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Mixed partial dervative w.r.t. tensor in Pytorch

Question:
Is there any working method to calculate gradient of (non-scalar) tensor function?
Example
Given n by n symmetric matrices X, Y and matrix function Z(X, Y) = torch.mm(X.mm(X), Y) calculate d(dZ/dX)/dY.
Expected answer
d(dZ/dX)/dY = d(2*XY)/dY = 2*X
Attempts
Because torch's .backward() works only for scalar variables I've tried to calculate derivative by applying torch.autograd.grad() to each element of tensor Z, but this approach is not correct, because it gives d(X^2)/dX = X + 2*D where D is a diagonal matrix with diagonal values of X. For me it's a bit weird that torch has an ability to build a computational graph, but can't track tensor through it as a variable to get tensor derivative.
Edit
Question was not very clear, so I decided to give more details.
My aim is to get partial derivative of loss function, which involves two matrices as variables. It looks like that:
loss = torch.linalg.norm(my_formula(X, Y) , ord='fro')
And I need to find
d^2(loss)/d(Y^2)
d/dX[d(loss)/dY]
Torch is capable of calculating 1. by using .backward() two times, but it's problematic to find 2. because torch.autograd.grad() expects scalar input and not the tensor

TL;DR
For function f which takes a matrix and gives a scalar:
Find first order derivative, let's name it dX
Take trace: Tr(dX)
To get mixed partial derivative just use the trace from above: d/dY[df/dX] = d/dY[Tr(df/dX)]
Intro
At the moment of posting the question I was not really that good at theory of matrix derivatives, but now I know much more all thanks to this Yandex ml book (unfortunately, I didn't find the english equivalent). This is an attempt to give a full answer to my question.
Basic Theory
Forgive me, Lord, for ugly representation of latex
Let's say you have a function which takes matrix X and returns it's squared Frobenius norm: f(X) = ||X||_F^2
It is a well-known fact that: ||X||_F^2 = Tr(X X^T)
Let's define derivative as shown in same book: D[f] at X_0 = f(X + H) - f(X)
We are ready to find dg(X)/dX:
df(X)/dX = dTr(X X^T)/dX =
(using Trace's feature)
= Tr(d/dX[X X^T]) = Tr(dX/dX X^T + X d[X^T]/dX ) =
(then we should use the definition of derivative from above)
= Tr(HX^T + XH^T) = Tr(HX^T) + Tr(XH^T) =
(now the main trick is to get all matrices H on the right side and get something like
Tr(g(X) H) or Tr(g(X) H^T), where g(X) will be the derivative we are looking for)
= Tr(HX^T) + Tr(XH^T) = Tr(XH^T) + Tr(XH^T) = Tr(2*XH^T)
That means: df(X)/dX = 2X
Second order derivative
Now, after we found out how to get matrix derivatives, let's try to find second order derivative of the same function f(X):
d/dX[df(X)/dX] = d/dX[Tr(2XH_1^T)] = Tr(d/dX[2XH_1^T]) =
= Tr(2I H_2 H_1^T)
We found out that d/dX[df(X)/dX] = 2I where I stands for Identity matrix. But how will it help us to find derivatives in Pytorch?
Trace is the trick
As we can see from the formulas, both first and second order derivatives have Trace inside them, but when we take first order derivative we just instantly get matrix as a result. To get a higher order derivative we just need to take the derivative of trace of first order derivative:
d/dY[df/dX] = d/dY[Tr(df/dX)]
The thing is I was using JAX autograd library when this trick came to my mind, so the code with a function f(X,Y) will look like this:
def scalarized_dy(X, Y):
dY = grad(f, argnums=1)(X, Y)
return jnp.trace(dY)
dYX = grad(scalarized_dy, argnums=0)(X, Y)
dYY = grad(scalarized_dy, argnums=1)(X, Y)
In case of Pytorch I guess we will need to look after tensors' gradients (let loss be a function with X and Y as arguments):
loss = f(X, Y)
loss.backward(create_graph = True)
dX = torch.trace(X.grad)
dX.backward()
dXX = X.grad
dXY = Y.grad
Epilogue
I thought that the question itself is in some way interesting. Also, it took me several months to figure things out, so I decided to give my current point of view on this problem. I will not mark my answer as correct yet in hope that I will get some kind of feedback or, perhaps, even better answers or ideas.

How to implement the following formula for derivatives in python?

I'm trying to implement the following formula in python for X and Y points
I have tried following approach
def f(c):
"""This function computes the curvature of the leaf."""
tt = c
n = (tt[0]*tt[3] - tt[1]*tt[2])
d = (tt[0]**2 + tt[1]**2)
k = n/d
R = 1/k # Radius of Curvature
return R
There is something incorrect as it is not giving me correct result. I think I'm making some mistake while computing derivatives in first two lines. How can I fix that?
Here are some of the points which are in a data frame:
pts = pd.DataFrame({'x': x, 'y': y})
x y
0.089631 97.710199
0.089831 97.904541
0.090030 98.099313
0.090229 98.294513
0.090428 98.490142
0.090627 98.686200
0.090827 98.882687
0.091026 99.079602
0.091225 99.276947
0.091424 99.474720
0.091623 99.672922
0.091822 99.871553
0.092022 100.070613
0.092221 100.270102
0.092420 100.470020
0.092619 100.670366
0.092818 100.871142
0.093017 101.072346
0.093217 101.273979
0.093416 101.476041
0.093615 101.678532
0.093814 101.881451
0.094013 102.084800
0.094213 102.288577
pts_x = np.gradient(x_c, t) # first derivatives
pts_y = np.gradient(y_c, t)
pts_xx = np.gradient(pts_x, t) # second derivatives
pts_yy = np.gradient(pts_y, t)
After getting the derivatives I am putting the derivatives x_prim, x_prim_prim, y_prim, y_prim_prim in another dataframe using the following code:
d = pd.DataFrame({'x_prim': pts_x, 'y_prim': pts_y, 'x_prim_prim': pts_xx, 'y_prim_prim':pts_yy})
after having everything in the data frame I am calling function for each row of the data frame to get curvature at that point using following code:
# Getting the curvature at each point
for i in range(len(d)):
temp = d.iloc[i]
c_temp = f(temp)
curv.append(c_temp)

You do not specify exactly what the structure of the parameter pts is. But it seems that it is a two-dimensional array where each row has two values x and y and the rows are the points in your curve. That itself is problematic, since the documentation is not quite clear on what exactly is returned in such a case.
But you clearly are not getting the derivatives of x or y. If you supply only one array to np.gradient then numpy assumes that the points are evenly spaced with a distance of one. But that is probably not the case. The meaning of x' in your formula is the derivative of x with respect to t, the parameter variable for the curve (which is separate from the parameters to the computer functions). But you never supply the values of t to numpy. The values of t must be the second parameter passed to the gradient function.
So to get your derivatives, split the x, y, and t values into separate one-dimensional arrays--lets call them x and y and t. Then get your first and second derivatives with
pts_x = np.gradient(x, t) # first derivatives
pts_y = np.gradient(y, t)
pts_xx = np.gradient(pts_x, t) # second derivatives
pts_yy = np.gradient(pts_y, t)
Then continue from there. You no longer need the t values to calculate the curvatures, which is the point of the formula you are using. Note that gradient is not really designed to calculate the second derivatives, and it absolutely should not be used to calculate third or higher-order derivatives. More complex formulas are needed for those. Numpy's gradient uses "second order accurate central differences" which are pretty good for the first derivative, poor for the second derivative, and worthless for higher-order derivatives.

I think your problem is that x and y are arrays of double values.
The array x is the independent variable; I'd expect it to be sorted into ascending order. If I evaluate y[i], I expect to get the value of the curve at x[i].
When you call that numpy function you get an array of derivative values that are the same shape as the (x, y) arrays. If there are n pairs from (x, y), then
y'[i] gives the value of the first derivative of y w.r.t. x at x[i];
y''[i] gives the value of the second derivative of y w.r.t. x at x[i].
The curvature k will also be an array with n points:
k[i] = abs(x'[i]*y''[i] -y'[i]*x''[i])/(x'[i]**2 + y'[i]**2)**1.5
Think of x and y as both being functions of a parameter t. x' = dx/dt, etc. This means curvature k is also a function of that parameter t.
I like to have a well understood closed form solution available when I program a solution.
y(x) = sin(x) for 0 <= x <= pi
y'(x) = cos(x)
y''(x) = -sin(x)
k = sin(x)/(1+(cos(x))**2)**1.5
Now you have a nice formula for curvature as a function of x.
If you want to parameterize it, use
x(t) = pi*t for 0 <= t <= 1
x'(t) = pi
x''(t) = 0
See if you can plot those and make your Python solution match it.

Derivatives blow up in python

I am trying to find higher order derivatives of a dataset (x,y). x and y are 1D arrays of length N.
Let's say I generate them as :
xder0=np.linspace(0,10,1000)
yder0=np.sin(xder0)
I define the derivative function which takes in 2 array (x,y) and returns (x1, y1) where y1 is the derivative calculated at each index as : (y[i+1]-y[i])/(x[i+1]-x[i]). x1 is just the mean of x[i+1] and x[i]
Here is the function that does it:
def deriv(x,y):
delx =np.zeros((len(x)-1), dtype=np.longdouble)
ydiff=np.zeros((len(x)-1), dtype=np.longdouble)
for i in range(len(x)-1):
delx[i] =(x[i+1]+x[i])/2.0
ydiff[i] =(y[i+1]-y[i])/(x[i+1]-x[i])
return delx, ydiff
Now to calculate the first derivative, I call this function as:
xder1, yder1 = deriv(xder0, yder0)
Similarly for second derivative, I call this function giving first derivatives as input:
xder2, yder2 = deriv(xder1, yder1)
And it goes on:
xder3, yder3 = deriv(xder2, yder2)
xder4, yder4 = deriv(xder3, yder3)
xder5, yder5 = deriv(xder4, yder4)
xder6, yder6 = deriv(xder5, yder5)
xder7, yder7 = deriv(xder6, yder6)
xder8, yder8 = deriv(xder7, yder7)
xder9, yder9 = deriv(xder8, yder8)
Something peculiar happens after I reach order 7. The 7th order becomes very noisy! Earlier derivatives are all either sine or cos functions as expected. However 7th order is a noisy sine. And hence all derivatives after that blow up.
Any idea what is going on?

This is a well known stability issue with numerical interpolation using equally-spaced points. Read the answers at http://math.stackexchange.com.
To overcome this problem you have to use non-equally-spaced points, like the roots of Lagendre polynomial. The instability occurs due to the unavailability of information at the boundaries, thus more concentration of points at the boundaries is required, as per the roots of say Lagendre polynomials or others with similar properties such as Chebyshev polynomial.

Find intersection of A(x) and B(y) in complex plane plus corr. x and y

suppose I have the following Problem:
I have a complex function A(x) and a complex function B(y). I know these functions cross in the complex plane. I would like to find out the corresponding x and y of this intersection point, numerically ( and/or graphically). What is the most clever way of doing that?
This is my starting point:
import matplotlib.pyplot as plt
import numpy as np
from numpy import sqrt, pi
x = np.linspace(1, 10, 10000)
y = np.linspace(1, 60, 10000)
def A_(x):
return -1/( 8/(pi*x)*sqrt(1-(1/x)**2) - 1j*(8/(pi*x**2)) )
A = np.vectorize(A_)
def B_(y):
return 3/(1j*y*(1+1j*y))
B = np.vectorize(B_)
real_A = np.real(A(x))
imag_A = np.imag(A(x))
real_B = np.real(B(y))
imag_B = np.imag(B(y))
plt.plot(real_A, imag_A, color='blue')
plt.plot(real_B, imag_B, color='red')
plt.show()
I don't have to plot it necessarily. I just need x_intersection and y_intersection (with some error that depends on x and y).
Thanks a lot in advance!
EDIT:
I should have used different variable names. To clarify what i need:
x and y are numpy arrays and i need the index of the intersection point of each array plus the corresponding x and y value (which again is not the intersection point itself, but some value of the arrays x and y ).

Here I find the minimum of the distance between the two curves. Also, I cleaned up your code a bit (eg, vectorize wasn't doing anything useful).
import matplotlib.pyplot as plt
import numpy as np
from numpy import sqrt, pi
from scipy import optimize
def A(x):
return -1/( 8/(pi*x)*sqrt(1-(1/x)**2) - 1j*(8/(pi*x**2)) )
def B(y):
return 3/(1j*y*(1+1j*y))
# The next three lines find the intersection
def dist(x):
return abs(A(x[0])-B(x[1]))
sln = optimize.minimize(dist, [1, 1])
# plotting everything....
a0, b0 = A(sln.x[0]), B(sln.x[1])
x = np.linspace(1, 10, 10000)
y = np.linspace(1, 60, 10000)
a, b = A(x), B(y)
plt.plot(a.real, a.imag, color='blue')
plt.plot(b.real, b.imag, color='red')
plt.plot(a0.real, a0.imag, "ob")
plt.plot(b0.real, b0.imag, "xr")
plt.show()
The specific x and y values at the intersection point are sln.x[0] and sln.x[1], since A(sln.x[0])=B(sln.x[1]). If you need the index, as you also mention in your edit, I'd use, for example, numpy.searchsorted(x, sln.x[0]), to find where the values from the fit would insert into your x and y arrays.
I think what's a bit tricky with this problem is that the space for graphing where the intersection is (ie, the complex plane) does not show the input space, but one has to optimize over the input space. It's useful for visualizing the solution, then, to plot the distance between the curves over the input space. That can be done like this:
data = dist((X, Y))
fig, ax = plt.subplots()
im = ax.imshow(data, cmap=plt.cm.afmhot, interpolation='none',
extent=[min(x), max(x), min(y), max(y)], origin="lower")
cbar = fig.colorbar(im)
plt.plot(sln.x[0], sln.x[1], "xw")
plt.title("abs(A(x)-B(y))")
From this it seems much more clear how optimize.minimum is working -- it just rolls down the slope to find the minimum distance, which is zero in this case. But still, there's no obvious single visualization that one can use to see the whole problem.
For other intersections one has to dig a bit more. That is, #emma asked about other roots in the comments, and there I mentioned that there's no generally reliable way to find all roots to arbitrary equations, but here's how I'd go about looking for other roots. Here I won't lay out the complete program, but just list the changes and plots as I go along.
First, it's obvious that for the domain shown in my first plot that there's only one intersection, and that there are no intersection in the region to the left. The only place there could be another intersection is to the right, but for that I'll need to allow the sqrt in the def of B to get a negative argument without throwing an exception. An easy way to do this is to add 0j to the argument of the sqrt, like this, sqrt(1+0j-(1/x)**2). Then the plot with the intersection becomes
I plotted this over a broader range (x=np.linspace(-10, 10, 10000) and y=np.linspace(-400, 400, 10000)) and the above is the zoom of the only place where anything interesting is going on. This shows the intersection found above, plus the point where it looks like the two curves might touch (where the red curve, B, comes to a point nearly meeting the blue curve A going upward), so that's the new interesting thing, and the thing I'll look for.
A bit of playing around with limits, etc, show that B is coming to a point asymptotically, and the equation of B is obvious that it will go to 0 + 0j for large +/- y, so that's about all there is to say for B.
It's difficult to understand A from the above plot, so I'll look at the real and imaginary parts independently:
So it's not a crazy looking function, and the jumping between Re=const and Im=const is just the nature of sqrt(1-x-2), which is pure complex for abs(x)<1 and pure real for abs(x)>1.
It's pretty clear now that the other time the curves are equal is at y= +/-inf and x=0. And, quick look at the equations show that A(0)=0+0j and B(+/- inf)=0+0j, so this is another intersection point (though since it occurs at B(+/- inf), it's sort-of ambiguous on whether or not it would be called an intersection).
So that's about it. One other point to mention is that if these didn't have such an easy analytic solution, like it wasn't clear what B was at inf, etc, one could also graph/minimize, etc, by looking at B(1/y), and then go from there, using the same tools as above to deal with the infinity. So using:
def dist2(x):
return abs(A(x[0])-B(1./x[1]))
Where the min on the right is the one initially found, and the zero, now at x=-0 and 1./y=0 is the other one (which, again, isn't interesting enough to apply an optimizer here, but it could be interesting in other equations).
Of course, it's also possible to estimate this by just finding the minimum of the data that goes into the above graph, like this:
X, Y = np.meshgrid(x, y)
data = dist((X, Y))
r = np.unravel_index(data.argmin(), data.shape)
print x[r[1]], y[r[0]]
# 2.06306306306 1.8008008008 # min approach gave 2.05973231 1.80069353
But this is only approximate (to the resolution of data) and involved many more calculations (1M compared to a few hundred). I only post this because I think it might be what the OP originally had in mind.

Briefly, two analytic solutions are derived for the roots of the problem. The first solution removes the parametric representation of x and solves for the roots directly in the (u, v) plane, where for example A(x): u(x) + i v(y) gives v(u) = f(u). The second solution uses a polar representation, e.g. A(x) is given by r(x) exp(i theta(x)), and offers a better understanding of the behavior of the square root as x passes through unity towards zero. Possible solutions occurring at the singular points are explored. Finally, a bisection root finding algorithm is constructed as a Python iterator to invert certain solutions. Summarizing, the one real root can be found as a solution to either of the following equations:
and gives:
x0 = -2.059732
y0 = +1.800694
A(x0) = B(y0) = (-0.707131, -i 0.392670)
As in most problems there are a number of ways to proceed. One can use a "black box" and hopefully find the root they are looking for. Sometimes an answer is all that is desired, and with a little understanding of the functions this may be an adequate way forward. Unfortunately, it is often true that such an approach will provide less insight about the problem then others.
For example, algorithms find it difficult locating roots in the global space. Local roots may be found with other roots lying close by and yet undiscovered. Consequently, the question arises: "Are all the roots accounted for?" A more complete understanding of the functions, e.g. asymptotic behaviors, branch cuts, singular points, can provide the global perspective to better answer this, as well as other important questions.
So another possible solution would be building one's own "black box." A simple bisection routine might be a starting point. Robust if the root lies in the initial interval and fairly efficient. This encourages us to look at the global behavior of the functions. As the code is structured and debugged the various functions are explored, new insights are gained, and the algorithm has become a tool towards a more complete solution to the problem. Perhaps, with some patience, a closed-form solution can be found. A Python iterator is constructed and listed below implementing a bisection root finding algorithm.
Begin by putting the functions A(x) and B(x) in a more standard form:
C(x) = u(x) + i v(x)
and here the complex number i is brought out of the denominator and into the numerator, casting the problem into the form of functions of a complex variable. The new representation simplifies the original functions considerably. The real and imaginary parts are now clearly separated. An interesting graph is to plot A(x) and B(x) in the 3-dimensional space (u, v, x) and then visualize the projection into the u-v plane.
import numpy as np
from numpy import real, imag
import matplotlib.pyplot as plt
ax = fig.gca(projection='3d')
s = np.linspace(a, b, 1000)
ax.plot(f(s).real, f(s).imag, z, color='blue')
ax.plot(g(s).real, g(s).imag, z, color='red')
ax.plot(f(s).real, f(s).imag, 0, color='black')
ax.plot(g(s).real, g(s).imag, 0, color='black')
The question arises: "Can the parametric representation be replaced so that a relationship such as,
A(x): u(x) + i v(x) gives v(u) = f(u)
is obtained?" This will provide A(x) as a function v(u) = f(u) in the u-v plane. Then, if for
B(x): u(x) + i v(x) gives v(u) = g(u)
a similar relationship can be found, the solutions can be set equal to one another,
f(u) = g(u)
and the root(s) computed. In fact, it is convenient to look for a solution in the square of the above equation. The worst case is that an algorithm will have to be built to find the root, but at this point the behavior of our functions are better understood. For example, if f(u) and g(u) are polynomials of degree n then it is known that there are n roots. The best case is that a closed-form solution might be a reward for our determination.
Here is more detail to the solution. For A(x) the following is derived:
and v(u) = f(u) is just v(u) = constant. Similarly for B(x) a slightly more complex form is required:
Look at the function g(u) for B(x). It is imaginary if u > 0, but the root must be real since f(u) is real. This means that u must be less then 0, and there is both a positive and negative real branch to the square root. The sign of f(u) then allows one to pick the negative branch as the solution for the root. So the fact that the solution must be real is determined by the sign of u, and the fact that the real root is negative specifies what branch of the square root to choose.
In the following plot both the real (u < 0) and complex (u > 0) solutions are shown.
The camera looks toward the origin in the back corner, where the red and blue curves meet. The z-axis is the magnitude of f(u) and g(u). The x and y axes are the real/complex values of u respectively. The blue curves are the real solution with (3 - |u|). The red curves are the complex solution with (3 + |u|). The two sets meet at u = 0. The black curve is f(u) equal to (-pi/8).
There is a divergence in g(u) at |u| = 3 and this is associated with x = 0. It is far removed from the solution and will not be considered further.
To obtain the roots to f = g it is easier to square f(u) and equate the two functions. When the function g(u) is squared the branches of the square root are lost, much like squaring the solutions for x**2 = 4. In the end the appropriate root will be chosen by the sign of f(u) and so this is not an issue.
So by looking at the dependence of A and B, with respect to the parametric variable x, a representation for these functions was obtained where v is a function of u and the roots found. A simpler representation can be obtained if the term involving c in the square root is ignored.
The answer gives all the roots to be found. A cubic equation has at most three roots and one is guaranteed to be real. The other two may be imaginary or real. In this case the real root has been found and the other two roots are complex. Interestingly, as c changes these two complex roots may move into the real plane.
In the above figure the x-axis is u and the y axis is the evaluated cubic equation with constant c. The blue curve has c as (pi/8) squared. The red curve uses a larger and negative value for c, and has been translated upwards for purposes of demonstration. For the blue curve there is an inflection point near (0, 0.5), while the red curve has a maximum at (-0.9, 2.5) and a minimum at (0.9, -0.3).
The intersection of the cubic with the black line represents the roots given by: u**3 + c u + 3c = 0. For the blue curve there is one intersection and one real root with two roots in the complex plane. For the red curve there are three intersections, and hence 3 roots. As we change the value of the constant c (blue to red) the one real root undergoes a "pitchfork" bifurcation, and the two roots in the complex plane move into the real space.
Although the root (u_t, v_t) has been located, obtaining the value for x requires that (u, v) be inverted. In the present example this is a trivial matter, but if not, a bisection routine can be used to avoid the algebraic difficulties.
The parametric representation also leads to a solution for the real root, and it rounds out the analysis with an independent verification of the first result. Second, it answers the question about what happens at the singularity in the square root. Third, it gives a greater understanding of the multiplicity of roots.
The steps are: (1) convert A(x) and B(x) into polar form, (2) equate the modulus and argument giving two equations in two unknowns, (3) make a substitution for z = x**2. Converting A(x) to polar form:
Absolute value bars are not indicated, and it should be understood that the moduli r(x) and s(x) are positive definite as their names imply. For B(x):
The two equations in two unknowns:
Finally, the cubic solution is sketched out here where the substitution z = x**2 has been made:
The solution for z = x**2 gives x directly, which allows one to substitute into both A(x) and B(x). This is an exact solution if all terms are maintained in the cubic solution, and there is no error in x0, y0, A(x0), or B(x0). A simpler representation can be found by considering terms proportional to 1/d as small.
Before leaving the polar representation consider the two singular points where: (1) abs(x) = 1, and (2) x = 0. A complicating factor is that the arguments behave something like 1/x instead of x. It is worthwhile to look at a plot of the arctan(a) and then ask yourself how that changes if its argument is replaced by 1/a. The following graphs will then look less foreign.
Consider the polar representation of B(x). As x approaches 0 the modulus and argument tend toward infinity, i.e. the point is infinitely far from the origin and lies along the y-axis. Approaching 0 from the negative direction the point lies along the negative y-axis with varphi = (-pi/2), while approaching from the other direction the point lies along the positive y-axis with varphi = (+pi/2).
A somewhat more complicated behavior is exhibited by A(x). A(x) is even in x since the modulus is positive definite and the argument involves only x**2. There is a symmetry across the y-axis that allows us to only consider the x > 0 plane.
At x = 1 the modulus is just (pi/8), and as x continues to approach 0 so does r(x). The behavior of the argument is more complex. As x approaches unity from large positive values the argument is diverging towards +inf and so theta is approaching (+pi/2). As x passes through 1 the argument becomes complex. At x equals 0 the argument has reached its minimum value of -i. For complex arguments the arctan is given by:
The following are plots of the arguments for A(x) and B(x). The x-axis is the value of x, and the y-axis is the value of the angle in units of pi. In the first plot theta is shown in blue curves, and as x approaches 1 the angle approaches (+pi/2). Theta is real because abs(x) >= 1, and notice it is symmetric across the y-axis. The black curve is varphi and as x approaches 0 it approaches plus or minus (pi/2). Notice it is an odd function in x.
In the second plot A(x) is shown where abs(x) < 1 and the argument becomes complex. Near x = 1 theta is equal to (+pi/2), the blue curve, minus a small imaginary part, the red curve. As x approaches zero theta is equal to (+pi/2) minus a large imaginary part. At x equals 0 the argument is equal to -i and theta = (+pi/2) minus an infinite imaginary part, i.e ln(0) = -inf:
The values for x0 and y0 are determined by the set of equations that equate modulus and argument of A(x) and B(x), and there are no other roots. If x0 = 0 was a root, then it would fall out of these equations. The same holds for x0 = 1. In fact, if one uses approximations in the argument of A(x) about these points, and then substitutes into the equation for the modulus, the equality cannot be maintained there.
Here is another perspective: consider the set of equations where x is assumed large and call it x_inf. The equation for the argument then gives x_inf = y_inf, where 1 is neglected with respect to x_inf squared. Upon substitution into the second equation a cubic is obtained in x_inf. Will this give the correct answer? Yes, if x0 is actually large, and in this case you might get away with it since x0 is approximately 2. The difference between the sqrt(4) and the sqrt(5) is around 10%. But does this mean that x_inf = 100 is a solution? No it does not: x_inf is only a solution if it equals x0.
The initial reason for examining the problem in the first place was to find a context for building a root-finding bisection routine as a Python iterator. This can be used to find any of the roots discussed here, and looks something like this:
class Bisection:
def __init__(self, a, b, func, max_iter):
self.max_iter = max_iter
self.count_iter = 0
self.a = a
self.b = b
self.func = func
fa = func(self.a)
fb = func(self.b)
if fa*fb >= 0.0:
raise ValueError
def __iter__(self):
self.x1 = self.a
self.x2 = self.b
self.xmid = self.x1 + ((self.x2 - self.x1)/2.0)
return self
def __next__(self):
f1 = self.func(self.x1)
f2 = self.func(self.x2)
error = abs(f1 - f2)
fmid = self.func(self.xmid)
if fmid == 0.0:
return self.xmid
if f1*fmid < 0:
self.x2 = self.xmid
else:
self.x1 = self.xmid
self.xmid = self.x1 + ((self.x2 - self.x1)/2.0)
f1 = self.func(self.x1)
fmid = self.func(self.xmid)
self.count_iter += 1
if self.count_iter >= self.max_iter:
raise StopIteration
return self.xmid
The routine does only a minimal amount in the way of catching exceptions and was used to find x for the given solution in the u-v plane. The arguments a and b give the lower and upper brackets for the root to be found. The argument func is the function for the root to be found. This might look like: u0 - B(x).real. The constant max_iterations tells the iterator to stop after a given number of bisections has been attempted.

Fast 3D interpolation of atmospheric data in Numpy/Scipy

I am trying to interpolate 3D atmospheric data from one vertical coordinate to another using Numpy/Scipy. For example, I have cubes of temperature and relative humidity, both of which are on constant, regular pressure surfaces. I want to interpolate the relative humidity to constant temperature surface(s).
The exact problem I am trying to solve has been asked previously here, however, the solution there is very slow. In my case, I have approximately 3M points in my cube (30x321x321), and that method takes around 4 minutes to operate on one set of data.
That post is nearly 5 years old. Do newer versions of Numpy/Scipy perhaps have methods that handle this faster? Maybe new sets of eyes looking at the problem have a better approach? I'm open to suggestions.
EDIT:
Slow = 4 minutes for one set of data cubes. I'm not sure how else I can quantify it.
The code being used...
def interpLevel(grid,value,data,interp='linear'):
"""
Interpolate 3d data to a common z coordinate.
Can be used to calculate the wind/pv/whatsoever values for a common
potential temperature / pressure level.
grid : numpy.ndarray
The grid. For example the potential temperature values for the whole 3d
grid.
value : float
The common value in the grid, to which the data shall be interpolated.
For example, 350.0
data : numpy.ndarray
The data which shall be interpolated. For example, the PV values for
the whole 3d grid.
kind : str
This indicates which kind of interpolation will be done. It is directly
passed on to scipy.interpolate.interp1d().
returns : numpy.ndarray
A 2d array containing the *data* values at *value*.
"""
ret = np.zeros_like(data[0,:,:])
for yIdx in xrange(grid.shape[1]):
for xIdx in xrange(grid.shape[2]):
# check if we need to flip the column
if grid[0,yIdx,xIdx] > grid[-1,yIdx,xIdx]:
ind = -1
else:
ind = 1
f = interpolate.interp1d(grid[::ind,yIdx,xIdx], \
data[::ind,yIdx,xIdx], \
kind=interp)
ret[yIdx,xIdx] = f(value)
return ret
EDIT 2:
I could share npy dumps of sample data, if anyone was interested enough to see what I am working with.

Since this is atmospheric data, I imagine that your grid does not have uniform spacing; however if your grid is rectilinear (such that each vertical column has the same set of z-coordinates) then you have some options.
For instance, if you only need linear interpolation (say for a simple visualization), you can just do something like:
# Find nearest grid point
idx = grid[:,0,0].searchsorted(value)
upper = grid[idx,0,0]
lower = grid[idx - 1, 0, 0]
s = (value - lower) / (upper - lower)
result = (1-s) * data[idx - 1, :, :] + s * data[idx, :, :]
(You'll need to add checks for value being out of range, of course).For a grid your size, this will be extremely fast (as in tiny fractions of a second)
You can pretty easily modify the above to perform cubic interpolation if need be; the challenge is in picking the correct weights for non-uniform vertical spacing.
The problem with using scipy.ndimage.map_coordinates is that, although it provides higher order interpolation and can handle arbitrary sample points, it does assume that the input data be uniformly spaced. It will still produce smooth results, but it won't be a reliable approximation.
If your coordinate grid is not rectilinear, so that the z-value for a given index changes for different x and y indices, then the approach you are using now is probably the best you can get without a fair bit of analysis of your particular problem.
UPDATE:
One neat trick (again, assuming that each column has the same, not necessarily regular, coordinates) is to use interp1d to extract the weights doing something like follows:
NZ = grid.shape[0]
zs = grid[:,0,0]
ident = np.identity(NZ)
weight_func = interp1d(zs, ident, 'cubic')
You only need to do the above once per grid; you can even reuse weight_func as long as the vertical coordinates don't change.
When it comes time to interpolate then, weight_func(value) will give you the weights, which you can use to compute a single interpolated value at (x_idx, y_idx) with:
weights = weight_func(value)
interp_val = np.dot(data[:, x_idx, y_idx), weights)
If you want to compute a whole plane of interpolated values, you can use np.inner, although since your z-coordinate comes first, you'll need to do:
result = np.inner(data.T, weights).T
Again, the computation should be practically immediate.

This is quite an old question but the best way to do this nowadays is to use MetPy's interpolate_1d funtion:
https://unidata.github.io/MetPy/latest/api/generated/metpy.interpolate.interpolate_1d.html

There is a new implementation of Numba accelerated interpolation on regular grids in 1, 2, and 3 dimensions:
https://github.com/dbstein/fast_interp
Usage is as follows:
from fast_interp import interp2d
import numpy as np
nx = 50
ny = 37
xv, xh = np.linspace(0, 1, nx, endpoint=True, retstep=True)
yv, yh = np.linspace(0, 2*np.pi, ny, endpoint=False, retstep=True)
x, y = np.meshgrid(xv, yv, indexing='ij')
test_function = lambda x, y: np.exp(x)*np.exp(np.sin(y))
f = test_function(x, y)
test_x = -xh/2.0
test_y = 271.43
fa = test_function(test_x, test_y)
interpolater = interp2d([0,0], [1,2*np.pi], [xh,yh], f, k=5, p=[False,True], e=[1,0])
fe = interpolater(test_x, test_y)