查找序列中的零岛

本文介绍了查找序列中的零岛的处理方法，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

想象一下你有一个很长的序列.找到序列全为零(或更准确地说，序列下降到接近零值abs(X)

Imagine you have a very long sequence. What is the most efficient way of finding the intervals where the sequence is all zeros (or more precisely the sequence drops to near-zero values abs(X)<eps):

为简单起见，我们假设以下顺序:

For simplicity, lets assume the following sequence:

sig = [1 1 0 0 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0];

我正在尝试获取以下信息:

I'm trying to get the following information:

startIndex EndIndex Duration 3 6 4 12 12 1 14 16 3 25 26 2 30 30 1

然后使用这些信息，我们找到持续时间 >= 到某个指定值(比如 3)的区间，并返回所有这些区间组合中的值的索引:

then using this information, we find the intervals with duration >= to some specified value (say 3), and returning the indices of the values in all these intervals combined:

indices = [3 4 5 6 14 15 16];

最后一部分与上一个问题有关:

That last part is related to a previous question:

MATLAB:矢量化数组创建从开始/结束索引列表

这是我目前所拥有的:

sig = [1 1 0 0 0 0 1 1 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0]; len = length(sig); thresh = 3; %# align the signal with itself successively shifted by one %# v will thus contain 1 in the starting locations of the zero interval v = true(1,len-thresh+1); for i=1:thresh v = v & ( sig(i:len-thresh+i) == 0 ); end %# extend the 1's till the end of the intervals for i=1:thresh-1 v(find(v)+1) = true; end %# get the final indices v = find(v);

我希望对代码进行矢量化/优化，但我愿意接受其他解决方案.我必须强调，空间和时间效率非常重要，因为我正在处理大量长生物信号.

I'm looking to vectorize/optimize the code, but I'm open to other solutions. I have to stress that space and time efficiencies are very important, since I'm processing a large number of long bio-signals.

推荐答案

这些是我以矢量化方式解决您的问题所采取的步骤，从给定的矢量 sig 开始:

These are the steps I would take to solve your problem in a vectorized way, starting with a given vector sig:

• 首先，对向量设置阈值以获得零和一的向量tsig(信号绝对值下降到足够接近零的零，其他地方的零):

• First, threshold the vector to get a vector tsig of zeros and ones (zeroes where the absolute value of the signal drops close enough to zero, ones elsewhere):

tsig = (abs(sig) >= eps); %# Using eps as the threshold

接下来，使用函数DIFF 和 查找:

dsig = diff([1 tsig 1]); startIndex = find(dsig < 0); endIndex = find(dsig > 0)-1; duration = endIndex-startIndex+1;

然后，找到持续时间大于或等于某个值(例如 3，来自您的示例)的零字符串:

Then, find the strings of zeroes with a duration greater than or equal to some value (such as 3, from your example):

stringIndex = (duration >= 3); startIndex = startIndex(stringIndex); endIndex = endIndex(stringIndex);

最后，使用 我对链接问题的回答中的方法 用于生成您的最终索引集:

Finally, use the method from my answer to the linked question to generate your final set of indices:

indices = zeros(1,max(endIndex)+1); indices(startIndex) = 1; indices(endIndex+1) = indices(endIndex+1)-1; indices = find(cumsum(indices));