本文介绍了获取布尔数组中至少 n 个连续 False 值的第一个块的索引的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
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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