How to solve product scalar without using numpy library (python)

Question

Hello, I’m trying to do this operation without using the Numpy library, which in itself is already a tragedy.

I get from the user two vectors of equal sizes, and need to do the scalar product operation between them. Remembering which scalar product is a. b = a1.B1 + a2.B2 .... + an.bn

I think there’s something missing that I don’t fully understand...

For example, if I put vector A as [1, 2, 3] and vector B as [3, 2, 1], the result of multiplication between them should vector C = [3, 4, 3]. But the way this code was written what I have is: [3, 2, 1, 6, 4, 2, 9, 6, 3] Since I’m not able to multiply the elements of the vectors, it doesn’t even make sense to try to add the elements of the vector C to get the final product. follows the code:

vetorA = []
vetorB = []

numA = int(input('Digite o tamanho do vetor do A: '))
while len(vetorA) < numA:
    vetorA.append(int(input('Digite um numero para preencher o vetor A: ')))

numB = int(input('Digite o tamanho do vetor do B: '))
while len(vetorB) < numB:
    vetorB.append(int(input('Digite um numero para preencher o vetor B: ')))

if len(vetorA) == len(vetorB):
    vetorC = []
    for itemA in vetorA:
        for itemB in vetorB:
            new = itemA * itemB
            vetorC.append(new)
    print(vetorC)

Answer 1

You can use a list comprehension:

vetorC = [elemA * elemB for elemA, elemB in zip(vetorA, vetorB)]

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The function zip returns the elements of vetorA and vetorB in parallel for each iteration, then

list(zip([1, 2, 3, 4], [5, 6, 7, 8]))

returns

[(1, 5), (2, 6), (3, 7), (4, 8)]

The list comprehension creates a list by building an item for each iteration. In this case, the item to be built is the value elemA * elemB. In the documentation there is an explanation explaining and demonstrating the use of list understandings.