I want to run the script in DolphinDB:
T = 1..25
X = 2..26
Y1 = 3..27
Y2 = 4..28
y_matrix = matrix(table(Y1 as y1, Y2 as y2))
y_matrix.rename!(1..25,`y1`y2).setIndexedMatrix!()
x_series = indexedSeries(1..25,X)
tmbeta(T=1..25, Y=y_matrix[0], X=x_series[0],window=10)
It returns an error: tmbeta(1 .. 25, y_matrix[0], x_series[0], 10) => Usage: tmbeta(T, X, Y, window). X must be a vector with the same length as T.
I’d like to calculate the coefficient estimate of an ordinary-least-squares regression of Y on X. What is the correct usage of tmbeta?
For function tmbeta, the parameter X (y_matrix[0] in the script) must be a vector. The error is caused by the incorrect data type of y_matrix[0].
typestr(y_matrix[0])
FAST INT MATRIX
As the y_matrix is an indexed matrix in DolphinDB, you can use mbeta instead of tmbeta as no index (T) is needed.
mbeta(X=x_series,Y=y_matrix,window=10)
The correct way to apply function tmbeta is:
y1_matrix=matrix(table(Y1 as y1, Y2 as y2))
tmbeta(T,X=x_series[0],Y=y1_matrix[0],window=10)
Here y1_matrix[0] is a vector:
typestr(y1_matrix[0])
FAST INT VECTOR
Related
I am trying to implement the secant method using Octave, and in particular I made a function called secant("function","x0","x1","tolerance"). I used the while-loop for calculate the root of a function and I think the problem is in the loop.
But the result that I get, it is not what I expect. The correct answer is x=0,49438.
My code is the following:
tol = 10.^(-5); # tolerance
x0 = 0.4; # initial point
x1 = 0.6; # initial point
syms x;
fun = #(x) exp(sin(x)) - 2/(1+x.^2); # f(x) = exp(sin(x)) - 2/(1+x^2)
fprintf("Function f(x)\n");
fun
fprintf("\n");
fprintf("initial points: %f\n",x0,x1);
function[raiz,errel,niter] = secant(fun,x0,x1,tol)
niter = 0; # steps number
errel = []; # the vector of relative errors
y0 = fun(x0); # f(x0)
y1 = fun(x1); # f(x1)
ra = 0.0; # ra is the variable of the function's root
while abs(ra-x1)>= tol
niter += 1;
ra = x1 - ((x1-x0)./(y1-y0)).*y0; # formula of the secant method
if abs((ra-x1))<tol
raiz = ra;
return;
endif
x0 = x1; y0 = y1; x1 = ra;
y1 = fun(ra);
errel(niter) = abs(ra-x1)/abs(ra); # Calcule of relative error
endwhile
if abs((ra-x1)/ra)<tol
fprintf ('The method is over\n');
fprintf ('\n');
endif
raiz = ra;
endfunction
[r,er,tot] = secant(fun,x0,x1,tol)
I appreciate the help you can give me
It makes little sense to use the secant root in the loop condition. At first it is not defined, and inside the loop it is shifted into the support points for the next secant. Note that at the end ra and x1 contain the same value, making the loop condition trivial, in a wrong way.
Next, the secant has the equation
y(x) = y1 + (y1-y0)/(x1-x0)*(x-x_1)
check that with this formula y(x0)=y0 and y(x1)=y1. Thus the secant root is to be found at
x = x1 - (x1-x0)/(y1-y0)*y1
Finally, at no point are symbolic expressions used, defining x as symbolic variable is superfluous.
The break-out test inside the loop is also redundant and prevents a coherent state after the loop. Just remove it and remember that x1 contains the last approximation.
With all this I get an execution log as follows:
Function f(x)
fun =
#(x) exp (sin (x)) - 2 / (1 + x .^ 2)
initial points: 0.400000
initial points: 0.600000
The method is over
r = 0.494379048216965
er =
2.182723270633349e-01 3.747373180587413e-03 5.220701832080676e-05 1.899377363987941e-08
tot = 4
I'm trying to wrap Octave interpolation function in a function body,
function FUN = inter(p);
FUN = interpn (x1, x2, x3, x4, x5, A, p(1), p(2), p(3), p(4), p(5), "spline");
end
The reason why I'm doing this is because I'm using a package which function needs a string name function which would be in this case packageFunction("inter", argument1);
The issue is calling now for instance like,
disp("value = "), inter([10 2 4 3 4])
doesn't work; Doesn't see the vectors error: 'x1' undefined ,
Of course the vectors xi and matrix A are defined above the function body. Would appreciate advice on this,
thanks, Damir
------------- in file example1.m
[a b c] = fminuit('gaussian','mnplot',[10 166 33],[x;y;dy])
------------- in file gaussian.m
function f = gaussian(par,data);
%theoretical function
f = par(1)/(sqrt(2*pi)*par(3)) * exp(-.5*((data(1,:)-
par(2))./par(3)).^2);
if (size(data,1)==2), %chi-square, error = 1
f = sum((data(2,:) - f).^2);
elseif (size(data,1)>2), %chi-square, error = 3rd row of data
f = sum(((data(2,:) - f)./data(3,:)).^2);
end
Given you are using an old function that requires a string as the function, the first solution below will not work. This is, however, the right way to do it. Changing the old function to use function handles instead of strings would be my preferred solution. However, you can also use an alternative solution further down below, which uses global variables. This is not the recommended approach (we should strive to avoid globals), but will solve your near-term problems.
Correct approach: use an anonymous function
You should use an anonymous function, these can capture variables when they're defined:
inter = #(p)interpn (x1, x2, x3, x4, x5, A, p(1), p(2), p(3), p(4), p(5), "spline");
Now inter(p) works just as if inter had been declared as a normal function. But the values of x1, x2, etc as they were defined when inter was defined will be stored inside inter.
As stated, the function you pass inter to must be written to accept function handles.
Bad, quick solution: use global variables
First, create a file inter.m with the following contents:
function FUN = inter(p);
global x1 x2 x3 x4 x5 A
FUN = interpn (x1, x2, x3, x4, x5, A, p(1), p(2), p(3), p(4), p(5), "spline");
end
Next, in your function of script that calls inter, again declare the global variables (currently MATLAB warns that you should declare them as globals before giving them a value, in future versions this will be required):
global x1 x2 x3 x4 x5 A
x1 = ...
x2 = ...
% etc
inter([10 2 4 3 4])
% or:
fminuit('inter',...)
I need help to know a function can extract residual from regression equation.
I need that function to make 2-stage credit model. I want to extract a residual from first stage model(regression) and apply the residual to second stage model(y value).
It will be very helpful if there is proper function in SAS 9.4.
thank you
Look at the documentation around PROC REG.
proc reg data=inData;
model y = x1 x2 x3;
output out=ouData r=resid;
run;
quit;
This takes data from the INDATA data set, regresses Y on X1, X2, and X3, and outputs the residuals in OUTDATA.
If you want to get fancy, you can do it 2-stage least squares with proc model.
proc model data=have;
exo x1 x2 x3;
endo y1 y2;
y1 = b1 + b2*y2 + b3*x1 + b4*x2;
y2 = b5 + b6*y1 + b7*x3;
fit y1 y2 / 2sls;
instruments _exog_;
run;
I can't make matrices with variables in it for some reason. I get following message.
>>> A= [a b ;(-1-a) (1-b); (1+a) b]
error: horizontal dimensions mismatch (2x3 vs 1x1)
Why is it? Please show me correct way if I'm wrong.
In Matlab you first need to assign a variable before you can use it,
a = 1;
b = a+1;
This will thus give an error,
clear;
b = a+1; % ERROR! Undefined function or variable 'a
Matlab does never accept unassigned variables. This is because, on the lowest level, you do not have a. You will have machine code which is assgined the value of a. This is handled by the JIT compiler in Matlab, so you do not need to worry about this though.
If you want to use something as the variable which you have in maths you can specifically express this to matlab. The object is called a sym and the syntax that define the sym x to a variable xis,
syms x;
That said, you can define a vector or a matrix as,
syms a b x y; % Assign the syms
A = [x y]; % Vector
B = A= [a b ;(-1-a) (1-b); (1+a) b]; % Matrix.
The size of a matrix can be found with size(M) or for dim n size(M,n). You can calcuate the matrix product M3=M1*M2 if and only if M1 have the size m * n and M2 have the size n * p. The size of M3 will then be m * p. This will also mean that the operation A^N = A * A * ... is only allowed when m=n so to say, the matrix is square. This can be verified in matlab by the comparison,
syms a b
A = [a,1;56,b]
if size(A,1) == size(A,2)
disp(['A is a square matrix of size ', num2str(size(A,1)]);
else
disp('A is not square');
end
These are the basic rules for assigning variables in Matlab as well as for matrix multiplication. Further, a google search on the error error: 'x' undefined does only give me octave hits. Are you using octave? In that case I cannot guarantee that you can use sym objects or that the syntaxes are correct.
I am having a problem graphing a 3d function - when I enter data, I get a linear graph and the values don't add up if I perform the calculations by hand. I believe the problem is related to using matrices.
INITIAL_VALUE=999999;
INTEREST_RATE=0.1;
MONTHLY_INTEREST_RATE=INTEREST_RATE/12;
# ranges
down_payment=0.2*INITIAL_VALUE:0.1*INITIAL_VALUE:INITIAL_VALUE;
term=180:22.5:360;
[down_paymentn, termn] = meshgrid(down_payment, term);
# functions
principal=INITIAL_VALUE - down_payment;
figure(1);
plot(principal);
grid;
title("Principal (down payment)");
xlabel("down payment $");
ylabel("principal $ (amount borrowed)");
monthly_payment = (MONTHLY_INTEREST_RATE*(INITIAL_VALUE - down_paymentn))/(1 - (1 + MONTHLY_INTEREST_RATE)^-termn);
figure(2);
mesh(down_paymentn, termn, monthly_payment);
title("monthly payment (principal(down payment)) / term months");
xlabel("principal");
ylabel("term (months)");
zlabel("monthly payment");
The 2nd figure like I said doesn't plot like I expect. How can I change my formula for it to render properly?
I tried your script, and got the following error:
error: octave_base_value::array_value(): wrong type argument `complex matrix'
...
Your monthly_payment is a complex matrix (and it shouldn't be).
I guess the problem is the power operator ^. You should be using .^ for element-by-element operations.
From the documentation:
x ^ y
x ** y
Power operator. If x and y are both scalars, this operator returns x raised to the power y. If x is a scalar and y is a square matrix, the result is computed using an eigenvalue expansion. If x is a square matrix. the result is computed by repeated multiplication if y is an integer, and by an eigenvalue expansion if y is not an integer. An error results if both x and y are matrices.
The implementation of this operator needs to be improved.
x .^ y
x .** y
Element by element power operator. If both operands are matrices, the number of rows and columns must both agree.