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

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Emre Can Yilmaz am 21 Apr. 2022
Bearbeitet: per isakson am 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
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Emre Can Yilmaz am 21 Apr. 2022
Bearbeitet: Image Analyst am 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

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per isakson am 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
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Emre Can Yilmaz am 21 Apr. 2022
This was exactly the solution I wanted, thank you very much.
per isakson am 21 Apr. 2022
Bearbeitet: per isakson am 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"

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KSSV am 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)
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KSSV am 21 Apr. 2022
You can get the time step to equal, append NaN's at the end and then use reshape.
Emre Can Yilmaz am 21 Apr. 2022

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