Python 3：在没有 NumPy 的情况下将向量乘以矩阵
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Python 3: Multiply a vector by a matrix without NumPy
提问by JGraham353
I'm fairly new to Python and trying to create a function to multiply a vector by a matrix (of any column size). e.g.:
我对 Python 相当陌生，并试图创建一个函数来将向量乘以矩阵（任何列大小）。例如：
multiply([1,0,0,1,0,0], [[0,1],[1,1],[1,0],[1,0],[1,1],[0,1]])
[1, 1]
Here is my code:
这是我的代码：
def multiply(v, G):
result = []
total = 0
for i in range(len(G)):
r = G[i]
for j in range(len(v)):
total += r[j] * v[j]
result.append(total)
return result
The problem is that when I try to select the first row of each column in the matrix (r[j]) the error 'list index out of range' is shown. Is there any other way of completing the multiplication without using NumPy?
问题是，当我尝试选择矩阵 (r[j]) 中每列的第一行时，会显示错误“列表索引超出范围”。有没有其他方法可以在不使用 NumPy 的情况下完成乘法？
采纳答案by Kasramvd
The Numpythonic approach: (using numpy.dotin order to get the dot product of two matrices)
Numpythonic 方法：（numpy.dot为了得到两个矩阵的点积而使用）
In [1]: import numpy as np
In [3]: np.dot([1,0,0,1,0,0], [[0,1],[1,1],[1,0],[1,0],[1,1],[0,1]])
Out[3]: array([1, 1])
The Pythonic approach:
Pythonic 方法：
The length of your second forloop is len(v)and you attempt to indexing vbased on that so you got index Error . As a more pythonic way you can use zipfunction to get the columns of a list then use starmapand mulwithin a list comprehension:
您的第二个for循环的长度是len(v)，您尝试v根据它进行索引，因此您得到了索引 Error 。作为一种更pythonic的方式，您可以使用zip函数来获取列表的列，然后在列表理解中使用starmap和mul：
In [13]: first,second=[1,0,0,1,0,0], [[0,1],[1,1],[1,0],[1,0],[1,1],[0,1]]
In [14]: from itertools import starmap
In [15]: from operator import mul
In [16]: [sum(starmap(mul, zip(first, col))) for col in zip(*second)]
Out[16]: [1, 1]
回答by Simeon Visser
ris an element from Gso it's a row which only has two elements. That means you can't use index jto get a value from rbecause jgoes from 0 till the length of v, which is 6 in your example.
r是一个元素 fromG所以它是一个只有两个元素的行。这意味着您不能使用 indexj从中获取值，r因为j从 0 到 的长度v，在您的示例中为 6。
回答by ChrisFreeman
I needed solution where the first matrix could be 2-dimensional. Extending the solution from @Kasramvd to accept a two dimensional firstmatrix. Posted here for reference:
我需要第一个矩阵可以是二维的解决方案。扩展来自@Kasramvd 的解决方案以接受二维first矩阵。贴在这里供参考：
>>> first,second=[[1,0,0,1,0,0],[0,1,1,1,0,0]], [[0,1],[1,1],[1,0],[1,0],[1,1],[0,1]]
>>> from itertools import starmap
>>> from operator import mul
>>> [[sum(starmap(mul, zip(row, col))) for col in zip(*second)] for row in first]
[[1, 1], [3, 1]]
回答by stackPusher
I think the problem with your code was that you loop through the rows of the matrix rather than by the columns. Also you don't reset your 'total' variable after each vector*matrix column calculation. This is what you want:
我认为您的代码的问题在于您遍历矩阵的行而不是列。此外，您不会在每个 vector*matrix 列计算后重置“total”变量。这就是你想要的：
def multiply(v, G):
result = []
for i in range(len(G[0])): #this loops through columns of the matrix
total = 0
for j in range(len(v)): #this loops through vector coordinates & rows of matrix
total += v[j] * G[j][i]
result.append(total)
return result
回答by Pradeep Padmanaban
i have attached a code for matrix multiplication do follow the example format for one dimensional multiplication (lists of list)
我附上了矩阵乘法的代码，请遵循一维乘法的示例格式（列表列表）
def MM(a,b):
c = []
for i in range(0,len(a)):
temp=[]
for j in range(0,len(b[0])):
s = 0
for k in range(0,len(a[0])):
s += a[i][k]*b[k][j]
temp.append(s)
c.append(temp)
return c
a=[[1,2]]
b=[[1],[2]]
print(MM(a,b))
result is [[5]]
结果是 [[5]]
回答by Ilyas Maamri
There is a code that help u to multiply two matrix:
有一个代码可以帮助您将两个矩阵相乘：
A=[[1,2,3],[4,5,6],[7,8,9]] B=[[1,2,3],[4,5,6],[7,8,9]] matrix=[] def multiplicationLineColumn(line,column): try: sizeLine=len(line) sizeColumn=len(column) if(sizeLine!=sizeColumn): raise ValueError("Exception") res = sum([line[i] * column[i] for i in range(sizeLine)]) return res except ValueError: print("sould have the same len line & column") def getColumn(matrix,numColumn): size=len(matrix) column= [matrix[i][numColumn] for i in range(size)] return column def getLine(matrix,numLine): line = matrix[numLine] return line for i in range(len(A)): matrix.append([]) for j in range(len(B)): matrix[i].append(multiplicationLineColumn(getLine(A,i),getColumn(B,j))) print(matrix)
A=[[1,2,3],[4,5,6],[7,8,9]] B=[[1,2,3],[4,5,6],[7,8,9]] matrix=[] def multiplicationLineColumn(line,column): try: sizeLine=len(line) sizeColumn=len(column) if(sizeLine!=sizeColumn): raise ValueError("Exception") res = sum([line[i] * column[i] for i in range(sizeLine)]) return res except ValueError: print("sould have the same len line & column") def getColumn(matrix,numColumn): size=len(matrix) column= [matrix[i][numColumn] for i in range(size)] return column def getLine(matrix,numLine): line = matrix[numLine] return line for i in range(len(A)): matrix.append([]) for j in range(len(B)): matrix[i].append(multiplicationLineColumn(getLine(A,i),getColumn(B,j))) print(matrix)
