计算numpy数组中的元素连续满足条件的次数 [英] Count the number of times elements in a numpy array consecutively satisfy a condition

问题描述

我有一个numpy数组，如下所示:

I have a numpy array as follows:

import numpy as np
a = np.array([1, 4, 2, 6, 4, 4, 6, 2, 7, 6, 2, 8, 9, 3, 6, 3, 4, 4, 5, 8])

和常数b=6

我正在搜索数字c，该数字由a中的元素连续小于b的次数等于或大于2的次数定义.

I am searching for a number c which is defined by the number of times the elements in a are less than b 2 or more times consecutively.

因此在此示例中为c=3

我没有有效的代码，这就是为什么我在这里问这个问题.基于上一个问题，我可以使用np.sum(a<b)来获取a<b的次数.

I have no working code, that's why I am asking that here. Based on a previous question I can use np.sum(a<b) to get the number of times that a<b.

print(np.sum(a<b))
#12

现在，我想计算a连续两次小于b的次数.

Now I want to count the number of times where a is two or more times consecutively less than b.

以下是此示例a中的3个组的说明:

Here is an illustration of the 3 groups in for this sample a:

1, 4, 2, 6, 4, 4, 6, 2, 7, 6, 2, 8, 9, 3, 6, 3, 4, 4, 5, 8  # numbers in a
1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0  # (a<b)
^^^^^^^-----^^^^-----------------------------^^^^^^^^^^---  # (a<b) 2+ times consecutively
   1         2                                    3

推荐答案

您可以使用numpy遮罩和itertools.groupby.

from itertools import groupby

b = 6
sum(len(list(g))>=2 for i, g in groupby(a < b) if i)
#3

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