Separating data into one-second intervals, and finding the maximum data in each interval

5 次查看(过去 30 天)
I have a 2 column matrix with around1300 data per second and measurements in total between 40-80 seconds, the exact number of data is not certain. I'm trying to print the largest three data and the smallest three values in every second in the matrix I have. I think my algorithm knowledge is insufficient for this. Is there anyone who can help?
load("Green6000X.csv")
b12xtime=Green6000X(:,1);
b12xacc=Green6000X(:,2);
u=0:5:height(b12xtime);
v=zeros(138,1);
for i=1:height(b12xtime)
a=find(b12xtime(:,1)<=i);
b=find(b12xtime(:,1)<=i+1);
t(i,1)=height(a);
s(i,1)=height(b);
% val=abs(s-t);
for x=t(i):s(i)
c(x,1)=b12xacc(x,1);
d=max(c);
b=height(s);
n=height(t);
v(x,1)=d;
if(v(x-1,1)==v(x,1))
v(x-1,1)=0;
end
end
clear c;
end
column= find(v==0);
for i=1:length(column)
column= find(v==0);
v(column(1),:) = [];
end
  1 个评论
Emre Can Yilmaz
Emre Can Yilmaz 2022-4-21
编辑:Image Analyst 2022-4-21
This time I tried a different code for maximum value. It takes a very long time to run, about 30 minutes. I don't even know if your conclusion is correct.
load("Green6000X.csv")
b12xtime=Green6000X(:,1);
b12xacc=Green6000X(:,2);
lastValueOfTime=ceil(b12xtime(end));
for ii=0:lastValueOfTime
for i=1:height(b12xtime)
a=find(b12xtime(:,1)<=i);
b=find(b12xtime(:,1)<=i+1);
t(i,1)=height(a);
s(i,1)=height(b);
% val=abs(s-t);
for x=t(i):s(i)
c(x,1)=b12xacc(x,1);
d=max(c);
v(x,1)=d;
if(v(x-1,1)==v(x,1))
v(x-1,1)=0;
end
end
end
end
for i=1:length(column)
column= find(v==0);
v(column(1),:) = [];
end

请先登录,再进行评论。

采纳的回答

per isakson
per isakson 2022-4-21
%% load data
num = load('Green6000X.csv');
%% split data into chunks of one second
N = ceil( num(end,1) );
chunk = cell( 1, N );
ixb = 1;
for jj = 0 : N-2
ixe = find( num(:,1) >= jj+1, 1, 'first' );
chunk{jj+1} = num( ixb:ixe-1, : );
ixb = ixe;
end
chunk{N} = num(ixb:end,:);
%% calculate max for each chunk
v = nan( N, 1 );
for ii = 1 : N
v(ii) = max( chunk{ii}(:,2) );
end
%% first three and the last three values of each chunk
three = nan( N, 6 );
for ii = 1 : N
three(ii,:) = reshape( chunk{ii}([1:3,end-2:end],2), 1,[] );
end
  2 个评论
per isakson
per isakson 2022-4-21
编辑:per isakson 2022-4-21
Response to "Can we also write the largest three data and the smallest three in the part?"
Add the section below to the script
%% the largest three data and the smallest three of each chunk
min3max3 = nan( N, 6 );
for ii = 1 : N
min3max3(ii,:) = [ reshape( mink( chunk{ii}(:,2), 3 ), 1,[] ) ...
reshape( maxk( chunk{ii}(:,2), 3 ), 1,[] ) ];
end
I don't understand "in the part"

请先登录,再进行评论。

更多回答(1 个)

KSSV
KSSV 2022-4-21
As you said data is from 40-80 seconds and each second has 1300 data points, you can pick the first 40*1300 rows and reshape the data.
T = readtable('Green6000X.csv') ;
d = reshape(T.(2)(1:40*1300),1300,40) ;
% first three elements of each second
d(:,1:3)
% last three elements of each second
d(:,end-3:end)
% max in each row
max(d,[],2)
  3 个评论

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Matrix Indexing 的更多信息

产品


版本

R2021b

Community Treasure Hunt

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

Start Hunting!

Translated by