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

Question

Emre Can Yilmaz 2022-4-21

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1701235-separating-data-into-one-second-intervals-and-finding-the-maximum-data-in-each-interval

编辑： per isakson 2022-4-21

Green6000X.csv

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

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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

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Answer 1

per isakson 2022-4-21

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1701235-separating-data-into-one-second-intervals-and-finding-the-maximum-data-in-each-interval#answer_947215

%%  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

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Emre Can Yilmaz 2022-4-21

This was exactly the solution I wanted, thank you very much.

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"

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Answer 2

KSSV 2022-4-21

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1701235-separating-data-into-one-second-intervals-and-finding-the-maximum-data-in-each-interval#answer_947200

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)