计数长度变化值的连续OCCURENCES在numpy的阵列 [英] Count consecutive occurences of values varying in length in a numpy array

问题描述

说我有一串数字在numpy的阵列和我进行测试的基础上返回布尔数组的一个条件：

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

和这个布尔数组我要统计所有的真连续OCCURENCES的长度。例如，如果我有 [真，真，真，假，假，真，真，假，真] 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].

我能做到这一点使用此code：

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

但有已经为这个或蟒蛇，numpy的，SciPy的实施什么，等功能计数在列表或数组给定输入？

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

下面是一个使用和itertools（这可能不是最快的解决方案）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]

这篇关于计数长度变化值的连续OCCURENCES在numpy的阵列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！