CODEWITHSUNDEEP
python
chicken
chicken

Reputation: 23

basic operations on list of list

I am curious on understanding how to multiply values from one list with two other list but both list being in the same list.

For example,

a = [[2, 3, 1, 6], [1, 6, 1, 9], [1,`1, 1,2]]

and I want to get:

[[2, 18, 1, 54], [2, 3, 1, 12]],[[2, 18, 1, 54], [1, 6, 1, 18]], [[2, 3, 1, 12], [1, 6, 1, 18]]

where a[0] multiplies with the rest of the list and a[1] multiplies with the other lists and etc.

I have searched online but I don't understand it or there aren't any examples that really look at multiply or lists together in the context of nested lists.

I am sorry if it doesn't make sense, I am still pretty new with python and english isn't my first language.

Upvotes: 1

Views: 107

Answers (4)

Mad Physicist
Mad Physicist

Reputation: 114588

You can speed up the computation by using numpy and realizing that a * b == b * a.

First convert a to a numpy array:

a = np.array(a)

The output will be a 3D array:

output = np.empty((a.shape[0], a.shape[0] - 1, a.shape[1]), dtype=a.dtype)

Now get the indices in a that you want to multiply together:

r, c = np.triu_indices(a.shape[0], k=1)

Now do the multiplication:

output[r, c - 1] = a[r] * a[c]

The lower half of the matrix is just a copy of the upper, no need to multiply:

output[c, r] = output[r, c - 1]

This solution is vectorized, which means that all of the loops have been taken out of the python layer, and moved to a fast C implementation under the hood. You can see this with the operation a[r] * a[c], which performs all of your multiplications in a single line. The advantage of using vectorized operations is that even when your dataset becomes very large, the operations run in a reasonable amount of time.

Upvotes: 1

Sihag
Sihag

Reputation: 11

You can use zip for multiplying and a for loop for accessing the elements.

a = [[2, 3, 1, 6], [1, 6, 1, 9], [1,1, 1,2]]
b = a

result = []

for i in a:
    for j in b:
        if(j!=i):
            k = [n1 * n2 for n1,n2 in zip(i, j)] 
            result.append(k) 
          
print(result)

Upvotes: 1

I'mahdi
I'mahdi

Reputation: 24069

You can use numpy.array and * like below:

import numpy as np
a = [[2, 3, 1, 6], [1, 6, 1, 9], [1,1, 1,2]]

[list(np.array(a[i])*np.array(a[j])) for i in range(len(a)) for j in range(len(a)) if i!=j]

#expand version
# out = []
# for i in range(len(a)):
#     for j in range(len(a)):
#         if i!=j:
#             out.append(list(np.array(a[i])*np.array(a[j])))

output:

[[2, 18, 1, 54],
 [2, 3, 1, 12],
 [2, 18, 1, 54],
 [1, 6, 1, 18],
 [2, 3, 1, 12],
 [1, 6, 1, 18]]

In this answer I use numpy.array and * like below: (numpy do this for you)

np.array([2, 3, 1, 6]) * np.array([1, 6, 1, 9])
# [2*1, 3*6, 1*1, 6*9]

Upvotes: 2

Corralien
Corralien

Reputation: 120559

Use itertools.permutations.

from itertools import permutations

a = [[2, 3, 1, 6], [1, 6, 1, 9], [1, 1, 1,2]]

b = []
for l1, l2 in permutations(a, 2):
    b.append([i*j for i, j in zip(l1, l2)])

# OR with comprehension

b = [[i*j for i, j in zip(l1, l2)] for l1, l2 in permutations(a, 2)]

Output:

>>> b
[[2, 18, 1, 54],
 [2, 3, 1, 12],
 [2, 18, 1, 54],
 [1, 6, 1, 18],
 [2, 3, 1, 12],
 [1, 6, 1, 18]]

Upvotes: 1

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