A number contain in a series.

1 次查看（过去 30 天）
Silpa K
Silpa K 2019-9-19
评论： Silpa K 2019-9-20
I have a series 's'
s=(1:1,2:end)
I find its subsequences like
a=s(1:30)
b=s(30:60)
c=s(60:90)
d=s(90:120)
e=s(120:150)
f=s(150:180)
j=s(180:210)
h=s(210:250)
and I have a set of points 'k'.
I need to check any of the subsequence contain any points of the K.If it contain I need to add those subsequence into a variable 'A'.How can I do this.Please help me.
  1 个评论
Guillaume
Guillaume 2019-9-19
s = (1:1, 2:end)
is not valid matlab syntax. So, the first problem with your question is we don't really not what your series is.

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

thoughtGarden
thoughtGarden 2019-9-19
编辑：thoughtGarden 2019-9-20
You haven't provided enough information to be certain of what you want, but makeing some assumptions, this should work.
clear;clc;
% "series" s, which is an array
s = 1:1:250;
% Build "subsequences"
for ii = 1:7
subSequence(ii,:) = s((ii-1)*30 + 1:ii*30);
end
% Build "set of points 'k'"
k = randi(1000,1,10);
% A will contain all subsequences that contain any of the values of k
A = [];
% In each loop, determine if there is any overlap betweent he subsequence
% and the variable k. If there is, add the subsequency to the variable A.
% Otherwise, move on to the next subsequence.
for ii = 1:length(subSequence(:,1))
if(intersect(subSequence(ii,:),k))
A(end+1,:) = subSequence(ii,:);
else
%do nothing...
end
end
% Disp A, which in the case of this script might be empty as k is random
% values...
disp(A)
This builds an array containing all the subsequences that contain any values found in k. If that is what you are looking for, this works.
  23 个评论
显示 21更早的评论
Silpa K
Silpa K 2019-9-20
I need to display the subsequence name that contain k.
Silpa K
Silpa K 2019-9-20
[~,ii] = min(abs(s(:) - k(:)'));
out = s(unique(ii));
using this I get nearest elements,now also I can't getting the name of the subsequence

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Data Type Conversion 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by