I'm aware that we can us numpy to differentiate polynomials with the following:
f = numpy.poly1d([1, 0, 1])
f.deriv()
I've tried interpolating a set of data points and performing deriv() on the resulting polynomial.
from scipy import interpolate
x = [-2,-1,2]
y = [-2,1,-1]
f = interpolate.interp1d(x, y)
f.deriv()
But the object f is of different type.
Basically, how might I convert f to a numpy polynomial object ready for differentiation?
Thanks a lot!
The issue you're facing here is the way the interpolation actually works. Interpolation can at most guess some local function that matches the given points the best, but it can't exactly, and probably never is, except for perhaps some extreme easy cases(?), be exactly correct as the actual function given.
That said, you can approximate a function in a given range as a Taylor polynomial to a good degree. This should for relatively narrow ranges and a good guess of the initial function work sufficiently well for you.(?)
import numpy as np
from scipy import interpolate
x = [-2, -1, 2]
y = [-2, 1, -1]
f = interpolate.interp1d(x, y)
h = interpolate.approximate_taylor_polynomial(f, -1, 2, 2)
h
>>>> poly1d([-0.61111111, 1.16666667, -0.22222222])
h.deriv()
>>>> poly1d([-1.22222222, 1.16666667])
EDIT I Expanding the original answer for clarification:
I wanted to show that this approach works to a point. The above OP example used is really small MWE example and thus the results are less than convincing.
To show it's fairly close approximation I'll construct a polynomial. I'll get its values in range [-5, 5]. I'll use the range [-5, 5] and the returned values of polynomial in as interpolation arrays.
I'll approximate the interpolated function with the Taylor series expansion using the best "guesses" I have (since I constructed the original polynomial this is not really a guess tbh).
I'll compare the results in range [-5, 5] from the Taylor expansion with the original polynomial values in the range.
f = np.poly1d([1,0,1])
f([-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5])
>>> array([26, 17, 10, 5, 2, 1, 2, 5, 10, 17, 26])
x = [-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5]
y = [26, 17, 10, 5, 2, 1, 2, 5, 10, 17, 26]
f = interpolate.interp1d(x, y)
h = interpolate.approximate_taylor_polynomial(f, 0, 2, 5)
h(x)
>>> [26., 17.12, 10.21333333, 5.28, 2.32, 1.33333333, 2.32, 5.28, 10.21333333, 17.12, 26.]
f(x)
>>> [26., 17., 10., 5., 2., 1., 2., 5., 10., 17., 26.]
Here are some examples to show how the guesses get better and better the higher order of Taylor expansion you use. Careful as the manual says the expansion is unstable once it reaches order of 30.
h = interpolate.approximate_taylor_polynomial(f, 0, 15, 5)
>>> [ 25.41043927, 17.18570392, 10.19122784, 5.09107466,
2.02363911, 1. , 2.02664952, 5.07915194,
10.22646919, 17.13871545, 26. ])
h = interpolate.approximate_taylor_polynomial(f, 0, 20, 5)
>>> [ 26. , 17.13481942, 10.10070835, 5.21247548,
2.13174692, 1.23098041, 2.13174692, 5.21247548,
10.10070835, 17.13481942, 25.9999999 ])
EDIT II Answers for questions in comments:
It's not a stupid question. I can see that Taylor series is confusing you. In math they usually show the mathematical definition for the Taylor series based on nth order derivatives of the original function in the point of expansion but don't show others so it might be confusing as how to apply it in a broader sense.
In essence it's the same as with derivatives:
f' = lim d->0 [ ( f(x+d)-f(x) )/d ]
which in numerical programing we just approximate with:
f' = f(x+d)-f(x)/d (there are other approx of the derivative as well)
and that's an ok approximation as long as the d remains really small. Taylor series of a function goes something like this:
0th order: h ~ f(a)
1st order: h ~ f(a) + f'(a)(x-a)
2nd order: h ~ f(a) + f'(a)(x-a) + f''(a)/2 * (x-a)^2 ...
...
so if we now introduce our derivative approximation into the series expansion:
1st deriv 2nd deriv
h ~ f(a) + [ (f(a+d)-f(a))/d ] (x-a) + [ ( f(a+d) - 2f(a) + f(a-d) )/d^2 ) ] * (x-a)^2 ....
so now you see why the function needs the point in which it needs to be evaluated. Now, this helps us only to get rid of the derivation of the original function we had.
So you see, we don't need to know the original function at all. All we have to be able to do is provide approximated values the original function would have had in the point of expansion. And that is exactly what interpolation gives you.
Interpolation takes in a set of points presumably attributed to some original function and then based on the behavior of those points tries to guess which points in between the given points would most likely be in the graph of the original function as well. So in an essence, interpolation tries to guess the values of would-be function whose original points we know, within the range of those original points.
Ok. But what do we do about the fact that Taylor exp. goes into infinity?
We just round it of:
h ~ 0th + 1st + 2nd + 3rd + ... + nth + P
and we call P the remainder. There are ways of estimating this remainder, some given by Taylor himself iirc.
Ok, but now why do we need a range in our function call?
What we did here is what's called a finite difference method. In essence it is just about as simple as that, in reality things can get a bit more complicated because you have to show that indeed you can do these things and not break the convergence of Taylor series. It turns out that you don't break the Taylor series BUT only for bounded continuous functions, which means that you can only approx functions on a certain interval.
Think of it this way. You can approximate a straight line with a Taylor series. Think of it as compounding more and more and more polynomial orders until their "waviness" cancels each other out. Like doing sin^2 + cos^2 which is always 1.
But if you stop the series expansion at some order, then suddenly you don't have anything stopping the series from diverging again. Because Taylor series is just one big polynomial it will either start going up into infinity or down into infinity. Look at the image bellow, it shows Taylor series approximating the original quadratic function f in the point of expansion 0 on a range of 10 around it; but plotted from -50 to 50.
Special interest is the 1st series order, which is just a straight line as you can see from the formulas above (green). Notice how as soon as the series cross -10 or 10 they start diverging from the actual function by a lot. In some cases the functions were similar enough to continue being close in value with the original function (i.e. 2nd order Taylor series is also a quadratic equation which is why it traces the original function very well).
Unfortunately because we do not have any prior knowledge about the original function in your case it's impossible to determine that some Taylor expansion estimates it perfectly. As far as we know we only approximated the function around 0. It might as well contain sine or cosine members for all we know.
As far as your question about f is concerned:
f is just some dummy function I started from. It's supposed to look like np.poly1d([1,0,1]) which is supposed to be something like f(x) = x**2 +1. I don't know where you got 2.1667 + 0.25x - 0.9167x**2 from.
I used f just to create x and y arrays, so that I can be sure that indeed those numbers belong to a function. What would be the point otherwise. I only used it once more in the end by doing f(x) to show how similar the numbers turn out.
Remember x is an array, and f(x) means "calculate the value of function f for every member of array x". Nothing more. It's just the value of the function f(x) = x**2+1 in the points [-5, -4, .... 4, 5].
All other works was just based on how to approximate a function by Taylor expansion when all you have is some fixed data set and no knowledge of the original function. And I showed that if you interpolate between the points and approximate the unknown function with Taylor expansion you can reconstruct a function that has meaningfully similar results on a bounded range of numbers x.
That approximated function is called h in my snippets and it looks something like:
h = 2.28194274e-08 + 5.37467022e-17 x - 1.98652602e-06 x^2 - 3.65181145e-15 x^3 + 7.38646849e-05 x^4 + 1.02224219e-13 x^5 + ... till 25th order would be reached
and to get its derivative in python all you would need to do is
h.deriv()
because its type is poly1d.
Related
I am testing the scipy.optimize function curve_fit(). I am testing on a Quadratic function, and I have assigned the x and y data manually for this question. I do get the expected answer for the values of my parameters for basically every guess I put in. However, I noticed that for guesses of the first parameter not close to 0 (particularly, after 1), I get a Covariance Matrix full of infinity. I am not sure why such a simple test is failing.
# python version: 3.9.7
# using a venv
# numpy version: 1.23.2
# scipy version: 1.9.0
import numpy as np
from scipy.optimize import curve_fit
# data taken from a quadratic function of: y = 3*x**2 + 2
x = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], dtype=np.float64)
y = np.array([2, 5, 14, 29, 50, 77, 110, 149, 194, 245, 302], dtype=np.float64)
# quadratic function
def func(x, a, b, c):
return a * x**2 + b * x + c
# test to reproduce success case - notice that we have success when changing the first value upto a value of 1.0
success = [0, 0, 0]
# test to reproduce failure case
failure = [4, 0, 0]
popt, pcov = curve_fit(func, x, y, p0=failure) # change p0 to success or failure
print(popt) # expected answer is [3, 0, 2]
print(pcov) # covariance matrix
I'm not sure why you're expecting a different covariance matrix. The documentation says:
If the Jacobian matrix at the solution doesn’t have a full rank, then ‘lm’ method returns a matrix filled with np.inf
As far as I understand the Jacobian matrix is estimated during the optimization, and depending on what initialization you use the above case might happen. Note that the result of popt still converges!
The covariance matrix is really only useful (and in general, can only be calculated) when each and every variable is optimized. That generally means the variable is moved away from its initial value and in a way so that the dependence of the fit quality (typically, chi-square) from changing the value of this variable can be determined.
It also turns out that if initial guesses are bad, the solution may not be found -- and some variables may not actually be moved from their initial values. I think that is what is happening for you.
An initial value of "0" is particularly troublesome, as the fit really does not know "how zero" that is. Is that "magnitude less than 1e-16" or "magnitude less than 1"? Even using initial values of [4, 0.01, 0.01] would get to a good solution.
An additional potential problem is that your "data" is exactly given by the model function and values. At "the right solution", the residual will be really very very close to zero, and converting the Jacobian matrix of derivatives (of misfit with respect to variables) to covariance can be numerically unstable. That would be very unlikely with real data, but you may want to add a small amount of noise to the data being modeled.
I expected the poly() and roots() functions to be each other's inverse. However, this isn't quite true:
# Polys coeffs
pol_c = np.poly([-1, 1, 1, 10]) # Get Polynomial coeffs for eqt with stated roots
# Roots from the poly equation
root_val = np.roots(pol_c)
# Roots from the poly equation, manually entered as integers
roots_v2 = np.roots([1,-11,9,11,-10])
print(pol_c)
print(root_val)
print(roots_v2)
Gives
[1. -11. 9. 11. -10.]
[10.+0.0000000e+00j -1.+0.0000000e+00j 1.+9.6357437e-09j
1.-9.6357437e-09j]
[10.+0.0000000e+00j -1.+0.0000000e+00j 1.+9.6357437e-09j
1.-9.6357437e-09j]
ie. the 3rd & 4th roots are (slightly) imaginary instead of real
My first thought was floating point error, but given that roots() outputs the same answer for float and int inputs that seems not to be the case. Plus I would expect poly() to give non-integer answers if floating point accuracy was limiting the solves.
The functions are inverses of each other, within some computational errors (which may be complex), and up to reordering of roots.
pol_c = np.poly([-1, 1, 1, 10])
root_val = np.roots(pol_c)
print(np.real_if_close(np.around(root_val, 6)))
prints [10. -1. 1. 1.] which is the same as we started with, in another order.
Of course, the order need not be the same: the original order of roots is lost when pol_c was formed, and there is no canonical order for the roots of polynomials (which are generally complex) anyway.
I want to solve the linear equation for n given points in n dimensional space to get the equation of hyper-plane.
for example, in two dimensional case, Ax + By + C = 0.
How can I get one solution if there are infinite solutions in a linear equations ?
I have tried scipy.linalg.solve() but it requires coefficient matrix A to be nonsingular.
I also tried sympy
A = Matrix([[0, 0, 1], [1, 1, 1]])
b = Matrix([0, 0])
linsolve((A, b), [x, y, z])
It returned me this
{(−y,y,0)}
I have to parse the result to determine which one is the free variable and then assign a number to it to get a solution.
Is there a more convenient way since I only want to get a specific solution ?
Using example from Andrew Ng's class (finding parameters for Linear Regression using normal equation):
With Python:
X = np.array([[1, 2104, 5, 1, 45], [1, 1416, 3, 2, 40], [1, 1534, 3, 2, 30], [1, 852, 2, 1, 36]])
y = np.array([[460], [232], [315], [178]])
θ = ((np.linalg.inv(X.T.dot(X))).dot(X.T)).dot(y)
print(θ)
Result:
[[ 7.49398438e+02]
[ 1.65405273e-01]
[ -4.68750000e+00]
[ -4.79453125e+01]
[ -5.34570312e+00]]
With Julia:
X = [1 2104 5 1 45; 1 1416 3 2 40; 1 1534 3 2 30; 1 852 2 1 36]
y = [460; 232; 315; 178]
θ = ((X' * X)^-1) * X' * y
Result:
5-element Array{Float64,1}:
207.867
0.0693359
134.906
-77.0156
-7.81836
Furthermore, when I multiple X by Julia's — but not Python's — θ, I get numbers close to y.
I can't figure out what I am doing wrong. Thanks!
Using X^-1 vs the pseudo inverse
pinv(X) which corresponds to the pseudo inverse is more broadly applicable than inv(X), which X^-1 equates to. Neither Julia nor Python do well using inv, but in this case apparently Julia does better.
but if you change the expression to
julia> z=pinv(X'*X)*X'*y
5-element Array{Float64,1}:
188.4
0.386625
-56.1382
-92.9673
-3.73782
you can verify that X*z = y
julia> X*z
4-element Array{Float64,1}:
460.0
232.0
315.0
178.0
A more numerically robust approach in Python, without having to do the matrix algebra yourself is to use numpy.linalg.lstsq to do the regression:
In [29]: np.linalg.lstsq(X, y)
Out[29]:
(array([[ 188.40031942],
[ 0.3866255 ],
[ -56.13824955],
[ -92.9672536 ],
[ -3.73781915]]),
array([], dtype=float64),
4,
array([ 3.08487554e+03, 1.88409728e+01, 1.37100414e+00,
1.97618336e-01]))
(Compare the solution vector with #waTeim's answer in Julia).
You can see the source of the ill-conditioning by printing the matrix inverse you're calculating:
In [30]: np.linalg.inv(X.T.dot(X))
Out[30]:
array([[ -4.12181049e+13, 1.93633440e+11, -8.76643127e+13,
-3.06844458e+13, 2.28487459e+12],
[ 1.93633440e+11, -9.09646601e+08, 4.11827338e+11,
1.44148665e+11, -1.07338299e+10],
[ -8.76643127e+13, 4.11827338e+11, -1.86447963e+14,
-6.52609055e+13, 4.85956259e+12],
[ -3.06844458e+13, 1.44148665e+11, -6.52609055e+13,
-2.28427584e+13, 1.70095424e+12],
[ 2.28487459e+12, -1.07338299e+10, 4.85956259e+12,
1.70095424e+12, -1.26659193e+11]])
Eeep!
Taking the dot product of this with X.T leads to a catastrophic loss of precision.
Notice that X is a 4x5 matrix or in statistical terms that you have fewer observations than parameters to estimate. Therefore, the least squares problem has infinitely many solutions with the sum of the squared errors exactly equal to zero. In this case, the normal equations don't help you much because the matrix X'X is singular. Instead, you should just find a solution to X*b=y.
Most numerical linear algebra systems are based on the FORTRAN package LAPACK which uses the a pivoted QR factorization for solving the problem X*b=y. Since there are infinitely many solutions, LAPACK's picks the solution with the smallest norm. In Julia, you can get this solution, simply by writing
float(X)\y
(Unfortunately, the float part is necessary right now, but that will change.)
In exact arithmetic, you should get the same solution as the one above with either of your proposed methods, but the floating point representation of you problem introduces small rounding errors and these errors will affect the calculated solution. The effect of the rounding errors on the solution is much larger when using the normal equations compared to using the QR factorization directly on X.
This holds true also in the usual case where X has more rows than columns so often it is recommended that you avoid the normal equations when solving least squares problems. However, when X has many more rows than columns, the matrix X'X is relatively small. In this case, it will be much faster to solve the problem with the normal equations instead of using the QR factorization. In many statistical problems, the extra numerical error is extremely small compared to the statical error so the loss of precision due to the normal equations can simply be ignored.
Given two vectors X and Y, I have to find their correlation, i.e. their linear dependence/independence. Both vectors have equal dimension. The result should be a floating point number from [-1.0 .. 1.0].
Example:
X=[-1, 2, 0]
Y=[ 4, 2, -0.3]
Find y = cor(X,Y) such that y belongs to [-1.0 .. 1.0].
It should be a simple construction involving a list-comprehension. No external library is allowed.
UPDATE: ok, if the dot product is enough, then here is my solution:
nX = 1/(sum([x*x for x in X]) ** 0.5)
nY = 1/(sum([y*y for y in Y]) ** 0.5)
cor = sum([(x*nX)*(y*nY) for x,y in zip(X,Y) ])
right?
Sounds like a dot product to me.
Solve the equation for the cosine of the angle between the two vectors, which is always in the range [-1, 1], and you'll have what you want.
It's equal to the dot product divided by the magnitudes of two vectors.
Since range is supposed to be [-1, 1] I think that the Pearson Correlation can be ok for your purposes.
Also dot-product would work but you'll have to normalize vectors before calculating it and you can have a -1,1 range just if you have also negative values.. otherwise you would have 0,1
Don't assume because a formula is algebraically correct that its direct implementation in code will work. There can be numerical problems with some definitions of correlation.
See How to calculate correlation accurately