Question

我有一个0和0的numpy数组：

y=[1,1,1,0,0,0,0,0,1,1,0,0,0,0,0,0,1,1,1,1]

我想计算一组（或零）的索引。因此，对于上面的示例，一组的结果应该类似于：

result=[(0,2), (8,9), (16,19)]

（怎么样）我可以用numpy做到吗？我没有发现像分组功能。

我尝试了np.ediff1d，但无法找到一个好的解决方案。并不是说数组可能会也可能不会以一组结果开始/结束：

import numpy as np

y = [1,1,1,0,0,0,0,0,1,1,0,0,0,0,0,0,1,1,1,1]
mask = np.ediff1d(y)
starts = np.where(mask > 0)
ends = np.where(mask < 0)

我在这里也找到了部分解决方案： Find index where elements change value numpy

但那个只给了我改变值的指数。

Answer 1

我们可以做这样的事情，适用于任何通用数组 -

def islandinfo(y, trigger_val, stopind_inclusive=True):
    # Setup "sentients" on either sides to make sure we have setup
    # "ramps" to catch the start and stop for the edge islands
    # (left-most and right-most islands) respectively
    y_ext = np.r_[False,y==trigger_val, False]

    # Get indices of shifts, which represent the start and stop indices
    idx = np.flatnonzero(y_ext[:-1] != y_ext[1:])

    # Lengths of islands if needed
    lens = idx[1::2] - idx[:-1:2]

    # Using a stepsize of 2 would get us start and stop indices for each island
    return zip(idx[:-1:2], idx[1::2]-int(stopind_inclusive)), lens

示例运行 -

In [320]: y
Out[320]: array([1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1])

In [321]: islandinfo(y, trigger_val=1)[0]
Out[321]: [(0, 2), (8, 9), (16, 19)]

In [322]: islandinfo(y, trigger_val=0)[0]
Out[322]: [(3, 7), (10, 15)]

或者，我们可以使用diff进行切片比较，然后使用2列重新整形以替换步长大小的切片，为自己提供单线 -

In [300]: np.flatnonzero(np.diff(np.r_[0,y,0])!=0).reshape(-1,2) - [0,1]
Out[300]: 
array([[ 0,  2],
       [ 8,  9],
       [16, 19]])

Numpy查找具有相同值的组的索引

1 个答案: