Python 实用宝典

Question 1

I know @ is for decorators, but what is @= for in Python? Is it just reservation for some future idea?

This is just one of my many questions while reading tokenizer.py.

Question 2

The @ (at) operator is intended to be used for matrix multiplication. No builtin Python types implement this operator.

The @ operator was introduced in Python 3.5. @= is matrix multiplication followed by assignment, as you would expect. They map to __matmul__, __rmatmul__ or __imatmul__ similar to how + and += map to __add__, __radd__ or __iadd__.

The operator and the rationale behind it are discussed in detail in PEP 465.

Question 3

@= and @ are new operators introduced in Python 3.5 performing matrix multiplication. They are meant to clarify the confusion which existed so far with the operator * which was used either for element-wise multiplication or matrix multiplication depending on the convention employed in that particular library/code. As a result, in the future, the operator * is meant to be used for element-wise multiplication only.

As explained in PEP0465, two operators were introduced:

A new binary operator A @ B, used similarly as A * B
An in-place version A @= B, used similarly as A *= B

Matrix Multiplication vs Element-wise Multiplication

To quickly highlight the difference, for two matrices:

A = [[1, 2],    B = [[11, 12],
     [3, 4]]         [13, 14]]

Element-wise multiplication will yield:

A * B = [[1 * 11,   2 * 12], 
         [3 * 13,   4 * 14]]

Matrix multiplication will yield:

A @ B  =  [[1 * 11 + 2 * 13,   1 * 12 + 2 * 14],
           [3 * 11 + 4 * 13,   3 * 12 + 4 * 14]]

Usage in Numpy

So far, Numpy used the following convention:

the * operator (and arithmetic operators in general) were defined as element-wise operations on ndarrays and as matrix-multiplication on numpy.matrix type.
method/function dot was used for matrix multiplication of ndarrays

Introduction of the @ operator makes the code involving matrix multiplications much easier to read. PEP0465 gives us an example:

# Current implementation of matrix multiplications using dot function
S = np.dot((np.dot(H, beta) - r).T,
            np.dot(inv(np.dot(np.dot(H, V), H.T)), np.dot(H, beta) - r))

# Current implementation of matrix multiplications using dot method
S = (H.dot(beta) - r).T.dot(inv(H.dot(V).dot(H.T))).dot(H.dot(beta) - r)

# Using the @ operator instead
S = (H @ beta - r).T @ inv(H @ V @ H.T) @ (H @ beta - r)

Clearly, the last implementation is much easier to read and interpret as an equation.

Question 4

@ is the new operator for Matrix Multiplication added in Python3.5

Reference: https://docs.python.org/3/whatsnew/3.5.html#whatsnew-pep-465

Example

C = A @ B

Question 5

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?

Question 6

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.

Question 7

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]])

Question 8

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

Question 9

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.]]

Question 10

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.

Question 11

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

Question 12

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.

Question 13

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.

Question 14

I recently moved to Python 3.5 and noticed the new matrix multiplication operator (@) sometimes behaves differently from the numpy dot operator. In example, for 3d arrays:

import numpy as np

a = np.random.rand(8,13,13)
b = np.random.rand(8,13,13)
c = a @ b  # Python 3.5+
d = np.dot(a, b)

The @ operator returns an array of shape:

c.shape
(8, 13, 13)

while the np.dot() function returns:

d.shape
(8, 13, 8, 13)

How can I reproduce the same result with numpy dot? Are there any other significant differences?

Question 15

The @ operator calls the array’s __matmul__ method, not dot. This method is also present in the API as the function np.matmul.

>>> a = np.random.rand(8,13,13)
>>> b = np.random.rand(8,13,13)
>>> np.matmul(a, b).shape
(8, 13, 13)

From the documentation:

matmul differs from dot in two important ways.

Multiplication by scalars is not allowed.

Stacks of matrices are broadcast together as if the matrices were elements.

The last point makes it clear that dot and matmul methods behave differently when passed 3D (or higher dimensional) arrays. Quoting from the documentation some more:

For matmul:

If either argument is N-D, N > 2, it is treated as a stack of matrices residing in the last two indexes and broadcast accordingly.

For np.dot:

For 2-D arrays it is equivalent to matrix multiplication, and for 1-D arrays to inner product of vectors (without complex conjugation). For N dimensions it is a sum product over the last axis of a and the second-to-last of b

Question 16

The answer by @ajcr explains how the dot and matmul (invoked by the @ symbol) differ. By looking at a simple example, one clearly sees how the two behave differently when operating on ‘stacks of matricies’ or tensors.

To clarify the differences take a 4×4 array and return the dot product and matmul product with a 3x4x2 ‘stack of matricies’ or tensor.

import numpy as np
fourbyfour = np.array([
                       [1,2,3,4],
                       [3,2,1,4],
                       [5,4,6,7],
                       [11,12,13,14]
                      ])


threebyfourbytwo = np.array([
                             [[2,3],[11,9],[32,21],[28,17]],
                             [[2,3],[1,9],[3,21],[28,7]],
                             [[2,3],[1,9],[3,21],[28,7]],
                            ])

print('4x4*3x4x2 dot:\n {}\n'.format(np.dot(fourbyfour,threebyfourbytwo)))
print('4x4*3x4x2 matmul:\n {}\n'.format(np.matmul(fourbyfour,threebyfourbytwo)))

The products of each operation appear below. Notice how the dot product is,

…a sum product over the last axis of a and the second-to-last of b

and how the matrix product is formed by broadcasting the matrix together.

4x4*3x4x2 dot:
 [[[232 152]
  [125 112]
  [125 112]]

 [[172 116]
  [123  76]
  [123  76]]

 [[442 296]
  [228 226]
  [228 226]]

 [[962 652]
  [465 512]
  [465 512]]]

4x4*3x4x2 matmul:
 [[[232 152]
  [172 116]
  [442 296]
  [962 652]]

 [[125 112]
  [123  76]
  [228 226]
  [465 512]]

 [[125 112]
  [123  76]
  [228 226]
  [465 512]]]

Question 17

Just FYI, @ and its numpy equivalents dot and matmul are all equally fast. (Plot created with perfplot, a project of mine.)

Code to reproduce the plot:

import perfplot
import numpy


def setup(n):
    A = numpy.random.rand(n, n)
    x = numpy.random.rand(n)
    return A, x


def at(data):
    A, x = data
    return A @ x


def numpy_dot(data):
    A, x = data
    return numpy.dot(A, x)


def numpy_matmul(data):
    A, x = data
    return numpy.matmul(A, x)


perfplot.show(
    setup=setup,
    kernels=[at, numpy_dot, numpy_matmul],
    n_range=[2 ** k for k in range(15)],
)

Question 18

In mathematics, I think the dot in numpy makes more sense

dot(a,b)_{i,j,k,a,b,c} =

since it gives the dot product when a and b are vectors, or the matrix multiplication when a and b are matrices

As for matmul operation in numpy, it consists of parts of dot result, and it can be defined as

>matmul(a,b)_{i,j,k,c} =

So, you can see that matmul(a,b) returns an array with a small shape, which has smaller memory consumption and make more sense in applications. In particular, combining with broadcasting, you can get

matmul(a,b)_{i,j,k,l} =

for example.

From the above two definitions, you can see the requirements to use those two operations. Assume a.shape=(s1,s2,s3,s4) and b.shape=(t1,t2,t3,t4)

To use dot(a,b) you need
1. t3=s4;
To use matmul(a,b) you need
1. t3=s4
2. t2=s2, or one of t2 and s2 is 1
3. t1=s1, or one of t1 and s1 is 1

Use the following piece of code to convince yourself.

Code sample

import numpy as np
for it in xrange(10000):
    a = np.random.rand(5,6,2,4)
    b = np.random.rand(6,4,3)
    c = np.matmul(a,b)
    d = np.dot(a,b)
    #print 'c shape: ', c.shape,'d shape:', d.shape

    for i in range(5):
        for j in range(6):
            for k in range(2):
                for l in range(3):
                    if not c[i,j,k,l] == d[i,j,k,j,l]:
                        print it,i,j,k,l,c[i,j,k,l]==d[i,j,k,j,l] #you will not see them

Question 19

Here is a comparison with np.einsum to show how the indices are projected

np.allclose(np.einsum('ijk,ijk->ijk', a,b), a*b)        # True 
np.allclose(np.einsum('ijk,ikl->ijl', a,b), a@b)        # True
np.allclose(np.einsum('ijk,lkm->ijlm',a,b), a.dot(b))   # True

Question 20

My experience with MATMUL and DOT

I was constantly getting “ValueError: Shape of passed values is (200, 1), indices imply (200, 3)” when trying to use MATMUL. I wanted a quick workaround and found DOT to deliver the same functionality. I don’t get any error using DOT. I get the correct answer

with MATMUL

X.shape
>>>(200, 3)

type(X)

>>>pandas.core.frame.DataFrame

w

>>>array([0.37454012, 0.95071431, 0.73199394])

YY = np.matmul(X,w)

>>>  ValueError: Shape of passed values is (200, 1), indices imply (200, 3)"

with DOT

YY = np.dot(X,w)
# no error message
YY
>>>array([ 2.59206877,  1.06842193,  2.18533396,  2.11366346,  0.28505879, …

YY.shape

>>> (200, )

Question 21

I have two matrices

a = np.matrix([[1,2], [3,4]])
b = np.matrix([[5,6], [7,8]])

and I want to get the element-wise product, [[1*5,2*6], [3*7,4*8]], equaling

[[5,12], [21,32]]

I have tried

print(np.dot(a,b))

and

print(a*b)

but both give the result

[[19 22], [43 50]]

which is the matrix product, not the element-wise product. How can I get the the element-wise product (aka Hadamard product) using built-in functions?

Question 22

For elementwise multiplication of matrix objects, you can use numpy.multiply:

import numpy as np
a = np.array([[1,2],[3,4]])
b = np.array([[5,6],[7,8]])
np.multiply(a,b)

Result

array([[ 5, 12],
       [21, 32]])

However, you should really use array instead of matrix. matrix objects have all sorts of horrible incompatibilities with regular ndarrays. With ndarrays, you can just use * for elementwise multiplication:

a * b

If you’re on Python 3.5+, you don’t even lose the ability to perform matrix multiplication with an operator, because @ does matrix multiplication now:

a @ b  # matrix multiplication

Question 23

just do this:

import numpy as np

a = np.array([[1,2],[3,4]])
b = np.array([[5,6],[7,8]])

a * b

Question 24

import numpy as np
x = np.array([[1,2,3], [4,5,6]])
y = np.array([[-1, 2, 0], [-2, 5, 1]])

x*y
Out: 
array([[-1,  4,  0],
       [-8, 25,  6]])

%timeit x*y
1000000 loops, best of 3: 421 ns per loop

np.multiply(x,y)
Out: 
array([[-1,  4,  0],
       [-8, 25,  6]])

%timeit np.multiply(x, y)
1000000 loops, best of 3: 457 ns per loop

Both np.multiply and * would yield element wise multiplication known as the Hadamard Product

%timeit is ipython magic

Question 25

Try this:

a = np.matrix([[1,2], [3,4]])
b = np.matrix([[5,6], [7,8]])

#This would result a 'numpy.ndarray'
result = np.array(a) * np.array(b)

Here, np.array(a) returns a 2D array of type ndarray and multiplication of two ndarray would result element wise multiplication. So the result would be:

result = [[5, 12], [21, 32]]

If you wanna get a matrix, the do it with this:

result = np.mat(result)

Python 实用宝典

标签归档：matrix-multiplication

Python中的’@ =’符号是什么？

问题：Python中的’@ =’符号是什么？

回答 0

回答 1

矩阵乘法与按元素乘法

在Numpy中使用

Matrix Multiplication vs Element-wise Multiplication

Usage in Numpy

回答 2

NumPy矩阵与数组类的乘法有何不同？

问题：NumPy矩阵与数组类的乘法有何不同？

回答 0

回答 1

回答 2

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回答 7

numpy dot（）和Python 3.5+矩阵乘法@之间的区别

问题：numpy dot（）和Python 3.5+矩阵乘法@之间的区别

回答 0

回答 1

回答 2

回答 3

> matmul（a，b）_ {i，j，k，c} =

代码样例

>matmul(a,b)_{i,j,k,c} =

Code sample

回答 4

回答 5

如何在numpy中获得按元素矩阵乘法（Hadamard积）？

问题：如何在numpy中获得按元素矩阵乘法（Hadamard积）？

回答 0

回答 1

回答 2

回答 3

有趣好用的Python教程

问题：Python中的’@ =’符号是什么？

回答 0

回答 1

矩阵乘法与按元素乘法

在Numpy中使用

Matrix Multiplication vs Element-wise Multiplication

Usage in Numpy

回答 2

问题：NumPy矩阵与数组类的乘法有何不同？

回答 0

回答 1

回答 2

回答 3

回答 4

回答 5

回答 6

回答 7

问题：numpy dot（）和Python 3.5+矩阵乘法@之间的区别

回答 0

回答 1

回答 2

回答 3

> matmul（a，b）_ {i，j，k，c} =<img decoding="async" alt="式" src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%5Csum_m%20a_%7Bi%2Cj%2Ck%2Cm%7Db_%7Bi%2Cj%2Cm%2Cc%7D%0A" width="474" height="400">

代码样例

>matmul(a,b)_{i,j,k,c} = <img decoding="async" src="https://chart.googleapis.com/chart?cht=tx&amp;chl=%5Csum_m%20a_%7Bi%2Cj%2Ck%2Cm%7Db_%7Bi%2Cj%2Cm%2Cc%7D%0A" alt="formula" width="474" height="400">

Code sample

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问题：如何在numpy中获得按元素矩阵乘法（Hadamard积）？

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回答 1

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回答 3

有趣好用的Python教程

> matmul（a，b）_ {i，j，k，c} =

>matmul(a,b)_{i,j,k,c} =