算在Matlab载体内含有连续少于3个零元素 [英] Count elements containing less than 3 consecutive zeros within vector in Matlab

问题描述

例如，我有以下矢量：

A = [34 35 36 0 78 79 0 0 0 80 81 82 84 85 86 102 0 0 0 103 104 105 106 0 0 107 201 0 202 203 204];

一个再$ P $中的每个元素psents在每一秒的值。我想算包含在连续的不到3个零元素=>我会为A.秒获得持续时间值

Each element within A represents a value at every second. I want to count the elements containing less than 3 consecutive zeros in A => I will obtain duration values in seconds for A.

在这种情况下，计数每3个连续的零之前停止和3的连续零后重新开始，依此类推。它是这样的：

In this case, the counting stops before every 3 consecutive zeros and starts over after the 3 consecutive zeros, and so on. It's like this:

A = [34->1s 35->2s 36->3s 0->4s 78->5s 79->6s stop 80->1s 81->2s 82->3s 84->4s 85->5s 86->6s 102->7s stop 103->1s 104->2s 105->3s 106->4s 0->5s 0->6s 107->7s 201->8s 0->9s 202->10s 203->11s 204->12s];

其结果将是这样的：

The result would like this:

Duration = [6 7 12]; in seconds

任何人有什么想法？

Anyone got any idea?

推荐答案

您可以 A 转换为字符 0 和 1 （例如）根据原始值是否为零或非零，使用 strsplit 和然后获得每个子的元素的数量。

You can convert A to characters '0' and '1' (for example) depending on whether the original value is zero or nonzero, use strsplit and then obtain the number of elements of each substring.

让 N = 3 是零用于分割的数量。然后：

Let N = 3 be the number of zeros for splitting. Then:

Duration = cellfun(@numel, strsplit(char((A>0)+'0'), repmat('0',1,N)));

请注意，上面的code确实根据的究竟 N 零分裂。例如， A = [1 2 3 0 0 0 0 4 5] 给持续时间= [3 3] ，由于第四零被分配给第二子

Note that the above code does the splitting based on exactly N zeros. For example, A = [1 2 3 0 0 0 0 4 5] gives Duration = [3 3], because the fourth zero is assigned to the second substring.

如果您想根据 N 以上零分割，使用常规的前pression：

If you want to split based on N or more zeros, use a regular expression:

Duration = cellfun(@numel, regexp(char((A>0)+'0'), [repmat('0',1,N) '+'], 'split'));

有关 A = [1 2 3 0 0 0 0 4 5] 这给了持续时间= [3 2] 。

这篇关于算在Matlab载体内含有连续少于3个零元素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！