I look for an efficient way to get a row-wise intersection of two two-dimensional numpy ndarrays. There is only one intersection per row. For example:
[[1, 2], ∩ [[0, 1], -> [1,
[3, 4]] [0, 3]] 3]
In the best case zeros should be ignored:
[[1, 2, 0], ∩ [[0, 1, 0], -> [1,
[3, 4, 0]] [0, 3, 0]] 3]
My solution:
import numpy as np
arr1 = np.array([[1, 2],
[3, 4]])
arr2 = np.array([[0, 1],
[0, 3]])
arr3 = np.empty(len(arr1))
for i in range(len(arr1)):
arr3[i] = np.intersect1d(arr1[i], arr2[i])
print(arr3)
# [ 1. 3.]
I have about 1 million rows, so the vectorized operations are most preferred. You are welcome to use other python packages.
You can use np.apply_along_axis . I wrote a solution that pads to the size of the arr1. Didn't test the efficiency.
import numpy as np
def intersect1d_padded(x):
x, y = np.split(x, 2)
padded_intersection = -1 * np.ones(x.shape, dtype=np.int)
intersection = np.intersect1d(x, y)
padded_intersection[:intersection.shape[0]] = intersection
return padded_intersection
def rowwise_intersection(a, b):
return np.apply_along_axis(intersect1d_padded,
1, np.concatenate((a, b), axis=1))
result = rowwise_intersection(arr1,arr2)
>>> array([[ 1, -1],
[ 3, -1]])
if you know you have only one element in the intersection you can use
result = rowwise_intersection(arr1,arr2)[:,0]
>>> array([1, 3])
You can also modify intersect1d_padded to return a scalar with the intersection value.
我不知道在numpy
有一种优雅的方法可以做到这一点,但是一个简单的列表理解可以做到这一点:
[list(set.intersection(set(_x),set(_y)).difference({0})) for _x,_y in zip(x,y)]
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