查找具有范围的两个列表的元素对

Question

Is there a fastest way to get elements of the two lists and get the pairs that have difference between -21,21

list1 = [0,10,20]
list2 = [0,10,20,30,40,50]

I have really many lists and they are really large.

what I want to get the indexes as

[(list1_index1,list2_index1), ...] as below.

or directly:

diff: 0, lists1_index = 0, lists2_index = 0
diff: -10, lists1_index = 0, lists2_index = 1
diff: -20, lists1_index = 0, lists2_index = 2
diff: 10, lists1_index = 1, lists2_index = 0
diff: 0, lists1_index = 1, lists2_index = 1
diff: -10, lists1_index = 1, lists2_index = 2
diff: -20, lists1_index = 1, lists2_index = 3

... and so on.

I want to that in Python 2.7.

thanks.

Answer 1

Assuming your range is inclusive of -21 and 21, what about using itertools.product() with a list comprehension:

>>> from itertools import product
>>> list1 = [0,10,20]
>>> list2 = [0,10,20,30,40,50]
>>> [(i1, i2) for (i1, x1), (i2, x2) in product(enumerate(list1), enumerate(list2)) if abs(x1 - x2) <= 21]
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4)]

Here each pair of elements is paired with their indices with enumerate() , and when the difference is less than or equal to 21, the indices of the pairs are returned.

Note: You can use abs(x - y) , since it gives the absolute value of a number. So if we get a difference of -20, it will give back 20.

Answer 2

My solution below, with benchmarking versus @RoadRunner.

from numba import jit

lst1 = np.random.randint(0, 10, 1000)
lst2 = np.random.randint(0, 10, 1000)

# jp_data_analysis
@jit(nopython=True)
def differencer(l1, l2, n):
    return [(x_i, y_i) for x_i, x in enumerate(lst1) for (y_i, y) in enumerate(lst2) if abs(x-y) <= n]            

# RoadRunner
def differencer2(l1, l2, n):
    return [(i1, i2) for (i1, x1), (i2, x2) in product(enumerate(l1), enumerate(l2)) if abs(x1 - x2) <= n]

assert set(differencer(lst1, lst2, 21)) == set(differencer2(lst1, lst2, 21))

%timeit differencer(lst1, lst2, 21)     # 411ms
%timeit differencer2(lst1, lst2, 21)    # 1.02s

Answer 3

You can simply do without any external library:

list1 = [0,10,20]
list2 = [0,10,20,30,40,50]

print([(j,k) for j,i in enumerate(list1) for k,l in enumerate(list2) if (i-l) in list(range(-21,22))])

output:

[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (1, 3), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4)]