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

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Emre Can Yilmaz on 21 Apr 2022
Edited: per isakson on 21 Apr 2022
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?
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
Emre Can Yilmaz on 21 Apr 2022
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.
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

per isakson on 21 Apr 2022
%% 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 CommentsShowHide 1 older comment
per isakson on 21 Apr 2022
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"

KSSV on 21 Apr 2022
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.
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)
Emre Can Yilmaz on 21 Apr 2022

R2021b

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