I have a function replace_me which is defined like as: function w = replace_me(v,a,b,c). The first input argument v is a vector, while a, b, and c are all scalars. The function replaces every element of v that is equal to a with b and c. For example, the command
x = replace_me([1 2 3],2,4,5); returns x as [1 4 5 3].
The code that I have created is
function w = replace_me(v,a,b,c)
[row,column]=size(v);
new_col=column+1;
w=(row:new_col);
for n=(1:column)
if a==v(n)
v(n)=b;
o=n;
d=n-1;
u=n+1;
for z=1:d
w(z)=v(z);
end
for z=u:column
w(z+1)=v(z);
end
w(o)=b;
w(o+1)=c;
end
end
end
It works perfectly fine for x = replace_me([1 2 3],2,4,5); I get required output but when I try x = replace_me([1 2 3], 4, 4, 5) my function fails.
To resolve this problem I want to use an if else statements having conditions that if a is equal to any element of vector v we would follow the above equation else it returns back the vector.
I tried to use this as if condition but it didn't worked
if v(1:column)==a
Any ideas
I'm not entirely sure if I understand what you are trying to achieve, but form what I understand you're looking for something like this:
function [v] = replace_me(v,a,b,c)
v = reshape(v,numel(v),1); % Ensure that v is always a column vector
tol = 0.001;
aPos = find( abs(v-a) < tol ); % Used tol to avoid numerical issues as mentioned by excaza
for i=numel(aPos):-1:1 % Loop backwards since the indices change when inserting elements
index = aPos(i);
v = [v(1:index-1); b; c; v(index+1:end)];
end
end
function w = move_me(v,a)
if nargin == 2
w=v(v~=a);
w(end+1:end+(length(v)-length(w)))=a;
elseif isscalar(v)
w=v;
else
w=v(v~=0);
w(end+1)=0;
end
end
Related
so I'm new to Matlab and had to draw the Impulsefunction with y(n) is only 1 if n == 3, else 0. The following code works:
n = -5:5;
f = n; % allocate f
for i = 1 : length(n)
f(i) = dd1(n(i)-3);
end
stem(n, f);
function y = dd1(n)
y = 0;
if n == 0
y = 1;
end
end
But I feel like it's to complicated, so I tried the following:
n = -5:5
stem(n, fo)
function y = fo(n)
y = 0
if n == 3
y=1
end
end
This returns
Not enough input arguments.
Error in alternative>fo (line 5)
if n == 3
Error in alternative (line 2)
stem(n, fo)
I feel like I'm missing something trivial here.
if is no vector-wise operation but expects a single boolean (or at least a scalar that it can cast to a boolean).
But you can do this vector-wise:
lg = n == 3;
This produces a logical (MATLAB's name for boolean) array (because n is an array and not a vector), which is true where n is equal (==) to three. So you don't need a function, because you can make use of MATLAB's ability to work with vectors and arrays implicitly.
(for your code it would be f = (n-3) == 3)
A last hint: if you have a state-space system (ss-object), you can use the function step to get the step-response as a plot.
So, I'm trying to do the Gauss-Seidel method in Matlab and I found a code that does this but when I apply it to my matrices I get the Subscripted assignment dimension mismatch. error. I will show you my code in order to get a better idea.
%size of the matrix
n = 10;
%my matrices are empty in the beginning because my professor wants to run the algorithm for n = 100
and n = 1000. A's diagonal values are 3 and every other value is -1. b has the constants and the
first and last value will be 2,while every other value will be 1.
A = [];
b = [];
%assign the values to my matrices
for i=1:n
for j=1:n
if i == j
A(i,j) = 3;
else
A(i,j) = -1;
end
end
end
for i=2:n-1
b(i) = 1;
end
%here is the Gauss-Seidel algorithm
idx = 0;
while max(error) > 0.5 * 10^(-4)
idx = idx + 1;
Z = X;
for i = 1:n
j = 1:n; % define an array of the coefficients' elements
j(i) = []; % eliminate the unknow's coefficient from the remaining coefficients
Xtemp = X; % copy the unknows to a new variable
Xtemp(i) = []; % eliminate the unknown under question from the set of values
X(i) = (b(i) - sum(A(i,j) * Xtemp)) / A(i,i);
end
Xsolution(:,idx) = X;
error = abs(X - Z);
end
GaussSeidelTable = [1:idx;Xsolution]'
MaTrIx = [A X b]
I get the error for the Xsolution(:,idx) = X; part. I don't know what else to do. The code posted online works though, and the only difference is that the matrices are hardcoded in the m-file and A is a 5x5 matrix while b is a 5x1 matrix.
I am unable to run your code because some variables are not initialised, at least error and X. I assume the Problem is caused because Xsolution is already initialised from a previous run with a different size. Insert a Xsolution=[] to fix this.
Besides removing the error I have some suggestions to improve your code:
Use Functions, there are no "left over" variables from a previous run, causing errors like you got here.
Don't use the variable name error or i. error is a build-in function to throw errors and i is the imaginary unit. Both can cause hard to debug errors.
Initialise A with A=-1*ones(n,n);A(eye(size(A))==1)=3;, it's faster not to use a for loop in this case. To initialise b you can simply write b(1)=0;b(2:n-1)=1;
Use preallocation
the first time you run the code, Xsolution(:,idx) = X will create a Xsolution with the size of X.
the second time you run it, the existing Xsolution does not fit the size of new X.
this is another reason why you always want to allocate the array before using it.
Hey i have an issuse with plotting my own function in scilab.
I want to plot the following function
function f = test(n)
if n < 0 then
f(n) = 0;
elseif n <= 1 & n >= 0 then
f(n) = sin((%pi * n)/2);
else
f(n) = 1;
end
endfunction
followed by the the console command
x = [-2:0.1:2];
plot(x, test(x));
i loaded the function and get the following error
!--error 21
Invalid Index.
at line 7 of function lala called by :
plot(x, test(x))
Can you please tell me how i can fix this
So i now did it with a for loop. I don't think it is the best solution but i can't get the other ones running atm...
function f = test(n)
f = zeros(size(n));
t = length(n);
for i = 1:t
if n(i) < 0 then
f(i) = 0;
elseif n(i) <= 1 & n(i) >= 0
f(i) = sin((%pi * n(i)/2));
elseif n(i) > 1 then
f(i) = 1;
end
end
endfunction
I guess i need to find a source about this issue and get used with the features and perks matlab/scilab have to over :)
Thanks for the help tho
The original sin is
function f = test(n)
(...)
f(n) = (...)
(...)
endfunction
f is supposed to be the result of the function. Therefore, f(n) is not "the value that the function test takes on argument n", but "the n-th element of f". Scilab then handles this however it can; on your test case, it tries to access a non-integer index, which results in an error. Your loop solution solves the problem.
Replacing all three f(n) by f in your first formulation makes it into something that works... as long as the argument is a scalar (not an array).
If you want test to be able to accept vector arguments without making a loop, the problem is that n < 0 is a vector of the same size as n. My solution would use logical arrays for indexing each of the three conditions:
function f = test(n)
f = zeros(size(n));
negative = (n<0);//parentheses are optional, but I like them for readability
greater_than_1 = (n>1);
others = ~negative & ~greater_than_1;
f(isnegative)=0;
f(greater_than_1)=1;
f(others) = sin(%pi/2*n(others));
endfunction
I have a 30-vector, x where each element of x follows a standardised normal distribution.
So in Matlab,
I have:
for i=1:30;
x(i)=randn;
end;
Now I want to create 30*30=900 elements from vector, x to make a 900-vector, C defined as follows:
I am unable to do the loop for two variables (k and l) properly. I have:
for k=1:30,l=1:30;
C(k,l)=(1/30)*symsum((x(i))*(x(i-abs(k-l))),1,30+abs(k-l));
end
It says '??? Undefined function or method 'symsum' for input arguments of type
'double'.'
I hope to gain from this a 900-vector, C which I will then rewrite as a matrix. The reason I have using two indices k and l instead of one is because I eventually want these indices to denote the (k,l)-entry of such a matrix so it is important that that my 900-vector will be in the form of C = [ row 1 row 2 row 3 ... row 30 ] so I can use the reshape tool i.e.
C'=reshape(C,30,30)
Could anyone help me with the code for the summation and getting such a 900 vector.
Let's try to make this a bit efficient.
n = 30;
x = randn(n,1);
%# preassign C for speed
C = zeros(n);
%# fill only one half of C, since it's symmetric
for k = 2:n
for l = 1:k-1
%# shift the x-vector by |k-l| and sum it up
delta = k-l; %# k is always larger than l
C(k,l) = sum( x(1:end-delta).*x(1+delta:end) );
end
end
%# fill in the other half of C
C = C + C';
%# add the diagonal (where delta is 0, and thus each
%# element of x is multiplied with itself
C(1:n+1:end) = sum(x.^2);
It seems to me that you want a matrix C of 30x30 elements.
Given the formula that you provided I would do
x = randn(1,30)
C = zeros(30,30)
for k=1:30
for l=1:30
v = abs(k-l);
for i =1:30-v
C(k,l) = C(k,l) + x(i)*x(i+v);
end
end
end
if you actually need the vector you can obtain it from the matrix.
Say I have the following basic if-statement:
if (A ~= 0)
% do something like divide your favorite number by A
else
% do something like return NaN or infinity
end
The problem is that A is not a simple number but a vector. Matlab returns true if no element in A is 0. What I am looking for is a vectorized? way of perforimg the if-statement above for each element in A.
Actually, I simply want to do this as fast as possible.
Vectorized ifs don't exist, but there are some options. If you want to test for all or any elements true, use the all or any function.
Here's one example of conditionally modifying values of a matrix:
b = A ~= 0; % b is a boolean matrix pointing to nonzero indices
% (b could be derived from some other condition,
% like b = sin(A)>0
A(b) = f(A(b)) % do something with the indices that pass
A(~b) = g(A(~b)) % do something else with the indices that fail
B = zeros(size(A));
B(A~=0) = FAV./A(A~=0);
B(A==0) = NaN;
In general, to perform one operation on some elements of a matrix and another operation on the remaining elements, a one-line solution is:
Z = B .* X + ~B .* Y;
where B is a logical matrix. As an example,
Z = (A == 0) .* -1 + (A ~= 0) .* A;
copies A but assigns -1 everywhere that A is zero.
However, because the question deals with infinity or NaNs, it can be done even more succinctly:
Z = FAV ./ A; % produces inf where A == 0
Z = (A ~= 0) .* FAV ./ A; % produces NaN where A == 0
Are you looking for all non-zero elements? You can do this a couple of ways.
nonzero = find(A); % returns indicies to all non-zero elements of A
y = x./A(nonzero); % divides x by all non-zero elements of A
% y will be the same size as nonzero
Or for a one-liner, you can use a conditional in place of indicies
y = x./A(A~=0); % divides x by all non-zero elements of A
What you need to do is identify the elements you want to operate on. I would use FIND. I store the results in VI (Valid Indicies) and use that to populate the matrix.
clear
clc
den = [2 0 2; 0 2 0; -2 -2 -2]
num = ones(size(den));
frac = nan(size(den));
vi = (den ~=0)
frac(vi) = num(vi)./den(vi)
vi = (den == 0)
frac(vi) = nan %just for good measure...