Python中具有周期性边界的数组中的连续值 [英] Consecutive values in array with periodic boundaries in Python

问题描述

我有一些二维数组，其中填充了0和1:

I have some 2D-arrays filled with 0 and 1:

import numpy as np

a = np.random.randint(2, size=(20, 20))
b = np.random.randint(2, size=(20, 20))
c = np.random.randint(2, size=(20, 20))
d = np.random.randint(2, size=(20, 20)) 

，我想计算具有周期性边界的连续出现次数. 这意味着(为了清楚起见，以1D表示):

and I want to count the consecutive occurrence of the ones with periodic boundaries. That means (in 1D for clearness):

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

应该给我5(后三个元素+前两个元素).
应该在第三个(如果从0开始，则是第二个)轴上比较/计数2D数组，例如首先将数组堆叠在axis=2中，然后应用与1D相同的算法.但是我不确定这是否是最简单的方法.

should give me 5(last three elements + first two).
The 2D-arrays should be compared/counted in the third (second if you start with 0) axis, like first stacking the arrays in axis=2 and then applying the same algorithm like for 1D. But I am not sure if this is the most simple way.

推荐答案

这是2D的ndarrays a和更高的dim数组的一种方式，旨在提高性能-

Here's one way for ndarrays a of 2D and higher dim arrays, meant for performance efficiency -

def count_periodic_boundary(a):
    a = a.reshape(-1,a.shape[-1])
    m = a==1    
    c0 = np.flip(m,axis=-1).argmin(axis=-1)+m.argmin(axis=-1)
    z = np.zeros(a.shape[:-1]+(1,),dtype=bool)
    p = np.hstack((z,m,z))
    c = (p[:,:-1]<p[:,1:]).sum(1)
    s = np.r_[0,c[:-1].cumsum()]
    l = np.diff(np.flatnonzero(np.diff(p.ravel())))[::2]
    d = np.maximum(c0,np.maximum.reduceat(l,s))    
    return np.where(m.all(-1),a.shape[-1],d)

样品运行-

In [75]: np.random.seed(0)
    ...: a = np.random.randint(2, size=(5, 20))

In [76]: a
Out[76]: 
array([[0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1],
       [0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0],
       [0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1],
       [1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0],
       [0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0]])

In [77]: count_periodic_boundary(a)
Out[77]: array([7, 4, 5, 2, 6])


In [72]: np.random.seed(0)
    ...: a = np.random.randint(2, size=(2, 5, 20))

In [73]: a
Out[73]: 
array([[[0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1],
        [0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0],
        [0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 0, 1],
        [1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0],
        [0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0]],

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

In [74]: count_periodic_boundary(a)
Out[74]: array([7, 4, 5, 2, 6, 2, 5, 4, 2, 1])

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