P1{io,jo} = NumericalFourierForce{io,1}{jo,1}(1:L/2+1);
The assignment on that line would fail if P1 is an existing variable that is not a cell array or table or string array (or possibly some other object types.) In particular it would fail if P1 is an existing numeric array. The assignment would not fail of P1 does not already exist, in which case P1 would be constructed as a cell array. In each case -- cell array, table, string array -- the line would fail if P1 exists already and is not a "container" object.
P1(2:end-1) = 2*P1(2:end-1);
When you use () indexing to access a container object, the result is a container object. If we assume that P1 is a cell array of numeric arrays, then when you index using () you just get a smaller cell array of numeric arrays. But you cannot do arithmetic on cell arrays, only on their contents.
You could replace
P1(2:end-1) = 2*P1(2:end-1);
with
P1(2:end-1) = cellfun(@(C) 2*C, P1(2:end-1), 'uniform', 0);
but I recommend that you reconsider that! You are using linear indexing into a 2D array that appears to be dynamically growing. After you create P1{2,1} then P1 would be 2 x 2 with P1{2,2} not yet initialized, and P1(2:end-1) would be P1(2,1), P1(1,2) so you would double those. Then you create P1{2,2} and P1(2:end-1) would again be P1(2,1) and P1(1,2) and you would again double those.
The doubling of multiple elements should not be in the for loop.
Maybe something like,
numrow = size(NumericalFourierForce,1);
for io = 1 : numrow
for jo =1:1:2
temp = NumericalFourierForce{io,1}{jo,1}(1:L/2+1);
if io ~= 1 && io ~= numrow
temp = temp * 2;
end
P1{io,jo} = temp;
end
end
this does not do exactly the same thing as what you had programmed, but it seems a bit more plausible than what you were doing.
However, if this is what you wanted to program, then...
numrow = size(NumericalFourierForce,1);
for io = 1 : numrow
for jo =1:1:2
P1{io,jo} = NumericalFourierForce{io,1}{jo,1}(1:L/2+1);
end
end
P1(2:end,:) = cellfun(@(C) 2*C, P1(2:end,:), 'uniform', 0);
Best Answer