在2D列表中互换0和1的最快方法

Question

Assume that I have a matrix:

[[1, 1, 1, 0, 0,], [0, 1, 0, 1], [1, 0, 1]]

and I would like it to change to:

[[0, 0, 0, 1, 1,], [1, 0, 1, 0], [0, 1, 0]]

What could possibly be the fastest way to deal with this case?

Currently, I use for loop in another for loop as following which is obviously too slow.

for my_row in my_mat:
   for my_val in my_row:
      my_val = 1 if my_val == 0 else 0

Answer 1

I don't think it's slow, but it isn't the fastest. Here are a few faster alternatives.

Subtraction (super simple)

>>> [[1 - j for j in i] for i in lst]
[[0, 0, 0, 1, 1], [1, 0, 1, 0], [0, 1, 0]]

This eliminates the need for an if check. Although, this would only make sense if you have a list of 0/1s and you only want to flip those values.

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Bit Flips with XOR

>>> [[j ^ 1 for j in i] for i in lst]
[[0, 0, 0, 1, 1], [1, 0, 1, 0], [0, 1, 0]]

XOR operations are fast in general, so this is a good alternative, if you have positive values beyond one.

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not Inversion

>>> [[int(not j) for j in i] for i in lst]
[[0, 0, 0, 1, 1], [1, 0, 1, 0], [0, 1, 0]]

Note that this squashes non-zero values to 1. The not converts the integer to a truthy/falsy value, and the subsequent int converts False to 0 , or True to 1 .

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If you're interested in performance of all the methods suggested here,

lst = np.random.choice(2, (1000, 1000)).tolist()

%timeit [[int(not j) for j in i] for i in lst]
%timeit [[j ^ 1 for j in i] for i in lst]
%timeit [[1 - j for j in i] for i in lst]

10 loops, best of 3: 175 ms per loop
10 loops, best of 3: 89.8 ms per loop
10 loops, best of 3: 61.1 ms per loop

Answer 2

The fastest "algo" would be to leave your matrix untouched. Just remember in a separate flag that every value you read (or write) has to be inverted. Done.

But if you physically need to invert each value in the matrix, then there's only one "algo" for that - the one you already discovered. The rest is not about "algo", it is about the most efficient implementation of that "algo".