Subindex a list of lists into block matrices X = [ [A, B], [C, D]]

Question

Let

X = [[2,3, 5, 6], 
     [4,5, 9, 10], 
     [6,1, 3, 9], 
     [3,7, 11, 12]]

How do I output the sub-matrices such that X = [ [A, B], [C, D]]?

    A = [[2, 3], 
         [4,5]]
    B = [[5,6], 
        [9, 10]] 
    C = [[6, 1], 
         [3,7]]
    D = [[3,9], 
        [11, 12]]

@Paddy This is trivial in Matlab. For Python, I've tried A = x[:2][:2][:2] but to no avail.. — Ryan J. Shrott, May 25 '17 at 19:19
It's better to cast them as Numpy arrays. And then just need to give the start index as well. — Antimony, May 25 '17 at 19:20
Well Numpy allows you to slice them like `X[0:2, 0:2]` but regular lists don't. What's the reason you don't want to use Numpy? And yes, my bad, you can index them as `[:2]` as well. — Antimony, May 25 '17 at 19:22
A = [row[0:2] for row in X[0:2] First splice X to get the right rows, then splice the rows to get the right columns. They both happen to go from 0 to 2 in the case of A, and they both happen to be the same, but that may not always be the case. — Kenny Ostrom, May 25 '17 at 19:24
Honestly, if you want functionality similar to MATLAB, especially slicing 2D arrays, Numpy is the way to go in Python. MATLAB was built especially for things like vector algebra, but Python wasn't. — Antimony, May 25 '17 at 19:28
Similar question: https://stackoverflow.com/questions/15650538/sub-matrix-of-a-list-of-lists-without-numpy — user2314737, May 25 '17 at 19:28
Well, it's still trivial. But I don't know how you want to break up, e.g. a 5x7 matrix. Are you going to pad? — Kenny Ostrom, May 25 '17 at 20:02

Vadim · Answer 1 · 2017-05-25T19:34:00.700

2

Without numpy you can do it like this....

X = [[2,3, 5, 6], 
         [4,5, 9, 10], 
         [6,1, 3, 9], 
         [3,7, 11, 12]]

for i in X:
   print([i[ :len(i)//2],i[len(i)//2:]])

edited May 25 '17 at 19:34

answered May 25 '17 at 19:26

Vadim

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You have a syntax error -- an extra parenthesis. Also, the output is `[[5, 6], [2, 3]]` for A, which isn't what the OP wants. – rassar May 25 '17 at 19:31
@rassar good catch, thanks – Vadim May 25 '17 at 19:33

score 2 · Answer 2 · answered May 25 '17 at 19:26

You can try this, assuming the X is always a 4X4 Matrix:

X = [[2,3, 5, 6], 
 [4,5, 9, 10], 
 [6,1, 3, 9], 
 [3,7, 11, 12]]

new_matrix = [[X[i][:2], X[i+1][:2]] for i in range(0, len(X), 2)]
new_matrix.extend([[X[i][2:], X[i+1][2:]] for i in range(0, len(X), 2)]) 
print(new_matrix)

score 2 · Answer 3 · answered May 25 '17 at 19:27

2

A solution using Numpy would look like this:

A = X[0:2, 0:2]
B = X[0:2, 2:4]
C = X[2:4, 0:2]
D = X[2:4, 2:4]

answered May 25 '17 at 19:27

Antimony

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score 1 · Answer 4 · answered May 25 '17 at 19:27

1

For the first sub-matrix you can use

A = [row[:2] for row in X[:2]]

(and similarly for the others)

answered May 25 '17 at 19:27

user2314737

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score 1 · Accepted Answer · answered May 25 '17 at 19:33

Here is a solution without using numpy.

A = [X[i][:2] for i in range(2)]
B = [X[i][2:] for i in range(2)]
C = [X[i][:2] for i in range(2,4)]
D = [X[i][2:] for i in range(2,4)]

>>> A
[[2, 3], [4, 5]]
>>> B
[[5, 6], [9, 10]]
>>> C
[[6, 1], [3, 7]]
>>> D
[[3, 9], [11, 12]]

score 1 · Answer 6 · answered May 25 '17 at 20:28

1

One liner version, though its a bit less readable

>>> (A,B),(C,D)= [([X[i][:2],X[i+1][:2]],[X[i][2:],X[i+1][2:]]) for i in range(0,len(X),2)]
>>>
>>? A
[[2, 3], [4, 5]]
>>? B
[[5, 6], [9, 10]]
>>? C
[[6, 1], [3, 7]]
>>? D
[[3, 9], [11, 12]]
>>?

answered May 25 '17 at 20:28

Skycc

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Subindex a list of lists into block matrices X = [ [A, B], [C, D]]

6 Answers6