计算 numpy 数组中连续出现的长度不同的值 [英] Count consecutive occurences of values varying in length in a numpy array

问题描述

假设我在一个 numpy 数组中有一堆数字，我根据返回布尔数组的条件测试它们:

Say I have a bunch of numbers in a numpy array and I test them based on a condition returning a boolean array:

np.random.seed(3456)
a = np.random.rand(8)
condition = a>0.5

使用这个布尔数组，我想计算 True 连续出现的所有长度.例如，如果我有 [True,True,True,False,False,True,True,False,True] 我想找回 [3,2,1].

And with this boolean array I want to count all of the lengths of consecutive occurences of True. For example if I had [True,True,True,False,False,True,True,False,True] I would want to get back [3,2,1].

我可以使用此代码做到这一点:

I can do that using this code:

length,count = [],0
for i in range(len(condition)):

    if condition[i]==True:
        count += 1
    elif condition[i]==False and count>0:
        length.append(count)
        count = 0

    if i==len(condition)-1 and count>0:
        length.append(count)

    print length

但是是否已经为此或python、numpy、scipy 等实现了任何功能来计算给定输入的列表或数组中连续出现的长度?

But is there anything already implemented for this or a python,numpy,scipy, etc. function that counts the length of consecutive occurences in a list or array for a given input?

推荐答案

这是一个使用 itertools 的解决方案(它可能不是最快的解决方案):

Here's a solution using itertools (it's probably not the fastest solution):

import itertools
condition = [True,True,True,False,False,True,True,False,True]
[ sum( 1 for _ in group ) for key, group in itertools.groupby( condition ) if key ]

Out:
[3, 2, 1]

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