Insert all in matrix

for all iteration the function ricerca sometimes gives 7 values, sometimes 2

how can i insert all the values in a matrix. It's all ok for the first 8 iterations where the functions ricerca gaves always 7 values. with the 8th iteration where from ricerca I get less values it gives me the error

"Unable to perform assignment because the indices on the left side are not compatible with the size of the right side."

i=1;

for S=0.05:0.05:1
    LIV(i,:)=ricerca(V0,W,S,N)
    i=i+1
end

Answer 1

[Adam Danz]

You can either store the outputs in a cell array like this

i=1;
for S=0.05:0.05:1
    LIV{i}=ricerca(V0,W,S,N)
%      ^   ^ 
    i=i+1
end

or you can pad the shorter vectors with NaN values so they match the correct size

i=1;
for S=0.05:0.05:1
    results = ricerca(V0,W,S,N); 
    results = [results, nan(1,8-numel(results))];   %pad with NaN at the end
    LIV(i,:) = results;
    i=i+1
end

Don't forget to pre-allocate your LIV variable!

[Simo]

with the first method how can I then use the values in LIV matrix? For examples plot the first values or the second (how i write in the code below)

the second method gaves me

Unrecognized function or variable 'resutls'.

Error in Ex2_18_bella (line 45)

LIV(i,:) = resutls;

what do you mean by preallocate?

S=0.05:0.05:1
plot(S,LIV(:,1),S,LIV(:,2))

[Adam Danz]

"with the first method how can I then use the values in LIV matrix"

First, please see a small change I made (I changed LIV{i,:} to LIV{i}).

To access the results from iteration 'n', LIV{n}.

"he second method gaves me Unrecognized function or variable 'resutls'."

That's because of a typo in my answer. I've edited it. I changed "resutls" to "results".

[Simo]

I've also thinked this way, but it says "Dimensions of arrays being concatenated are not consistent."

A=0.05:0.05:1;
for S=0.5:0.05:1
    LIV=ricerca(V0,W,S,N);
    A=[A;LIV];
end

[Adam Danz]

'A' is 1x20 vector. You said the output of ricerca was either 8 or 2 elements long. Matrices need to have the same number of columns across all rows and the same number of rows across all columns.

You could store it in a cell array

i=1;
for S=0.05:0.05:1
    LIV{1,i}=ricerca(V0,W,S,N)
    LIV{2,i} = A;  % <----
    i=i+1
end