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

6 vues (au cours des 30 derniers jours)
Emre Can Yilmaz le 21 Avr 2022
Modifié(e) : per isakson le 21 Avr 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
1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
Emre Can Yilmaz le 21 Avr 2022
Modifié(e) : Image Analyst le 21 Avr 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

Connectez-vous pour commenter.

Réponse acceptée

per isakson le 21 Avr 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 commentairesAfficher AucuneMasquer Aucune
Emre Can Yilmaz le 21 Avr 2022
This was exactly the solution I wanted, thank you very much.
per isakson le 21 Avr 2022
Modifié(e) : per isakson le 21 Avr 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"

Connectez-vous pour commenter.

Plus de réponses (1)

KSSV le 21 Avr 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)
3 commentairesAfficher 1 commentaire plus ancienMasquer 1 commentaire plus ancien
KSSV le 21 Avr 2022
You can get the time step to equal, append NaN's at the end and then use reshape.
Emre Can Yilmaz le 21 Avr 2022

Connectez-vous pour commenter.

Catégories

En savoir plus sur Large Files and Big Data dans Help Center et File Exchange

R2021b

Community Treasure Hunt

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

Start Hunting!

Translated by