# How to fill array with value 0 if function returns empty []

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Sanders A. il 21 Feb 2022
Commentato: Sanders A. il 22 Feb 2022
I am creating a matrix for computing the natural frequency of beams with different boundary conditions. I have my assumed eigenfunction, I solve it at a boundary condition and then I want to collect all the terms attached to coefficients so I can collect thema ll and solve for the determinant which gives the characteristic equation to find the natural frequencies. However, the -C(T==c1)- terms return empty arrays which "disappear" when bringing them all together.
Is there an efficient way to replace these empty values with 0 as I show in the last line of my code?
I have tinkered around with cell arrays but they don't play nice with isempty without making loops.
syms x c1 c2 c3 c4 ki real
phi = c1*cos(ki*x) + c2*cosh(ki*x) + c3*sin(ki*x) + c4*sinh(ki*x);
phi_0 = simplify(subs( phi,x,0));
[C,T] = coeffs( phi_0,[c4 c3 c2 c1],'All'); %why is this backwards?
A = sym(zeros(4,4));
A(1,:) = [C(T==c1),C(T==c2),C(T==c3),C(T==c4)]; % Fails because RHS=[ki, ki] when it 'should' be [0, 0, ki, ki]
Unable to perform assignment because the size of the left side is 1-by-4 and the size of the right side is 1-by-2.

Error in sym/privsubsasgn (line 1229)
L_tilde2 = builtin('subsasgn',L_tilde,struct('type','()','subs',{varargin}),R_tilde);

Error in sym/subsasgn (line 1060)
C = privsubsasgn(L,R,inds{:});
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### Risposta accettata

DGM il 21 Feb 2022
Modificato: DGM il 21 Feb 2022
I hardly ever use symbolic tools, so I'm sure someone knows a better way. Just paying attention to the output alone, this is one way to do it with cell arrays:
syms x c1 c2 c3 c4 ki real
phi = c1*cos(ki*x) + c2*cosh(ki*x) + c3*sin(ki*x) + c4*sinh(ki*x);
phi_0 = simplify(subs( phi,x,0));
[C,T] = coeffs( phi_0,[c4 c3 c2 c1],'All'); %why is this backwards?
% using a cell array
A = cell(4,4);
A(1,:) = {C(T==c1),C(T==c2),C(T==c3),C(T==c4)}
A = 4×4 cell array
{[1 ]} {[1 ]} {0×1 sym } {0×1 sym } {0×0 double} {0×0 double} {0×0 double} {0×0 double} {0×0 double} {0×0 double} {0×0 double} {0×0 double} {0×0 double} {0×0 double} {0×0 double} {0×0 double}
% after whatever loop ostensibly fills the rest of A
mask = cellfun(@isempty,A) % find empties
0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
A(mask) = {sym(0)}; % replace with zero
A = reshape(vertcat(A{:}),4,4) % convert to sym array
A =
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Sanders A. il 22 Feb 2022
That worked wonderfully! Thank you! Now my system is much more robust than it was before as I don't have to manually find out which terms are null and replace them with 0.

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### Più risposte (1)

Bob Thompson il 21 Feb 2022
I admit I don't entirely understand the math here, so there may be a more elegant solution. My MATLAB access is also currently down, so nothing here has been tested.
That said, you could just use some logical indexing to find the empty values and replace them ahead of time.
syms x c1 c2 c3 c4 ki real
phi = c1*cos(ki*x) + c2*cosh(ki*x) + c3*sin(ki*x) + c4*sinh(ki*x);
phi_0 = simplify(subs( phi,x,0));
[C,T] = coeffs( phi_0,[c4 c3 c2 c1],'All'); %why is this backwards?
A = sym(zeros(4,4));
C(isempty(C)&T==c1) = 0; % Replace empty values with 0
A(1,:) = [C(T==c1),C(T==c2),C(T==c3),C(T==c4)]; % Fails because RHS=[ki, ki] when it 'should' be [0, 0, ki, ki]
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Sanders A. il 22 Feb 2022
I am not sure what your second to last line is doing and it does not seem to have quite the desired effect as C is unchanged by the action of that line. The last line continues to fail for the same reason as before RHS is a 1x2.
I truly appreciate the suggestion though!!

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