Sub matrix of a list of lists (without numpy)

Question

Suppose I have a matrix composed of a list of lists like so:

>>> LoL=[list(range(10)) for i in range(10)]
>>> LoL
[[0, 1, 2, 3, 4, 5, 6, 7,         


        

Answer 1

Do this,

submat = [ [ mat[ i ][ j ] for j in range( index1, index2 ) ] for i in range( index3, index4 ) ]

the submat will be the rectangular (square if index3 == index1 and index2 == index4) chunk of your original big matrix.

Answer 2

In [74]: [row[2:5] for row in LoL[1:4]]
Out[74]: [[2, 3, 4], [2, 3, 4], [2, 3, 4]]

---

You could also mimic NumPy's syntax by defining a subclass of list:

class LoL(list):
    def __init__(self, *args):
        list.__init__(self, *args)
    def __getitem__(self, item):
        try:
            return list.__getitem__(self, item)
        except TypeError:
            rows, cols = item
            return [row[cols] for row in self[rows]]

lol = LoL([list(range(10)) for i in range(10)])
print(lol[1:4, 2:5])

also yields

[[2, 3, 4], [2, 3, 4], [2, 3, 4]]

Using the LoL subclass won't win any speed tests:

In [85]: %timeit [row[2:5] for row in x[1:4]]
1000000 loops, best of 3: 538 ns per loop
In [82]: %timeit lol[1:4, 2:5]
100000 loops, best of 3: 3.07 us per loop

but speed isn't everything -- sometimes readability is more important.

Answer 3

I dont know if its easier, but let me throw an idea to the table:

from itertools import product
r = (1+1, 4+1)
s = (2+1, 5+1)
array = [LoL[i][j] for i,j in product(range(*r), range(*s))]

This is a flattened version of the submatrix you want.

Answer 4

For one, you can use slice objects directly, which helps a bit with both the readability and performance:

r = slice(1,4)
s = slice(2,5)
[LoL[i][s] for i in range(len(LoL))[r]]

And if you just iterate over the list-of-lists directly, you can write that as:

[row[s] for row in LoL[r]]