查找值超过阈值的最长连续数组 [英] find longest continuous array with values exceeding a threshold

问题描述

如何在matlab中使用查找"来查找超过给定阈值的最长的序列值.例如，如果我有:

How can I use 'find' in matlab to find the longest possible sequence of values that exceed a given threshold. Fro example, if I have:

X = [18 3 1 11 8 10 11 3 9 14 6 1 4 3 15 21 23 24];

并要查找大于例如8的值，请输入:

and want to find the values that are larger than, say, 8, I would type:

find(X > 8)

但是，我想找出数组中最长的连续序列大于给定值的位置.在这里，答案将是:

However, I want to find where in the array is the longest continuous sequence of values larger than a given value. Here, the answer would be:

15 21 23 24

有没有办法在Matlab中实现这一目标?

Is there a way to achieve this in matlab?

推荐答案

让您的数据定义为

X = [18 3 1 11 8 10 11 3 9 14 6 1 4 3 15 21 23 24]; %// input array
t = 8; %// threshold

方法1:使用 diff

Y = X>t; %// convert to zeros and ones
z = diff([false Y false]); %// compute changes. `false`s are needed to "close" the sequence
s = find(z>0); %// find indices of starts of runs
e = find(z<0)-1; %// find ends 
[~, w] = max(e-s); %// w is the index of the longest run
result_start = s(w); %// index of start of longest subsequence
result_end = e(w); %// index of end of longest subsequence
result_values = X(s(w):e(w)); %// values of longest subsequence

以示例X和t，

result_start =
    15
result_end =
    18
result_values =
    15    21    23    24

方法2:使用 regexp

Y = char(double(X>t) + '0'); %// convert to string of zeros and ones
[s, e] = regexp(Y, '1+', 'start', 'end'); %// find starts and ends of runs
[~, w] = max(e-s); %// w is the index of the longest run
result_start = s(w); %// index of start of longest subsequence
result_end = e(w); %// index of end of longest subsequence
result_values = X(s(w):e(w)); %// values of longest subsequence

结果与上面相同.

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