[英]How can i find the intersection of two multidimensional arrays faster?
there are two multidimensional boolean arrays with a different number of rows.有两个具有不同行数的多维布尔数组。 I want to quickly find indexes of True values in common rows.
我想快速找到公共行中 True 值的索引。 I wrote the following code but it is too slow.
我写了下面的代码,但它太慢了。 Is there a faster way to do this?
有没有更快的方法来做到这一点?
a=np.random.choice(a=[False, True], size=(100,100))
b=np.random.choice(a=[False, True], size=(1000,100))
for i in a:
for j in b:
if np.array_equal(i, j):
print(np.where(i))
Let's start with an edition to the question that makes sense and usually prints something:让我们从一个有意义的问题的版本开始,通常会打印一些东西:
a = np.random.choice(a=[False, True], size=(2, 2))
b = np.random.choice(a=[False, True], size=(4, 2))
print(f"a: \n {a}")
print(f"b: \n {b}")
matches = []
for i, x in enumerate(a):
for j, y in enumerate(b):
if np.array_equal(x, y):
matches.append((i, j))
And the solution using scipy.cdist
which compares all rows in a
against all rows in b
, using hamming distance for Boolean vector comparison:使用
scipy.cdist
的解决方案将a
中的所有行与b
中的所有行进行比较,使用汉明距离进行布尔向量比较:
import numpy as np
import scipy
from scipy import spatial
d = scipy.spatial.distance.cdist(a, b, metric='hamming')
cdist_matches = np.where(d == 0)
mathces_values = [(a[i], b[j]) for (i, j) in matches]
cdist_values = a[cdist_matches[0]], b[cdist_matches[1]]
print(f"matches_inds = \n{matches}")
print(f"matches = \n{mathces_values}")
print(f"cdist_inds = \n{cdist_matches}")
print(f"cdist_matches =\n {cdist_values}")
out:出去:
a:
[[ True False]
[False False]]
b:
[[ True True]
[ True False]
[False False]
[False True]]
matches_inds =
[(0, 1), (1, 2)]
matches =
[(array([ True, False]), array([ True, False])), (array([False, False]), array([False, False]))]
cdist_inds =
(array([0, 1], dtype=int64), array([1, 2], dtype=int64))
cdist_matches =
(array([[ True, False],
[False, False]]), array([[ True, False],
[False, False]]))
See this for a pure numpy implementation if you don't want to import scipy
如果您不想
import scipy
请参阅此纯 numpy 实现
The comparision of each row of a to each row of b can be made by making the shape of a broadcastable to the shape of b with the use of np.newaxis
and np.tile
可以通过使用
np.newaxis
和np.tile
使 a 的形状可广播到 b 的形状来比较 a 的每一行与 b 的每一行
import numpy as np
a=np.random.choice(a=[True, False], size=(2,5))
b=np.random.choice(a=[True, False], size=(10,5))
broadcastable_a = np.tile(a[:, np.newaxis, :], (1, b.shape[0], 1))
a_equal_b = np.equal(b, broadcastable_a)
indexes = np.where(a_equal_b)
indexes = np.stack(np.array(indexes[1:]), axis=1)
if you want to compare NDarrays element-wise, I would do something like that:如果你想比较 NDarrays 元素,我会做这样的事情:
import numpy as np
# data
a = np.random.choice(a = [False, True], size = (100,100))
b = np.random.choice(a = [False, True], size = (1000,100))
# extract matching coordinates
match = np.where((a == b[:100,:]) == True)
match = list(zip(*match))
# first 20 coordinates match
print("Number of matches:", len(match))
print(match[:20])
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