从列表或元组列表中选择随机对象的哪种方法更有效？

Question

I have got a list of 2d coordinates with this structure:

coo = [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0)]

Where coo[0] is the first coordinate stored in a tuple.

I would like to choose two different random coordinates. I can of course use this method:

import numpy  as np
rndcoo1 = coo[np.random.randint(0,len(coo))]
rndcoo2 = coo[np.random.randint(0,len(coo))]
if rndcoo1 != rndcoo2:
     #do something

But because I have to repeat this operation 1'000'000 times I was wondering if there is a faster method to do that. np.random.choice() can't be used for 2d array is there any alternative that I can use?

Answer 1

import random
result = random.sample(coo, 2)

will give you the expected output. And it is (probably) as fast as you can get with Python.

Answer 2

Listed in this post is a vectorized approach that gets us a number of such random choices for a number of iterations in one go without looping through those many times of iterations. The idea uses np.argpartition and is inspired by this post .

Here's the implementation -

def get_items(coo, num_items = 2, num_iter = 10):
    idx = np.random.rand(num_iter,len(coo)).argpartition(num_items,axis=1)[:,:2]
    return np.asarray(coo)[idx]

Please note that we would return a 3D array with the first dimension being the number of iterations, second dimension being the number of choices to be made at each iteration and the last dimension is the length of each tuple.

A sample run should present a bit more clearer picture -

In [55]: coo = [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0)]

In [56]: get_items(coo, 2, 5)
Out[56]: 
array([[[2, 0],
        [1, 1]],

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

       [[0, 2],
        [2, 0]],

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

       [[0, 2],
        [1, 1]]])

Runtime test comparing a loopy implementation with random.sample as listed in @freakish's post -

In [52]: coo = [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0)]

In [53]: %timeit [random.sample(coo, 2) for i in range(10000)]
10 loops, best of 3: 34.4 ms per loop

In [54]: %timeit get_items(coo, 2, 10000)
100 loops, best of 3: 2.81 ms per loop

Answer 3

Is coo just an example, or are your coordinates actually equally spaced? If so, you can just sample M 2D-coordinates like this:

import numpy

N = 100
M = 1000000
coo = numpy.random.randint(0, N, size=(M, 2))

Of course you can also bias and scale the distribution using addition and multiplication to account for different step sizes and offsets.

If you run into memory limitations with large M s, you can of course sample smaller sizes, or just one array of 2 values with size=2 .