Splitting an array every n rows
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Vincent den Ronden
am 26 Nov. 2018
Bearbeitet: Stephen23
am 26 Nov. 2018
Hi everyone,
fairly new at matlab and i am having a problem with arrays.
My goal: I have an array(mx3) of data that i want to separate it every N rows so i can save that (sub)array as a separate file.
I am having trouble with how the loop should be setup:
p=1;
n = ceil(length(Accell)/N); % number of samples
y=N;
for i = 1:n
for q = 1:j
M{i,q} = Accell(p:y,q);
end
end
A obvious problem occurs; the variable p and y stay constant so the row won't change. Accell is my Data en N is the number of rows I want to separate.
so ideally M would be:
M{1,:} = Accel(1:80,:)
M{2,:} = Accel(81:160,:)
etc etc
After this loop I extract the rows of M into a array and save it.
for x=1:size_M % Number of samples (rows) in M
i=1;
while exist(['Saved Data\', num2str(classificatie), '\', num2str(i.', 'D%03d'), '.mat'], 'file')==2
i=i+1;
end
inputs = [M{x,1},M{x,2},M{x,3}];
save((['Saved Data\', num2str(classificatie), '\', num2str(i.', 'D%03d'), '.mat']), 'inputs')
% set(handles.debug, 'string','Measurement values have been saved successful')
end
Right now the code saves the first 80 rows of the data (Accell) in n (samples) files. I know it's very convoluted and I am very open to improvements.
Thank you in advance!
-Vincent
2 Kommentare
KALYAN ACHARJYA
am 26 Nov. 2018
Can you elaborate your problem with some example?
Input (consider example as representation) and what result you expect?
Akzeptierte Antwort
Stephen23
am 26 Nov. 2018
Bearbeitet: Stephen23
am 26 Nov. 2018
"so ideally M would be:"
M{1,:} = Accel(1:80,:)
M{2,:} = Accel(81:160,:)
etc etc
>> N = 80;
>> A = rand(192,3); % fake data, # of rows != multiple of N.
>> S = size(A);
>> R = N*ones(1,ceil(S(1)/N)); % number of rows in each cell.
>> R(end) = 1+mod(S(1)-1,N); % number of rows in last cell.
>> M = mat2cell(A,R,S(2));
And checking:
>> cellfun('size',M,1)
ans =
80
80
32
Or, if you really want to use a loop, you can use end inside the array indexing:
R = ceil(S(1)/N);
M = cell(R,1);
for k = 1:R
B = (k-1)*N+1;
E = k*N;
M{k} = A(B:min(end,E),:);
end
And checking:
>> cellfun('size',M,1)
ans =
80
80
32
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