I would like to define n-th order tensor X in Julia.
When n is small, we can define i[1]×i[2]×…×[n] tensor as
X = rand(i[1],i[2],i[3])
or
X = Array{Float64, 3}(undef, i[1], i[2], i[3])
These examples are in case of n=3.
But when n is large and list i is given, how can I define high order tensor??
EDIT
I found heuristics answer.
using TensorToolbox
i = [3,4,2,5,4,1,3]
X = diagt(i)
Is this best practice?
Maybe you are looking for this?
rand(i...)
or that?
Array{Float64}(undef, i...)
Related
I got an apparently quite common Julia error when trying to use AD with forward.diff. The error messages vary a bit (sometimes matching function name sometimes Float64)
MethodError: no method matching logL_multinom(::Vector{ForwardDiff.Dual{ForwardDiff.Tag{typeof(logL_multinom), Real}, Real, 7}})
My goal: Transform a probability vector to be unbounded (θ -> y), do some stuff (namely HMC sampling) and transform back to the simplex space whenever the unnormalized posterior (logL_multinom()) is evaluated. DA should be used to overome problems for later, more complex, models than this.
Unfortunately, neither the Julia documentation, not the solutions from other questions helped me figure the particular problem out. Especially, it seems to work when I do the first transformation (y -> z) outside of the function, but the first transformation is a 1-to-1 mapping via logistic and should not cause any harm to differentiation.
Here is an MWE:
using LinearAlgebra
using ForwardDiff
using Base
function logL_multinom(y)
# transform to constrained
K = length(y)+1
k = collect(1:(K-1))
# inverse logit:
z = 1 ./ (1 .+ exp.(-y .- log.(K .- k))) # if this is outside, it works
θ = zeros(eltype(y),K) ; x_cumsum = zeros(eltype(y),K-1)
typeof(θ)
for i in k
x_cumsum[i] = 1-sum(θ)
θ[i] = (x_cumsum[i]) * z[i]
end
θ[K] = x_cumsum[K-1] - θ[K-1]
#log_dens_correction = sum( log(z*(1-z)*x_cumsum) )
dot(colSums, log.(θ))
end
colSums = [835, 52, 1634, 3469, 3053, 2507, 2279, 1115]
y0 = [-0.8904013824298864, -0.8196709647741431, -0.2676845405543302, 0.31688184351556026, -0.870860684394019,0.15187821053559714,0.39888119498547964]
logL_multinom(y0)
∇L = y -> ForwardDiff.gradient(logL_multinom,y)
∇L(y0)
Thanks a lot and especially some further readings/ explanations for the problem are appreciated since I'll be working with it moreoften :D
Edit: I tried to convert the input and any intermediate variable into Real / arrays of these, but nothing helped so far.
I wanted to fit a geometric mapping parameter with some input/output (x,y) points. The model is very simple:
xp = x .+ k.*x.*(x.^2+y.^2)
yp = y .+ k.*y.*(x.^2+y.^2)
k is the only parameter, (x,y) is an input point and (xp,yp) is an output point.
I formulated the input/output data array as:
x = [x for x=-2.:2. for y=-2.:2.]
y = [y for x=-2.:2. for y=-2.:2.]
in_data = [x y]
out_data = [xp yp]
However I'm confused about how to turn this into the LsqFit model, I tried:
k0=[0.]
#. model(x,p) = [x[:,1]+p[1]*x[:,1]*(x[:,1]^2+x[:,2]^2) x[:,2]+p[1]*x[:,2]*(x[:,1]^2+x[:,2]^2)]
ret = curve_fit(model, in_data, out_data, k0)
but got an error:
DimensionMismatch("dimensions must match: a has dims (Base.OneTo(25),
Base.OneTo(2)), must have singleton at dim 2")
So the question is: is it possible to use LsqFit for multi-variate output? (even though this particular problem can be solved analytically)
OK Just figured out the correct way to do this. The vector output variable needs to be stacked together to form a 1D array. So the only changes needed is:
out_data = [xp; yp]
Background
I read here that newton method fails on function x^(1/3) when it's inital step is 1. I am tring to test it in julia jupyter notebook.
I want to print a plot of function x^(1/3)
then I want to run code
f = x->x^(1/3)
D(f) = x->ForwardDiff.derivative(f, float(x))
x = find_zero((f, D(f)),1, Roots.Newton(),verbose=true)
Problem:
How to print chart of function x^(1/3) in range eg.(-1,1)
I tried
f = x->x^(1/3)
plot(f,-1,1)
I got
I changed code to
f = x->(x+0im)^(1/3)
plot(f,-1,1)
I got
I want my plot to look like a plot of x^(1/3) in google
However I can not print more than a half of it
That's because x^(1/3) does not always return a real (as in numbers) result or the real cube root of x. For negative numbers, the exponentiation function with some powers like (1/3 or 1.254 and I suppose all non-integers) will return a Complex. For type-stability requirements in Julia, this operation applied to a negative Real gives a DomainError. This behavior is also noted in Frequently Asked Questions section of Julia manual.
julia> (-1)^(1/3)
ERROR: DomainError with -1.0:
Exponentiation yielding a complex result requires a complex argument.
Replace x^y with (x+0im)^y, Complex(x)^y, or similar.
julia> Complex(-1)^(1/3)
0.5 + 0.8660254037844386im
Note that The behavior of returning a complex number for exponentiation of negative values is not really different than, say, MATLAB's behavior
>>> (-1)^(1/3)
ans =
0.5000 + 0.8660i
What you want, however, is to plot the real cube root.
You can go with
plot(x -> x < 0 ? -(-x)^(1//3) : x^(1//3), -1, 1)
to enforce real cube root or use the built-in cbrt function for that instead.
plot(cbrt, -1, 1)
It also has an alias ∛.
plot(∛, -1, 1)
F(x) is an odd function, you just use [0 1] as input variable.
The plot on [-1 0] is deducted as follow
The code is below
import numpy as np
import matplotlib.pyplot as plt
# Function f
f = lambda x: x**(1/3)
fig, ax = plt.subplots()
x1 = np.linspace(0, 1, num = 100)
x2 = np.linspace(-1, 0, num = 100)
ax.plot(x1, f(x1))
ax.plot(x2, -f(x1[::-1]))
ax.axhline(y=0, color='k')
ax.axvline(x=0, color='k')
plt.show()
Plot
That Google plot makes no sense to me. For x > 0 it's ok, but for negative values of x the correct result is complex, and the Google plot appears to be showing the negative of the absolute value, which is strange.
Below you can see the output from Matlab, which is less fussy about types than Julia. As you can see it does not agree with your plot.
From the plot you can see that positive x values give a real-valued answer, while negative x give a complex-valued answer. The reason Julia errors for negative inputs, is that they are very concerned with type stability. Having the output type of a function depend on the input value would cause a type instability, which harms performance. This is less of a concern for Matlab or Python, etc.
If you want a plot similar the above in Julia, you can define your function like this:
f = x -> sign(x) * abs(complex(x)^(1/3))
Edit: Actually, a better and faster version is
f = x -> sign(x) * abs(x)^(1/3)
Yeah, it looks awkward, but that's because you want a really strange plot, which imho makes no sense for the function x^(1/3).
Let f be a continuous real function defined on the interval [a,b]. I want to aproximate this function by a piecewise quadratic polynomial. I already created a matrix that summarizes these polynomials. Let's say that I'm considering a uniform partition of the interval into N pieces ( therefore N+1 points).
I have a matrix A of size N times 3, where the k row represents the quadratic polynomial associated with the k-interval of this partition in the natural form ( the row [a b c] represents the polynomial a+bx+cx^2). I already created a method to find this matrix (obviously it depends on the choice of my interpolation points inside of each interval but that it doesn't matter for this question).
I'm trying to plot the corresponding function but I'm having some problems. I used the same idea given in Similar question. This is what I wrote
x=zeros(N+1,1);
%this is the set of points defining the uniform partition
for i=1:N+1
x(i)=a+(i-1)*((b-a)/(N));
end
%this is the length of my linspace for plotting the functions
l=100
And now I plot the functions:
figure;
hold on;
%first the original function
u=linspace(a,b,l*N);
v=arrayfun( f , u);
plot(u,v,'b')
% this is for plotting the other functions
for k=1:N
x0=linspace(x(k),x(k+1));
y0=arrayfun(#(t) [1,t,t^2]*A(k,:)',x0);
plot(x0, y0, 'r');
end
The problem is that the for is plotting the same function f and I don't know why. I tried with multiple different functions. I'm pretty sure that my matrix A is correct.
Please write a minimal working example that can be run as standalone code or copy/pasted from people here to check where you might have a bug -- often in the process of reducing your code to its bare principles in this manner, you end up figuring out what is the problem yourself in the first place. But, in any case, I have written one myself and cannot replicate the problem.
figure;
hold on;
# arbitrary values for Minimal Working Example
N = 10;
x = [10:10:110]; # (N+1, 1)
A = randn( N, 3 ); # (3 , N)
a = 100; b = 200; l = 3;
f = #(t) t.^2 .* sin(t);
%first the original function
u = linspace(a,b,l*N);
v = arrayfun( f , u);
plot(u,v,'b')
for k = 1 : N
x0 = linspace( x(k), x(k+1) )
y0 = arrayfun( #(t) ([1, t, t.^2]) * (A(k, :).'), x0 )
x0, y0
plot(x0, y0, 'r');
endfor
hold off;
Output:
Are you doing something different?
I have a scalar function which is obtained by iterative calculations. I wish to differentiate(find the directional derivative) of the values with respect to a matrix elementwise. How should I employ the finite difference approximation in this case. Does diff or gradient help in this case. Note that I only want numerical derivatives.
The typical code that I would work on is:
n=4;
for i=1:n
for x(i)=-2:0.04:4;
for y(i)=-2:0.04:4;
A(:,:,i)=[sin(x(i)), cos(y(i));2sin(x(i)),sin(x(i)+y(i)).^2];
B(:,:,i)=[sin(x(i)), cos(x(i));3sin(y(i)),cos(x(i))];
R(:,:,i)=horzcat(A(:,:,i),B(:,:,i));
L(i)=det(B(:,:,i)'*A(:,:,i)B)(:,:,i));
%how to find gradient of L with respect to x(i), y(i)
grad_L=tr((diff(L)/diff(R)')*(gradient(R))
endfor;
endfor;
endfor;
I know that the last part for grad_L would syntax error saying the dimensions don't match. How do I proceed to solve this. Note that gradient or directional derivative of a scalar functionf of a matrix variable X is given by nabla(f)=trace((partial f/patial(x_{ij})*X_dot where x_{ij} denotes elements of matrix and X_dot denotes gradient of the matrix X
Both your code and explanation are very confusing. You're using an iteration of n = 4, but you don't do anything with your inputs or outputs, and you overwrite everything. So I will ignore the n aspect for now since you don't seem to be making any use of it. Furthermore you have many syntactical mistakes which look more like maths or pseudocode, rather than any attempt to write valid Matlab / Octave.
But, essentially, you seem to be asking, "I have a function which for each (x,y) coordinate on a 2D grid, it calculates a scalar output L(x,y)", where the calculation leading to L involves multiplying two matrices and then getting their determinant. Here's how to produce such an array L:
X = -2 : 0.04 : 4;
Y = -2 : 0.04 : 4;
X_indices = 1 : length(X);
Y_indices = 1 : length(Y);
for Ind_x = X_indices
for Ind_y = Y_indices
x = X(Ind_x); y = Y(Ind_y);
A = [sin(x), cos(y); 2 * sin(x), sin(x+y)^2];
B = [sin(x), cos(x); 3 * sin(y), cos(x) ];
L(Ind_x, Ind_y) = det (B.' * A * B);
end
end
You then want to obtain the gradient of L, which, of course, is a vector output. Now, to obtain this, ignoring the maths you mentioned for a second, if you're basically trying to use the gradient function correctly, then you just use it directly onto L, and specify the grid X Y used for it to specify the spacings between the different elements in L, and collect its output as a two-element array, so that you capture both the x and y vector-components of the gradient:
[gLx, gLy] = gradient(L, X, Y);