I have a issue on concat 2 tensor,
say I have x and y:
x = torch.randn(35, 50)
y = torch.randn(35)
How do I concat every y value in to x[0] to make x has a shape 35,51?
I tried:
for i in y:
for a in range(x.shape[0]):
x[a] = torch.cat((x[a],i),0)
Still getting shape error. Any smart way of doing it?
This should work:
z = torch.cat([x,y.reshape(-1,1)], axis=1)
print(z.shape)
Output:
torch.Size([35, 51])
Related
Let's say that I want to solve the following problem.
minimize Tr(CY)
s.t. Y = xxT
x is 0 or 1.
where xxT indicates an outer product of n-1 dimension vector x. C is a n-1 by n-1 square matrix. To convert this problem to a problem with a single matrix variable, I can write down the code as follows by using cvxpy.
import cvxpy as cp
import numpy as np
n = 8
np.random.seed(1)
S = np.zeros(shape=(int(n), int(n)))
S[int(n-1), int(n-1)] = 1
C = np.zeros(shape=(n,n))
C[:n-1, :n-1] = np.random.randn(n-1, n-1)
X = cp.Variable((n,n), PSD=True)
constraints=[]
constraints.append(cp.trace(S # X) == 1)
for i in range(n-1):
Q = np.zeros(shape=(n,n))
Q[i,i] = 1
Q[-1,i] = -0.5
Q[i,-1] = -0.5
const = cp.trace(Q # X) == 0
constraints.append(const)
prob = cp.Problem(cp.Minimize(cp.trace(C # X)),constraints)
prob.solve(solver=cp.MOSEK)
print("X is")
print(X.value)
print("C is")
print(C)
To satisfy the binary constraint that the entries of the vector x should be one or zero, I added some constraints for the matrix variable X.
X = [Y x; xT 1]
Tr(QX) == 0
There are n-1 Q matrices which are forcing the vector x's entries to be 0 or 1.
However, when I ran this simple code, the constraints are violated severely.
Looking forward to see any suggestion or comments on this.
I've been working on numerical methods to solve polynomial and non-polynomial equations. I wanted to use numderivative to calculate the defined derivative of a function entered by the user with the following simple code:
clc
clear
x0 =input('Enter the x value: ') // x0 = 4
function y = f(x)
y = input('Enter your function: ') // y = sqrt(x)
endfunction
dd = numderivative(f,x0)
printf('The definite derivative value of f(x) in x = %d is %.6f',x0,dd)
The output is the following:
Enter the x value: 4
Enter your function: sqrt(x)
Enter your function: sqrt(x)
The definite derivative value of f(x) in x = 4 is 0.250000
This code asks for the function twice. I would like to know how to solve that problem. Thank in advance.
No it is not possible to enter a function, but you can enter the instructions of the function:
x0 =input('Enter the x value: ') // x0 = 4
instr = input('Enter the expression to derivate as a function of x: ',"s")//sqrt(x)
deff("y=f(x)","y="+instr)
dd = numderivative(f,x0)
printf('The definite derivative value of f(x) in x = %d is %.6f',x0,dd)
Im using SymPy in Julia. My purpose is to solve a homogeneous system of linear equations (Ax=0) with more unknowns than variables (A is not square).
Then, Im using the following code.
using SymPy
x, y, z, w = symbols("x y z w")
M = sympy.Matrix(((9, 2, 1,- 4, 0), (-4, -3, -1, -5, 0)))
s = linsolve(M, (x, y, z, w))
With this code Im able to get the correct solution. However, I´dont known how to manipulate that solution.
The final goal is to be able to get the solution in matrix form as lines representing (x and y) and column (z and w). (since x(z, w) and y(z,w)).
Thanks
If numerical (rather than symbolic) computations are acceptable, then this will get the job done:
julia> using LinearAlgebra
julia> M = rand(2,4)
2×4 Array{Float64,2}:
0.497965 0.704514 0.152799 0.69448
0.594486 0.695488 0.327688 0.710573
julia> Q,R = qr(M);
C = -R[1:2,1:2]\R[1:2,3:4]
xy(zw) = C*zw;
# Check that `[xy(zw); zw]` is indeed in the nullspace of `M`:
julia> zw = rand(2)
M*[xy(zw);zw]
2-element Array{Float64,1}:
-2.7755575615628914e-17
-1.1102230246251565e-16
I have the following code:
f=tan(x)*x**2
q=Wild('q')
s=f.match(tan(q))
s={q_ : x}
How to work with the result of the "wild"? How to not address the array, for example, s[0], s{0}?
Wild can be used when you have an expression which is the result of some complicated calculation, but you know it has to be of the form sin(something) times something else. Then s[q] will be the sympy expression for the "something". And s[p] for the "something else". This could be used to investigate both p and q. Or to further work with a simplified version of f, substituting p and q with new variables, especially if p and q would be complex expressions involving multiple variables.
Many more use cases are possible.
Here is an example:
from sympy import *
from sympy.abc import x, y, z
p = Wild('p')
q = Wild('q')
f = tan(x) * x**2
s = f.match(p*tan(q))
print(f'f is the tangent of "{s[q]}" multiplied by "{s[p]}"')
g = f.xreplace({s[q]: y, s[p]:z})
print(f'f rewritten in simplified form as a function of y and z: "{g}"')
h = s[p] * s[q]
print(f'a new function h, combining parts of f: "{h}"')
Output:
f is the tangent of "x" multiplied by "x**2"
f rewritten in simplified form as a function of y and z: "z*tan(y)"
a new function h, combining parts of f: "x**3"
If you're interested in all arguments from tan that appear in f written as a product, you might try:
from sympy import *
from sympy.abc import x
f = tan(x+2)*tan(x*x+1)*7*(x+1)*tan(1/x)
if f.func == Mul:
all_tan_args = [a.args[0] for a in f.args if a.func == tan]
# note: the [0] is needed because args give a tupple of arguments and
# in the case of tan you'ld want the first (there is only one)
elif f.func == tan:
all_tan_args = [f.args[0]]
else:
all_tan_args = []
prod = 1
for a in all_tan_args:
prod *= a
print(f'All the tangent arguments are: {all_tan_args}')
print(f'Their product is: {prod}')
Output:
All the tangent arguments are: [1/x, x**2 + 1, x + 2]
Their product is: (x + 2)*(x**2 + 1)/x
Note that neither method would work for f = tan(x)**2. For that, you'ld need to write another match and decide whether you'ld want to take the same power of the arguments.
I have two numbers X and Y and the following pseudocode:
i = 0
While X < Y:
X = X + complex_formula
i += 1
Print i
complex_formula is independent from the X and its previous value.
So, I was wondering if there is any way to calculate the i without doing the iterations.
Is complex_formula also independent of i and timing? If so then it's a constant and this is just simple math:
i = Ceiling( (Y - X)/complex_formula)
X = X + i*complex_formula