在numpy数组中查找False-True过渡 [英] Finding False-True transitions in a numpy array

问题描述

给出一个numpy数组:

Given a numpy array:

x = np.array([False, True, True, False, False, False, False, False, True, False])

如何查找值从False转换为True的次数? 对于上面的示例，答案是2.我不要在计数中包括从True到False的转换.

How do I find the number of times the values transitions from False to True? For the above example, the answer would be 2. I don't want to include transitions from True to False in the count.

从答案到我如何确定布尔数组中的值序列?，下面的代码将生成将要更改值的索引，这不是我想要的，因为其中包括True-False转换.

From the answers to How do I identify sequences of values in a boolean array?, the following produces the indices at which the values are about to change, which is not what I want as this includes True-False transitions.

np.argwhere(np.diff(x)).squeeze()
# [0 2 7 8]

我知道这可以通过遍历数组来完成，但是我想知道是否有更快的方法?

I know that this can be done by looping through the array, however I was wondering if there was a faster way to do this?

推荐答案

获取一次性切片-x[:-1](从第一个elem开始并在倒数第二个elem中结束)和x[1:](从第二个elem开始并进行到最后)，然后寻找第一个小于第二个的切片，即捕获[False, True]的模式，最后使用ndarray.sum()或np.count_nonzero()-

Get one-off slices - x[:-1] (starting from the first elem and ending in second last elem) and x[1:] (starting from the second elem and going on until the end), then look for the first slice being lesser than the second one, i.e. catch the pattern of [False, True] and finally get the count with ndarray.sum() or np.count_nonzero() -

(x[:-1] < x[1:]).sum()
np.count_nonzero(x[:-1] < x[1:])

另一种方法是寻找第一个切片为False，第二个切片为True，再次想到的是捕捉[False, True]-

Another way would be to look for the first slice being False and the second one as True, the idea again being to catch that pattern of [False, True] -

(~x[:-1] & x[1:]).sum()
np.count_nonzero(~x[:-1] & x[1:])

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