# How to extract sub-matrices from a big matrix?

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Mohamed Nedal on 21 Dec 2019
Edited: Stephen Cobeldick on 23 Dec 2019
Hello everyone,
I have two matrices, kindly find the attached file
soho ---> size = 23×12 , each row represents an event.
omni ---> size = 8760 × 17 , each row represent an hour (timeseries data).
The first three columns are, in order, "year", "month", "day".
I need to do the following operation:
if soho_day = omni_day
take that omni_day and the following 120 rows (hours) and put them as a separate matrix with a prefix (i.e., omniSUB_1).
and so on
I have these code blocks
% to find the matched date and store the that day along with the following 5 days (120 hours) in another matrix.
for n = 1:length(soho)
for m = 1:length(omni)
if datetime(soho(n,1),soho(n,2),soho(n,3),...
'Format','dd/MM/yyyy') == ...
datetime(omni(m,1),omni(m,2),omni(m,3),...
Format','dd/MM/yyyy')
for k = m:120
omniSUB{n,:} = omni(k,:);
end
end
end
end
and
% to create sub-matrices.
Prefix = 'omniSUB_';
for i = 1:Month_length
var_name = strcat(Prefix, num2str(i));
data_child = genvarname(var_name);
eval([data_child ' = omniSUB{i}']);
end
but I don't know how to put them together to do that operation. Please correct me.

Show 1 older comment
Mohamed Nedal on 22 Dec 2019
soho contains data about storms and omni contains time-series data of the variables affected by such storms.
So basically what I want to do is finding how these variables change over a time period of 5 days starting from the storm's starting time and separate each storm's data in a matrix to proceed further analysis.
That's why I need to separate each 5-day period in a separate matrix because I have another function that will loop over these matrices and do further analysis for each one.
For me, it's okay if the code is slow or complex as long as it gets the job done :)
Stephen Cobeldick on 22 Dec 2019
"That's why I need to separate each 5-day period in a separate matrix because I have another function that will loop over these matrices and do further analysis for each one."
Exactly as I wrote in my earlier comment, you just need to use indexing.
You can easily store the separated data in one array (e.g. a cell array) using basic MATLAB indexing, which will be much simpler and more efficient than what you are doing now, and then it is trivial to "loop over these matrices and do further analysis for each one" as you wrote. So far nothing you have shown or described requires your complex and inefficeint approach of accessing variable names dynamically.
Mohamed Nedal on 22 Dec 2019
I'll try the indexing approach. Thanks!

Ridwan Alam on 21 Dec 2019
Edited: Ridwan Alam on 21 Dec 2019
% to find the matched date and store the that day along with the following 5 days (120 hours) in another matrix.
for n = 1:size(soho,1)
omniRowInd = find(omni(:,1)==soho(n,1) & omni(:,2)==soho(n,2) & omni(:,3)==soho(n,3),1,'first');
if ~isempty(omniRowInd)
tempTable = omni(omniRowInd:min(omniRowInd+119,size(omni,1)),:);
eval(['omniSUB_' num2str(n) '=' 'tempTable;']);
end
end
Hope this helps!

Mohamed Nedal on 23 Dec 2019
Hi Stephen, I tried your code with much larger dataset, kindly find the attached file.
N = size(soho,1);
C = cell(1,N);
for k = 1:N
omniRowInd = find(omni(:,1)==soho(k,1) & omni(:,2)==soho(k,2) & omni(:,3)==soho(k,3),1,'first');
if ~isempty(omniRowInd)
C{k} = omni(omniRowInd:min(omniRowInd+119,size(omni,1)),:);
end
end
but the resulted C{k} is empty. Could you determine the problem here plz?
Stephen Cobeldick on 23 Dec 2019
"Could you determine the problem here plz? "
Are you checking all of the cells of the cell array, not just the last one?:
find(~cellfun(@isempty,C))
It is quite possible that the if condition is never true. You can check this by printing it in the loop, or collect it into a vector and checking it after the loop:
N = size(soho,1);
C = cell(1,N);
L = false(1,N); % logical vector
for k = 1:N
omniRowInd = find(omni(:,1)==soho(k,1) & omni(:,2)==soho(k,2) & omni(:,3)==soho(k,3),1,'first');
if ~isempty(omniRowInd)
L(k) = true; % if the condition is true
C{k} = omni(omniRowInd:min(omniRowInd+119,end),:);
end
end
% Check on which iteration/s the conditions was true:
any(L)
find(L)
Note how easy it is using indexing to access the new variable L. Trying to use eval to do something similar with lots of numbered variables would be much more complex.
Mohamed Nedal on 23 Dec 2019
Yes, you're right.
I found where's the error, it was a glitch in defining the date-time pairs and I corrected it.
Thank you!

R2017b

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