获取布尔数组中至少 n 个连续 False 值的第一个块的索引

get index of the first block of at least n consecutive False values in boolean array(获取布尔数组中至少 n 个连续 False 值的第一个块的索引)

问题描述

我有一个 numpy 布尔数组

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

我想获取第一次有 n_at_least 错误值的索引.比如这里

I would like to get the index of the first time there are at n_at_least false values. For instance here

`n_at_least`=1 -> desired_index=1

`n_at_least`=3 -> desired_index=4

我试过了

np.cumsum(~w)

每次遇到 False 值时都会增加.但是，当遇到 True 时，计数器不再从 0 开始，所以我只得到 False 元素的总数，而不是最后一个连续元素的计数.

which does increase every time a False value is encountered. However, when True is encountered the counter is not starting from 0 again so I only get the total count of False elements rather than the count of the last consecutive ones.

推荐答案

我认为对于这种线性搜索操作，python 实现是可以的.我的建议是这样的:

I think for this linear search operation a python implementation is ok. My suggestion looks like this:

def find_block(arr, n_at_least=1):
    current_index = 0
    current_count = 0
    for index, item in enumerate(arr):
         if item:
             current_count = 0
             current_index = index + 1
         else:
             current_count += 1
         if current_count == n_at_least:
             return current_index
    return None # Make sure this is outside for loop

运行此函数会产生以下输出:

Running this function yields the following outputs:

>>> import numpy
>>> w = numpy.array([True, False, True, True, False, False, False])
>>> find_block(w, n_at_least=1)
1
>>> find_block(w, n_at_least=3)
4
>>> find_block(w, n_at_least=4)
>>> # None

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