MATLAB: Looking for a command

How can I get from A=[21 3 NaN NaN] to A=[21 3 0 0] without using a for cycle?

Best Answer

A(isnan(A)) = 0;

Related Solutions

MATLAB: How to force nansum(NaN+ NaN) = NaN, not 0

Just keep track of where all elements are NaN, and replace their sum with NaN after using nansum. You should be able to adapt this example to help you, it sums along each column:

>> A = [1,2,NaN,NaN;3,NaN,4,NaN]
A =
     1     2   NaN   NaN
     3   NaN     4   NaN
>> B = [6,7,8,NaN;9,NaN,NaN,NaN]
B =
     6     7     8   NaN
     9   NaN   NaN   NaN
>> C = nansum(A+B,1)
C =
    19     9     0     0
>> C(all(isnan(A)&isnan(B),1)) = NaN
C =
    19     9     0   NaN

MATLAB: Solve command

Seems to be the same problem as here: http://www.mathworks.de/matlabcentral/answers/40281-substitution-after-differentiation-easy

Try this:

a = double(subs(a)); b = double(subs(b)); c = double(subs(c));

Like here:

[x,y] = meshgrid(-3:1:3);
z = 1.5*x - 3;
mesh(x,y,z);
p = -1.5*pi;
q = -pi/2;
r = 0;
for i = 1:1:10
  pause(0.1);
  p = p + (0.0173*5);
  q = q + (0.0173*5);
  r = r + (0.0173*5);
  a1 = 3*cos(p); b1 = 3*sin(p); c1 = -3;
  a2 = 3*cos(q); b2 = 3*sin(q); c2 = -3;
  a3 = 2*cos(r); b3 = 2*sin(r); c3 = 0;
  syms a b c d 
  eq1 = 'a1*a + b1*b +c1*c -d =0';
  eq2 = 'a2*a + b2*b + c2*c -d=0';
  eq3 = 'a3*a + b3*b +c3*c -d =0';
  [a,b,c] = solve(eq1,eq2,eq3,'a,b,c');
  a = a/d;b = b/d; c = c/d;
  a = double(subs(a)); b = double(subs(b)); c = double(subs(c));
  disp(a);disp(b);disp(c);
  z = -(a*x + b*y)/c;
  mesh(x,y,z);
end

Please note that I renamed one of your b1 to b2.

Best Answer

Related Solutions

MATLAB: How to force nansum(NaN+ NaN) = NaN, not 0

MATLAB: Solve command

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