计算numpy数组中每个其他元素的增量内有多少个元素 [英] count how many elements in a numpy array are within delta of every other element

问题描述

考虑数组x和增量变量d

np.random.seed([3,1415])
x = np.random.randint(100, size=10)
d = 10

对于x中的每个元素，我想计算每个元素中有多少其他元素在距离d distance 的范围内.

For each element in x, I want to count how many other elements in each are within delta d distance away.

所以x看起来像

print(x)

[11 98 74 90 15 55 13 11 13 26]

结果应该是

[5 2 1 2 5 1 5 5 5 1]

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我尝试过的
策略:

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what I've tried
Strategy:

• 利用广播带来外在差异
• 外差的绝对值
• 总和超过阈值

(np.abs(x[:, None] - x) <= d).sum(-1)

[5 2 1 2 5 1 5 5 5 1]

---

这很好.但是，它无法扩展.外部差异为O(n ^ 2)时间.如何获得不随二次时间缩放的相同解决方案?

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This works great. However, it doesn't scale. That outer difference is O(n^2) time. How can I get the same solution that doesn't scale with quadratic time?

推荐答案

这篇文章中列出了另外两个基于searchsorted strategy的 variant . com/a/41394596/3293881> OP's answer post .

Listed in this post are two more variants based on the searchsorted strategy from OP's answer post.

def pir3(a,d):  # Short & less efficient
    sidx = a.argsort()
    p1 = a.searchsorted(a+d,'right',sorter=sidx)
    p2 = a.searchsorted(a-d,sorter=sidx)
    return p1 - p2

def pir4(a, d):   # Long & more efficient
    s = a.argsort()

    y = np.empty(s.size,dtype=np.int64)
    y[s] = np.arange(s.size)

    a_ = a[s]
    return (
        a_.searchsorted(a_ + d, 'right')
        - a_.searchsorted(a_ - d)
    )[y]

更有效的方法是从 this post 中获取s.argsort()的有效思路.

The more efficient approach derives the efficient idea to get s.argsort() from this post.

运行时测试-

In [155]: # Inputs
     ...: a = np.random.randint(0,1000000,(10000))
     ...: d = 10


In [156]: %timeit pir2(a,d) #@ piRSquared's post solution
     ...: %timeit pir3(a,d)
     ...: %timeit pir4(a,d)
     ...: 
100 loops, best of 3: 2.43 ms per loop
100 loops, best of 3: 4.44 ms per loop
1000 loops, best of 3: 1.66 ms per loop

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