unlike octave (which I was using till recently), * doesn’t perform matrix multiplication, you need to use the function matrixmultipy(). I feel this makes the code very unreadable.

Does anybody share my views, and has found a solution?

Asked By: elexhobby

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Answers:

The main reason to avoid using the matrix class is that a) it’s inherently 2-dimensional, and b) there’s additional overhead compared to a “normal” numpy array. If all you’re doing is linear algebra, then by all means, feel free to use the matrix class… Personally I find it more trouble than it’s worth, though.

For arrays (prior to Python 3.5), use dot instead of matrixmultiply.

E.g.

import numpy as np
x = np.arange(9).reshape((3,3))
y = np.arange(3)

print np.dot(x,y)

Or in newer versions of numpy, simply use x.dot(y)

Personally, I find it much more readable than the * operator implying matrix multiplication…

For arrays in Python 3.5, use x @ y.

Answered By: Joe Kington

the key things to know for operations on NumPy arrays versus operations on NumPy matrices are:

  • NumPy matrix is a subclass of NumPy array

  • NumPy array operations are element-wise (once broadcasting is accounted for)

  • NumPy matrix operations follow the ordinary rules of linear algebra

some code snippets to illustrate:

>>> from numpy import linalg as LA
>>> import numpy as NP

>>> a1 = NP.matrix("4 3 5; 6 7 8; 1 3 13; 7 21 9")
>>> a1
matrix([[ 4,  3,  5],
        [ 6,  7,  8],
        [ 1,  3, 13],
        [ 7, 21,  9]])

>>> a2 = NP.matrix("7 8 15; 5 3 11; 7 4 9; 6 15 4")
>>> a2
matrix([[ 7,  8, 15],
        [ 5,  3, 11],
        [ 7,  4,  9],
        [ 6, 15,  4]])

>>> a1.shape
(4, 3)

>>> a2.shape
(4, 3)

>>> a2t = a2.T
>>> a2t.shape
(3, 4)

>>> a1 * a2t         # same as NP.dot(a1, a2t) 
matrix([[127,  84,  85,  89],
        [218, 139, 142, 173],
        [226, 157, 136, 103],
        [352, 197, 214, 393]])

but this operations fails if these two NumPy matrices are converted to arrays:

>>> a1 = NP.array(a1)
>>> a2t = NP.array(a2t)

>>> a1 * a2t
Traceback (most recent call last):
   File "<pyshell#277>", line 1, in <module>
   a1 * a2t
   ValueError: operands could not be broadcast together with shapes (4,3) (3,4) 

though using the NP.dot syntax works with arrays; this operations works like matrix multiplication:

>> NP.dot(a1, a2t)
array([[127,  84,  85,  89],
       [218, 139, 142, 173],
       [226, 157, 136, 103],
       [352, 197, 214, 393]])

so do you ever need a NumPy matrix? ie, will a NumPy array suffice for linear algebra computation (provided you know the correct syntax, ie, NP.dot)?

the rule seems to be that if the arguments (arrays) have shapes (m x n) compatible with the a given linear algebra operation, then you are ok, otherwise, NumPy throws.

the only exception i have come across (there are likely others) is calculating matrix inverse.

below are snippets in which i have called a pure linear algebra operation (in fact, from Numpy’s Linear Algebra module) and passed in a NumPy array

determinant of an array:

>>> m = NP.random.randint(0, 10, 16).reshape(4, 4)
>>> m
array([[6, 2, 5, 2],
       [8, 5, 1, 6],
       [5, 9, 7, 5],
       [0, 5, 6, 7]])

>>> type(m)
<type 'numpy.ndarray'>

>>> md = LA.det(m)
>>> md
1772.9999999999995

eigenvectors/eigenvalue pairs:

>>> LA.eig(m)
(array([ 19.703+0.j   ,   0.097+4.198j,   0.097-4.198j,   5.103+0.j   ]), 
array([[-0.374+0.j   , -0.091+0.278j, -0.091-0.278j, -0.574+0.j   ],
       [-0.446+0.j   ,  0.671+0.j   ,  0.671+0.j   , -0.084+0.j   ],
       [-0.654+0.j   , -0.239-0.476j, -0.239+0.476j, -0.181+0.j   ],
       [-0.484+0.j   , -0.387+0.178j, -0.387-0.178j,  0.794+0.j   ]]))

matrix norm:

>>>> LA.norm(m)
22.0227

qr factorization:

>>> LA.qr(a1)
(array([[ 0.5,  0.5,  0.5],
        [ 0.5,  0.5, -0.5],
        [ 0.5, -0.5,  0.5],
        [ 0.5, -0.5, -0.5]]), 
 array([[ 6.,  6.,  6.],
        [ 0.,  0.,  0.],
        [ 0.,  0.,  0.]]))

matrix rank:

>>> m = NP.random.rand(40).reshape(8, 5)
>>> m
array([[ 0.545,  0.459,  0.601,  0.34 ,  0.778],
       [ 0.799,  0.047,  0.699,  0.907,  0.381],
       [ 0.004,  0.136,  0.819,  0.647,  0.892],
       [ 0.062,  0.389,  0.183,  0.289,  0.809],
       [ 0.539,  0.213,  0.805,  0.61 ,  0.677],
       [ 0.269,  0.071,  0.377,  0.25 ,  0.692],
       [ 0.274,  0.206,  0.655,  0.062,  0.229],
       [ 0.397,  0.115,  0.083,  0.19 ,  0.701]])
>>> LA.matrix_rank(m)
5

matrix condition:

>>> a1 = NP.random.randint(1, 10, 12).reshape(4, 3)
>>> LA.cond(a1)
5.7093446189400954

inversion requires a NumPy matrix though:

>>> a1 = NP.matrix(a1)
>>> type(a1)
<class 'numpy.matrixlib.defmatrix.matrix'>

>>> a1.I
matrix([[ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028],
        [ 0.028,  0.028,  0.028,  0.028]])
>>> a1 = NP.array(a1)
>>> a1.I

Traceback (most recent call last):
   File "<pyshell#230>", line 1, in <module>
   a1.I
   AttributeError: 'numpy.ndarray' object has no attribute 'I'

but the Moore-Penrose pseudoinverse seems to works just fine

>>> LA.pinv(m)
matrix([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
        [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
        [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
        [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
        [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])

>>> m = NP.array(m)

>>> LA.pinv(m)
array([[ 0.314,  0.407, -1.008, -0.553,  0.131,  0.373,  0.217,  0.785],
       [ 1.393,  0.084, -0.605,  1.777, -0.054, -1.658,  0.069, -1.203],
       [-0.042, -0.355,  0.494, -0.729,  0.292,  0.252,  1.079, -0.432],
       [-0.18 ,  1.068,  0.396,  0.895, -0.003, -0.896, -1.115, -0.666],
       [-0.224, -0.479,  0.303, -0.079, -0.066,  0.872, -0.175,  0.901]])
Answered By: doug

This trick could be what you are looking for. It is a kind of simple operator overload.

You can then use something like the suggested Infix class like this:

a = np.random.rand(3,4)
b = np.random.rand(4,3)
x = Infix(lambda x,y: np.dot(x,y))
c = a |x| b
Answered By: Bitwise

There is a situation where the dot operator will give different answers when dealing with arrays as with dealing with matrices. For example, suppose the following:

>>> a=numpy.array([1, 2, 3])
>>> b=numpy.array([1, 2, 3])

Lets convert them into matrices:

>>> am=numpy.mat(a)
>>> bm=numpy.mat(b)

Now, we can see a different output for the two cases:

>>> print numpy.dot(a.T, b)
14
>>> print am.T*bm
[[1.  2.  3.]
 [2.  4.  6.]
 [3.  6.  9.]]
Answered By: Jadiel de Armas

Reference from http://docs.scipy.org/doc/scipy/reference/tutorial/linalg.html

…, the use of the numpy.matrix class is discouraged, since it adds nothing that cannot be accomplished with 2D numpy.ndarray objects, and may lead to a confusion of which class is being used. For example,

>>> import numpy as np
>>> from scipy import linalg
>>> A = np.array([[1,2],[3,4]])
>>> A
    array([[1, 2],
           [3, 4]])
>>> linalg.inv(A)
array([[-2. ,  1. ],
      [ 1.5, -0.5]])
>>> b = np.array([[5,6]]) #2D array
>>> b
array([[5, 6]])
>>> b.T
array([[5],
      [6]])
>>> A*b #not matrix multiplication!
array([[ 5, 12],
      [15, 24]])
>>> A.dot(b.T) #matrix multiplication
array([[17],
      [39]])
>>> b = np.array([5,6]) #1D array
>>> b
array([5, 6])
>>> b.T  #not matrix transpose!
array([5, 6])
>>> A.dot(b)  #does not matter for multiplication
array([17, 39])

scipy.linalg operations can be applied equally to numpy.matrix or to 2D numpy.ndarray objects.

Answered By: Yong Yang

In 3.5, Python finally got a matrix multiplication operator. The syntax is a @ b.

Answered By: Petr Viktorin

A pertinent quote from PEP 465 – A dedicated infix operator for matrix multiplication , as mentioned by @petr-viktorin, clarifies the problem the OP was getting at:

[…] numpy provides two different types with different __mul__ methods. For numpy.ndarray objects, * performs elementwise multiplication, and matrix multiplication must use a function call (numpy.dot). For numpy.matrix objects, * performs matrix multiplication, and elementwise multiplication requires function syntax. Writing code using numpy.ndarray works fine. Writing code using numpy.matrix also works fine. But trouble begins as soon as we try to integrate these two pieces of code together. Code that expects an ndarray and gets a matrix, or vice-versa, may crash or return incorrect results

The introduction of the @ infix operator should help to unify and simplify python matrix code.

Answered By: cod3monk3y

Function matmul (since numpy 1.10.1) works fine for both types and return result as a numpy matrix class:

import numpy as np

A = np.mat('1 2 3; 4 5 6; 7 8 9; 10 11 12')
B = np.array(np.mat('1 1 1 1; 1 1 1 1; 1 1 1 1'))
print (A, type(A))
print (B, type(B))

C = np.matmul(A, B)
print (C, type(C))

Output:

(matrix([[ 1,  2,  3],
        [ 4,  5,  6],
        [ 7,  8,  9],
        [10, 11, 12]]), <class 'numpy.matrixlib.defmatrix.matrix'>)
(array([[1, 1, 1, 1],
       [1, 1, 1, 1],
       [1, 1, 1, 1]]), <type 'numpy.ndarray'>)
(matrix([[ 6,  6,  6,  6],
        [15, 15, 15, 15],
        [24, 24, 24, 24],
        [33, 33, 33, 33]]), <class 'numpy.matrixlib.defmatrix.matrix'>)

Since python 3.5 as mentioned early you also can use a new matrix multiplication operator @ like

C = A @ B

and get the same result as above.

Answered By: Serenity