I'm trying to vectorize one function in Matlab, but I have a problem with assigning values.
function [val] = clenshaw(coeffs,x)
b=zeros(1,length(coeffs)+2);
for k=length(coeffs):-1:2
b(k)=coeffs(k)-b(k+2)+2*b(k+1).*x;
end
val=coeffs(1)-b(3)+b(2).*x;
The purpose of this function is to use Clenshaw's algorithm to compute a value of one polynomial with coefficients "coeffs" at point x.
It work fine when x is a single value, but I'd like it to work with vector of arguments too.
When I try to pass a vector I get an error:
Unable to perform assignment because the left
and right sides have a different number of
elements.
Error in clenshaw (line 7)
b(k)=coeffs(k)-b(k+2)+2*b(k+1).*x;
I understand that there is a problem, because I'm trying to assign vector to a scalar variable b(k).
I tried making b a matrix instead of a vector, however I still cannot get the return output I'd like to have which would be a vector of values of this function at points from vector x.
Thank you for helping and sorry if something isn't entirely clear, because English is not my native language.
The vectorized version of your function looks like this:
function [val] = clenshaw(coeffs,x)
b=zeros(length(x),length(coeffs)+2);
for k=length(coeffs):-1:2
b(:,k)=coeffs(k)-b(:,k+2)+2*b(:,k+1).*transpose(x);
end
val=coeffs(1)-b(:,3)+b(:,2).*transpose(x);
end
b needs to be a matrix. In your loop, you have to perform every operation per row of b. So you need to write b(:,k) instead of b(k). Since b(:,k) is a vector and not a scalar, you also have to be careful with the dimensions when using the .* operator. To get the correct results, you need to transpose x. The same goes for the calculation of val. If you don't like the transposition, just swap the rows and cols of b and you get this:
function [val] = clenshaw(coeffs,x)
b=zeros(length(coeffs)+2, length(x));
for k=length(coeffs):-1:2
b(k,:)=coeffs(k)-b(k+2,:)+2*b(k+1,:).*x;
end
val=coeffs(1)-b(3,:)+b(2,:).*x;
end
However, the first version returns a column vector and the second a row vector. So you might need to transpose the result if the vector type is important.
Related
I am new to Matlab and trying to define a simple function but keep running into an error. Details are:
1) V is a 31x1 vector;
2) The function mypi takes one input, which is a scalar (between 0 to 30). It finds the corresponding element in V vector and saves it in z.
3) Matrix A is a row vector with two elements 0 and z-10.
4) y, which is what I am interested in calculating is a linear function of the max of vector A.
Matlab, however, gives an error and is not recognizing element x in vector V. Can anyone please guide me how I should fix this problem? I will be grateful. Thank you.
function y=mypi(x)
z=V(x);
A=[0, z-10];
y=500+50*max(A);
end
you must pass V to mypi or make it visible to this function by defining it as global. But why bother passing both V and index x to this function instead of passing V(x) or z directly?
function y=mypi(z)
A=[0, z-10];
y=500+50*max(A);
end
and call it by mypi(V(x))
I am trying to write a function in MATLAB that takes 1x3 vectors as input. My code looks something like this:
function myFunction=([x1, x2, x3], [y1, y2, y3], [z1, z2, z3])
where all inputs are numbers, and then in the body of the function I perform some calculations indexing through the numerical values in the vectors. i want the vectors to be user input, so the user will enter the vectors and their components (x1, x2, etc.) into the function argument. However, I am getting an error saying "Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters." Therefore I believe I either have the syntax or something else wrong. I know MATLAB is supposed to be able to take vector input in functions, so please let me know what I am doing wrong. Thanks!
What you need to do is declare your function like this:
function myFunction(x,y,z)
% your function code here
end
Then within your function you can access the individual elements of the vectors using x(1), y(2), etc.
To call the function, including whatever number you like, you can enter on the Matlab command window (for example),
myFunction([1 2 3],[4 5 6],[7 8 9]) and the code in your function will be called with the x variable set to the vector [1,2,3], the y variable set to [4,5,6] and z to [7,8,9]. The use of commas to delineate values is optional. If your function then accesses y(2) it will get the second value of the y vector which will be 5 - it is important to note that indexing in Matlab is 1-based so the 1st element of x is obtained with x(1).
If you need to return values you can use:
function [a,b,c] = myFunction(x,y,z)
Then just assign the a, b or c in your code before the end statement.
See the offical Matlab documentation for more info.
I would add that much of the advantage of matlab is dealing with data in a vectorised form, so if you can avoid splitting out into separate elements I would do so. For example, if you need to add two vecors, you could do z = [x(1)+y(1), x(2)+y(2), x(3)+y(3)], but much better (more readable, more maintainable, faster) is z=x+y.
I am trying to replicate this formula:
I have gathered all variables in my workspace. However estimating vec(Theta') does not seem to work and so I am a little bit stuck.
Theta = A*B-C;
vTheta = vec(Theta');
A, B and C are defined. The problem is that MATLAB does not seem to know the function vec to do what I would like to do with Theta as in the formula.
How to fix this?
I don't know where you got that equation from, but vec is a function in R, maybe it's related to that? If you want to convert a matrix Theta into a vector, do
Theta(:)
Edit: If you need to transpose the matrix first, MATLAB might not let you do Theta'(:). Instead do it in two steps:
tmp = Theta'; tmp(:)
As written above the Colon Operator is the way vectorize defined variable.
Yet, sometime we want to vectorize a sub set of a variable.
Let's say we have a matrix - mA and we'd like to vectorize a sub section of it - mA(2:3, 4:7).
One way is to define a new variable and vectorize it:
vA = mA(2:3, 4:7);
vA = vA(:);
Yet, what if we only wanted to use this inside another expression and only once?
Could we escape the need to generate explicit variable?
Well, unfortunately MATLAB doesn't have the view() functionality like in Julia.
Yet if you want to avoid explicitly naming new variable (I'm not sure if MATLAB's JIT Engine can also void the memory allocation as Julia) you can do:
reshape(mA(2:3, 4:7), [], 1)
This will always yield a column vector.
You can also use:
reshape(mA(2:3, 4:7), 1, [])
To generate row vector.
For instance you can do:
reshape(mA(2:3, 4:7), 1, []) * reshape(mA(2:3, 4:7), [], 1, )
This will be the sum of squared values of those elements.
i have two problems in mathematica and want to do them in matlab:
measure := RandomReal[] - 0.5
m = 10000;
data = Table[measure, {m}];
fig1 = ListPlot[data, PlotStyle -> {PointSize[0.015]}]
Histogram[data]
matlab:
measure =# (m) rand(1,m)-0.5
m=10000;
for i=1:m
data(:,i)=measure(:,i);
end
figure(1)
plot(data,'b.','MarkerSize',0.015)
figure(2)
hist(data)
And it gives me :
??? The following error occurred
converting from function_handle to
double: Error using ==> double
If i do :
measure =rand()-0.5
m=10000;
data=rand(1,m)-0.5
then, i get the right results in plot1 but in plot 2 the y=axis is wrong.
Also, if i have this in mathematica :
steps[m_] := Table[2 RandomInteger[] - 1, {m}]
steps[20]
Walk1D[n_] := FoldList[Plus, 0, steps[n]]
LastPoint1D[n_] := Fold[Plus, 0, steps[n]]
ListPlot[Walk1D[10^4]]
I did this :
steps = # (m) 2*randint(1,m,2)-1;
steps(20)
Walk1D =# (n) cumsum(0:steps(n)) --> this is ok i think
LastPointold1D= # (n) cumsum(0:steps(n))
LastPoint1D= # (n) LastPointold1D(end)-->but here i now i must take the last "folding"
Walk1D(10)
LastPoint1D(10000)
plot(Walk1D(10000),'b')
and i get an empty matrix and no plot..
Since #Itamar essentially answered your first question, here is a comment on the second one. You did it almost right. You need to define
Walk1D = # (n) cumsum(steps(n));
since cumsum is a direct analog of FoldList[Plus,0,your-list]. Then, the plot in your code works fine. Also, notice that, either in your Mathematica or Matlab code, it is not necessary to define LastPoint1D separately - in both cases, it is the last point of your generated list (vector) steps.
EDIT:
Expanding a bit on LastPoint1D: my guess is that you want it to be a last point of the walk computed by Walk1D. Therefore, it would IMO make sense to just make it a function of a generated walk (vector), that returns its last point. For example:
lastPoint1D = #(walk) (walk(end));
Then, you use it as:
walk = Walk1D(10000);
lastPoint1D(walk)
HTH
You have a few errors/mistakes translating your code to Matlab:
If I am not wrong, the line data = Table[measure, {m}]; creates m copies of measure, which in your case will create a random vector of size (1,m). If that is true, in Matlab it would simply be data = measure(m);
The function you define gets a single argument m, therefor it makes no sense using a matrix notation (the :) when calling it.
Just as a side-note, if you insert data into a matrix inside a for loop, it will run much faster if you allocate the matrix in advance, otherwise Matlab will re-allocate memory to resize the matrix in each iteration. You do this by data = zeros(1,m);.
What do you mean by "in plot 2 the y=axis is wrong"? What do you expect it to be?
EDIT
Regarding your 2nd question, it would be easier to help you if you describe in words what you want to achieve, rather than trying to read your (error producing) code. One thing which is clearly wrong is using expression like 0:steps(n), since you use m:n with two scalars m and n to produce a vector, but steps(n) produces a vector, not a scalar. You probably get an empty matrix since the first value in the vector returned by steps(n) might be -1, and 0:-1 produces an empty vector.
I want to apply a function to all columns in a matrix with MATLAB. For example, I'd like to be able to call smooth on every column of a matrix, instead of having smooth treat the matrix as a vector (which is the default behaviour if you call smooth(matrix)).
I'm sure there must be a more idiomatic way to do this, but I can't find it, so I've defined a map_column function:
function result = map_column(m, func)
result = m;
for col = 1:size(m,2)
result(:,col) = func(m(:,col));
end
end
which I can call with:
smoothed = map_column(input, #(c) (smooth(c, 9)));
Is there anything wrong with this code? How could I improve it?
The MATLAB "for" statement actually loops over the columns of whatever's supplied - normally, this just results in a sequence of scalars since the vector passed into for (as in your example above) is a row vector. This means that you can rewrite the above code like this:
function result = map_column(m, func)
result = [];
for m_col = m
result = horzcat(result, func(m_col));
end
If func does not return a column vector, then you can add something like
f = func(m_col);
result = horzcat(result, f(:));
to force it into a column.
Your solution is fine.
Note that horizcat exacts a substantial performance penalty for large matrices. It makes the code be O(N^2) instead of O(N). For a 100x10,000 matrix, your implementation takes 2.6s on my machine, the horizcat one takes 64.5s. For a 100x5000 matrix, the horizcat implementation takes 15.7s.
If you wanted, you could generalize your function a little and make it be able to iterate over the final dimension or even over arbitrary dimensions (not just columns).
Maybe you could always transform the matrix with the ' operator and then transform the result back.
smoothed = smooth(input', 9)';
That at least works with the fft function.
A way to cause an implicit loop across the columns of a matrix is to use cellfun. That is, you must first convert the matrix to a cell array, each cell will hold one column. Then call cellfun. For example:
A = randn(10,5);
See that here I've computed the standard deviation for each column.
cellfun(#std,mat2cell(A,size(A,1),ones(1,size(A,2))))
ans =
0.78681 1.1473 0.89789 0.66635 1.3482
Of course, many functions in MATLAB are already set up to work on rows or columns of an array as the user indicates. This is true of std of course, but this is a convenient way to test that cellfun worked successfully.
std(A,[],1)
ans =
0.78681 1.1473 0.89789 0.66635 1.3482
Don't forget to preallocate the result matrix if you are dealing with large matrices. Otherwise your CPU will spend lots of cycles repeatedly re-allocating the matrix every time it adds a new row/column.
If this is a common use-case for your function, it would perhaps be a good idea to make the function iterate through the columns automatically if the input is not a vector.
This doesn't exactly solve your problem but it would simplify the functions' usage. In that case, the output should be a matrix, too.
You can also transform the matrix to one long column by using m(:,:) = m(:). However, it depends on your function if this would make sense.