if statement with multiple conditions within cells

Hello ,

I have two 75x1 cells called goodcase and badcase , each cell is composed by 75 logical vectors of the same size (241200 elements).

I want to find if the following case exists between the 2 cells:

one of the first 1200 elements of the logical vector in badcase is equal to 1 ,(let's call this element u) and one of the elements placed in the position u+1*1200 or the position u+2*1200 or u+3*1200 or u+4*1200 and so on until the highlighted numbers equal to 200 in the logical vector in goodcase is also equal to 1

I want to know the values of u and the values of the highlighted numbers that accomplish this case.

I tried with this code but It does not give me any result even when I have checked "by hand" that the case exists.

cases3=cell(75,1);%preallocating size of the statement
for gcellid=1:numel(goodcase)
cases3{gcellid}=zeros(1200,201);
end

for gcellid=1:numel(goodcase);
 ii=1:numel(1200);
 for y=1:200;
 if badcase{gcellid}(ii)==1 & goodcase{gcellid}(ii+y*1200)==1;
 cases3{gcellid}(ii,y)=1
 end    
 end
end

PP3=cellfun(@(x) find(x(:,:)==1),cases3,'un',false)%linear position of the cases I want

Answer 1

[Adam Danz]

Some of your question required interpretation. See the inline comments below and follow along line-by-line to understand how I interpretted the problem. If a line does not match your expectations, indicate the line and clarify the interpretation.

The final results are put into a table named results.

% Recreate inputs based on the following description:
% I have two 75x1 cells called goodcase and badcase , 
% each cell is composed by 75 logical vectors of the 
% same size (241200 elements).
rng(14159) % for reproducibility of results for this demo ONLY
goodcase = arrayfun(@(i){rand(1,241200)<0.05},1:75)';
badcase = arrayfun(@(i){rand(1,241200)<0.0001},1:75)';

% I want to find if the following case exists between the 2 cells:
% one of the first 1200 elements of the logical vector in badcase 
% is equal to 1 ,(let's call this element u) 
% [My interpretation is that u is a row or column number of the first TRUE in badcase{n}(1:1200)]
u = cellfun(@(x)find(x(1:1200),1,'first'),badcase,'UniformOutput',false);

% and  one of the elements 
% placed  in the position u+1*1200 or the position u+2*1200 or u+3*1200 
% or u+4*1200  and so on until the highlighted numbers equal to 200 in  
% the logical vector in goodcase is also equal to 1
% [My interpretation is that you're searching in goodcase for these values]
notIsEmpty = ~cellfun(@isempty,u); 
idx = cell(size(u)); 
idx(notIsEmpty) = cellfun(@(x,i)find(x(i+(1:200)*1200)),goodcase(notIsEmpty),u(notIsEmpty),'UniformOutput',false);

% I want to know the values of u and the values of the highlighted 
% numbers that  accomplish this case.
% [I'll organize them into a table]
results = table(u,idx,'VariableNames',{'u','idx'},'RowNames',cellstr(string(1:numel(u))));

% Let's look at he first few rows of results
% Empty values indicate a TRUE was not found in the first 1:1200 values of badcase{n}
head(results)

% ans =
%   8×2 table
%               u               idx     
%          ____________    _____________
% 
%     1    {1×0 double}    {0×0  double}
%     2    {1×0 double}    {0×0  double}
%     3    {[     609]}    {1×10 double}
%     4    {1×0 double}    {0×0  double}
%     5    {[     801]}    {1×11 double}
%     6    {1×0 double}    {0×0  double}
%     7    {1×0 double}    {0×0  double}
%     8    {1×0 double}    {0×0  double}

% We see that element #3 of badcase contains a TRUE in element #609. 
% To see the results for goodcase, results.idx{3}
% Element #3 in goodcase contains TRUE values for u = [3 40 49 53 67 74 85 104 179 190]

% return random number generation
rng shuffle 

[Paola Castiblanco]

This is what I was looking for.Thank you !

[Adam Danz]

Glad I could help!