Problem with Max function in a if loop

So, basically my max function at the end gives me a random number and does not apply my if and else if constraint :/

for i = 1:B    
JIM{i} = find(REG{1,i} > 0.5);
end
Ans = [JIM;REG;COEFF];%Merge REG and Coeff together so that we can keep an eye on variables name!
for i = 1:B
  for j = 1:C
      D{i} = find(Ans{3,i}.tStat(:).^2 > 3.8416);
  end
end
BLOG = [Ans;D];
for i = 1:B
  SIZE{i} = size(BLOG{4,i});
end
for i = 1:B
  if BLOG{1,i} ==1 
  elseif SIZE{1,i}(1,1) ==test.NumCoefficients
      H = max(BLOG{2,:});
  end
end
end

[Walter Roberson]

There is no such thing as an "if loop". There are "for loop" and "while loop" and "if statement"

[Walter Roberson]

Why would you expect max() of a value you pass in, to apply an "if constraint" ? max() is just going to find the maximum.

Where do you initialize H ? What if elseif does not happen to be true anywhere, then what are you expecting H to be set to afterwards? Are you aware that it looks like you can overwrite H on different iterations of the "for i" loop ?

[Simon]

I wanted the maximum value of the row and didn't know apparently how to set the if statements :/

I'm a noob and i'm just trying to survive my intership

[Walter Roberson]

Please show size(BLOG) class(BLOG{2,1}) size(BLOG{2,1})

Answer 1

[Hadi Hajieghrary]

As I know, the max function is not going to work on a cell data structure. Try to use a matrix. In your case just simply change '{}' to '()'.

[Walter Roberson]

The code has

max(BLOG{2,:})

which uses {} not (). Therefor it is not operating on a cell data structure, not unless BLOG{2,:} stored cell data structures. But BLOG(2,:) appears to be REG and the code has REG{1,i} in the context of a numeric array so BLOG{2,:} should be a list of numeric arrays.

Now, {} expansion of an array is like writing the elements out one by one, so

max(BLOG{2,:})

is like writing

max(BLOG{2,1}, BLOG{2,2}, BLOG{2,3}...)

and so on for however many entries are in BLOG. I'm not sure but I think it might be (B) entries.

max() applied to multiple matrices finds the maximum of each corresponding position, rather than finding the maximum over all of the values. I the maximum over all of the values is desired then the code would have to be

max([BLOG{2,:}])

The code is confusing to read; it seems quite likely to me that it could be written more clearly.

Answer 2

[Simon]

Ok thx guys. I will try to correct my code tommorrow at the office. Thank you very much. If any of you knows some trick to write more clearly pleasy feel free to educate me :) i know i need to learn alot on matlab!

[Walter Roberson]

I can't tell what the algorithm is intended to be.

I can say, though, that it seems confusing to package things up into Ans and BLOG and then pull out specific elements of those that appear to be intended to correspond to the parts that were packaged together. Instead of BLOG{2,:} why not refer to D{:} ?

[Simon]

I thought it would be good to make a matrix with 2 rows of constraints (1,[]) so that after i could just pick the the answer with the highest rsquared that respected the two constraint...

Answer 3

[Simon]

In my 'answer' matrix, row (1,:)is all [] or 1 whether the rsquared is higher than 0,5 or not. The second row (2,:) is the Rsquared of all 15 regressions. The third row (3,:) is the mdl.Coefficients of all 15 regression (including tstats) and the fourth row (4,:) is D{i} = find(Ans{3,i}.tStat(:).^2 > 3.8416, representing the number of coefficients that have a tStat over 1.96 or less than -1.96. I did the .^2 to make it simple (1.96^2 = 3.8416). My plan was to find a way to ONLY get one result :the highest Rsquared IF Rsquared > 0,5 and IF all coeff have tStats^2 over 3.8416.... That sound like incredibly hard to me XD

[Jan]

@Simon: Please post only answers in the answer section. Larger threads can be understood easier, if the standard structure is applied using the question, comments and answers specifically. Thanks.