Elementary product of two two-dimensional lists

Question

I cannot use Numpy or any other library function, since this is the question I need to do, I have to define my own path.

I am writing a function that takes two lists (2 dimensional) as arguments. The function should calculate the elemental product of both lists and store them in the third list and return this resulting list from the function. Example input lists:

LIST1:

[[2,3,5,6,7],[5,2,9,3,7]]  

list2:

[[5,2,9,3,7],[1,3,5,2,2]]

The function displays the following list:

[[10, 6, 45, 18, 49], [5, 6, 45, 6, 14]] 

This 2*5=10, 3*2=6, 5*9=45... etc.

, 2 () , , , , , () , 2-D ,

[[5,2,9,3,7],[1,3,5,2,2],[1,3,5,2,2]]

[[5,2,9,3,7],[1,3,5,2,2],[1,3,5,2,2],[5,2,9,3,7]]

.

def ElementwiseProduct(l,l2):
    i=0
    newlist=[] #create empty list to put prouct of elements in later
    newlist2=[]
    newlist3=[] #empty list to put both new lists which will have proudcts in them
    while i==0:
        a=0
        while a<len(l[i]):
            prod=l[i][a]*l2[i][a] #corresponding product of lists elements
            newlist.append(prod) #adding the products to new list
            a+=1
        i+=1
    while i==1:
        a=0
        while a<len(l[i]):
            prod=l[i][a]*l2[i][a] #corresponding product of lists elements
            newlist2.append(prod) #adding the products to new list
            a+=1
        i+=1
    newlist3.append(newlist)
    newlist3.append(newlist2)
    print newlist3

#2 dimensional list example
list1=[[2,3,5,6,7],[5,2,9,3,7]] 
list2=[[5,2,9,3,7],[1,3,5,2,2]]  
ElementwiseProduct(list1,list2)

Answer 1

zip , zip , , , :

list2 = [[5,2,9,3,7],[1,3,5,2,2]]
list1 = [[2,3,5,6,7],[5,2,9,3,7]]

result = [[a*b for a, b in zip(i, j)] for i, j in zip(list1, list2)]
print(result)
# [[10, 6, 45, 18, 49], [5, 6, 45, 6, 14]]

/ , itertools.izip_longest , , 0 :

from itertools import izip_longest

list1 = [[2,3,5,6]]
list2 = [[5,2,9,3,7],[1,3,5,2,2]]
result = [[a*b for a, b in izip_longest(i, j, fillvalue=0)] 
               for i, j in izip_longest(list1, list2, fillvalue=[])]
print(result)
# [[10, 6, 45, 18, 0], [0, 0, 0, 0, 0]]

fillvalue 0 1, , 0.

---

:

Answer 2

, , ( , 2):

def elementwiseProd(iterA, iterB):
    def multiply(a, b):
        try:
            iter(a)
        except TypeError:
            # You have a number
            return a * b
        return elementwiseProd(a, b)
    return [multiply(*pair) for pair in zip(iterA, iterB)]

. , . , , . , .

. , , , , (vs ), .

, , n- n :

def elementwiseApply(op, *iters):
    def apply(op, *items):
        try:
            iter(items[0])
        except TypeError:
            return op(*items)
        return elementwiseApply(op, *items)
    return [apply(op, *items) for items in zip(*iters)]

, operator.mul:

from operator import mul
list1=[[2,3,5,6,7], [5,2,9,3,7]] 
list2=[[5,2,9,3,7], [1,3,5,2,2]]
elementwiseApply(mul, list1, list2)

[[10, 6, 45, 18, 49], [5, 6, 45, 6, 14]]

Answer 3

Python , , . , , .

, , for. zip ( ) .

def elementwise_product(list1,list2):
    result = []
    for seq1, seq2 in zip(list1,list2):
        prods = []
        for u, v in zip(seq1, seq2):
            prods.append(u * v)
        result.append(prods)
    return result

list1=[[2,3,5,6,7], [5,2,9,3,7]] 
list2=[[5,2,9,3,7], [1,3,5,2,2]]

print(elementwise_product(list1,list2))

[[10, 6, 45, 18, 49], [5, 6, 45, 6, 14]]

, . , .

def elementwise_product(list1,list2):
    return [[u*v for u, v in zip(seq1, seq2)] 
        for seq1, seq2 in zip(list1,list2)]

Answer 4

numpy. , C , ,

numpy. (CMD, ),

pip install numpy

, Linux, sudo pip install numpy

import numpy as np

list1=np.array([[2,3,5,6,7],[5,2,9,3,7]]) #2 dimensional list example
list2=np.array([[5,2,9,3,7],[1,3,5,2,2]])

prod = np.multiply(list1,list2)
# or simply, as suggested by Mad Physicist,
prod = list1*list2