Develop Reference

Demonstrating the Universality of the Uniform using numpy - an issue with transformation

Recently I wanted to demonstrate generating a continuous random variable using the universality of the Uniform. For that, I wanted to use the combination of numpy and matplotlib. However, the generated random variable seems a little bit off to me - and I don't know whether it is caused by the way in which NumPy's random uniform and vectorized works or if I am doing something fundamentally wrong here.
Let U ~ Unif(0, 1) and X = F^-1(U). Then X is a real variable with a CDF F (please note that the F^-1 here denotes the quantile function, I also omit the second part of the universality because it will not be necessary).
Let's assume that the CDF of interest to me is:
then:
According to the universality of the uniform, to generate a real variable, it is enough to plug U ~ Unif(0, 1) in the F-1. Therefore, I've written a very simple code snippet for that:
U = np.random.uniform(0, 1, 1000000)
def logistic(u):
x = np.log(u / (1 - u))
return x
logistic_transform = np.vectorize(logistic)
X = logistic_transform(U)
However, the result seems a little bit off to me - although the histogram of a generated real variable X resembles a logistic distribution (which simplified CDF I've used) - the r.v. seems to be distributed in a very unequal way - and I can't wrap my head around exactly why it is so. I would be grateful for any suggestions on that. Below are the histograms of U and X.

You have a large sample size, so you can increase the number of bins in your histogram and still get a good number samples per bin. If you are using matplotlib's hist function, try (for exampe) bins=400. I get this plot, which has the symmetry that I think you expected:
Also--and this is not relevant to the question--your function logistic will handle a NumPy array without wrapping it with vectorize, so you can save a few CPU cycles by writing X = logistic(U). And you can save a few lines of code by using scipy.special.logit instead of implementing it yourself.

Double antiderivative computation in python

I have the following problem. I have a function f defined in python using numpy functions. The function is smooth and integrable on positive reals. I want to construct the double antiderivative of the function (assuming that both the value and the slope of the antiderivative at 0 are 0) so that I can evaluate it on any positive real smaller than 100.
Definition of antiderivative of f at x:
integrate f(s) with s from 0 to x
Definition of double antiderivative of f at x:
integrate (integrate f(t) with t from 0 to s) with s from 0 to x
The actual form of f is not important, so I will use a simple one for convenience. But please note that even though my example has a known closed form, my actual function does not.
import numpy as np
f = lambda x: np.exp(-x)*x
My solution is to construct the antiderivative as an array using naive numerical integration:
N = 10000
delta = 100/N
xs = np.linspace(0,100,N+1)
vs = f(xs)
avs = np.cumsum(vs)*delta
aavs = np.cumsum(avs)*delta
This of course works but it gives me arrays instead of functions. But this is not a big problem as I can interpolate aavs using a spline to get a function and get rid of the arrays.
from scipy.interpolate import UnivariateSpline
aaf = UnivariateSpline(xs, aavs)
The function aaf is approximately the double antiderivative of f.
The problem is that even though it works, there is quite a bit of overhead before I can get my function and precision is expensive.
My other idea was to interpolate f by a spline and take the antiderivative of that, however this introduces numerical errors that are too big for what I want to use the function.
Is there any better way to do that? By better I mean faster without sacrificing accuracy.
Edit: What I hope is possible is to use some kind of Fourier transform to avoid integrating twice. I hope that there is some convenient transform of vs that allows to multiply the values component-wise with xs and transform back to get the double antiderivative. I played with this a bit, but I got lost.
Edit: I figured out that by using the trapezoidal rule instead of a naive sum, increases the accuracy quite a bit. Using Simpson's rule should increase the accuracy further, but it's somewhat fiddly to do with numpy arrays.
Edit: As #user202729 rightfully complains, this seems off. The reason it seems off is because I have skipped some details. I explain here why what I say makes sense, but it does not affect my question.
My actual goal is not to find the double antiderivative of f, but to find a transformation of this. I have skipped that because I think it only confuses the matter.
The function f decays exponentially as x approaches 0 or infinity. I am minimizing the numerical error in the integration by starting the sum from 0 and going up to approximately the peak of f. This ensure that the relative error is approximately constant. Then I start from the opposite direction from some very big x and go back to the peak. Then I do the same for the antiderivative values.
Then I transform the aavs by another function which is sensitive to numerical errors. Then I find the region where the errors are big (the values oscillate violently) and drop these values. Finally I approximate what I believe are good values by a spline.
Now if I use spline to approximate f, it introduces an absolute error which is the dominant term in a rather large interval. This gets "integrated" twice and it ends up being a rather large relative error in aavs. Then once I transform aavs, I find that the 'good region' has shrunk considerably.
EDIT: The actual form of f is something I'm still looking into. However, it is going to be a generalisation of the lognormal distribution. Right now I am playing with the following family.
I start by defining a generalization of the normal distribution:
def pdf_n(params, center=0.0, slope=8):
scale, min, diff = params
if diff > 0:
r = min
l = min + diff
else:
r = min - diff
l = min
def retfun(m):
x = (m - center)/scale
E = special.expit(slope*x)*(r - l) + l
return np.exp( -np.power(1 + x*x, E)/2 )
return np.vectorize(retfun)
It may not be obvious what is happening here, but the result is quite simple. The function decays as exp(-x^(2l)) on the left and as exp(-x^(2r)) on the right. For min=1 and diff=0, this is the normal distribution. Note that this is not normalized. Then I define
g = pdf(params)
f = np.vectorize(lambda x:g(np.log(x))/x/area)
where area is the normalization constant.
Note that this is not the actual code I use. I stripped it down to the bare minimum.

You can compute the two np.cumsum (and the divisions) at once more efficiently using Numba. This is significantly faster since there is no need for several temporary arrays to be allocated, filled, read again and freed. Here is a naive implementation:
import numba as nb
#nb.njit('float64[::1](float64[::1], float64)') # Assume vs is contiguous
def doubleAntiderivative_naive(vs, delta):
res = np.empty(vs.size, dtype=np.float64)
sum1, sum2 = 0.0, 0.0
for i in range(vs.size):
sum1 += vs[i] * delta
sum2 += sum1 * delta
res[i] = sum2
return res
However, the sum is not very good in term of numerical stability. A Kahan summation is needed to improve the accuracy (or possibly the alternative Kahan–Babuška-Klein algorithm if you are paranoid about the accuracy and performance do not matter so much). Note that Numpy use a pair-wise algorithm which is quite good but far from being prefect in term of accuracy (this is a good compromise for both performance and accuracy).
Moreover, delta can be factorized during in the summation (ie. the result just need to be premultiplied by delta**2).
Here is an implementation using the more accurate Kahan summation:
#nb.njit('float64[::1](float64[::1], float64)')
def doubleAntiderivative_accurate(vs, delta):
res = np.empty(vs.size, dtype=np.float64)
delta2 = delta * delta
sum1, sum2 = 0.0, 0.0
c1, c2 = 0.0, 0.0
for i in range(vs.size):
# Kahan summation of the antiderivative of vs
y1 = vs[i] - c1
t1 = sum1 + y1
c1 = (t1 - sum1) - y1
sum1 = t1
# Kahan summation of the double antiderivative of vs
y2 = sum1 - c2
t2 = sum2 + y2
c2 = (t2 - sum2) - y2
sum2 = t2
res[i] = sum2 * delta2
return res
Here is the performance of the approaches on my machine (with an i5-9600KF processor):
Numpy cumsum: 51.3 us
Naive Numba: 11.6 us
Accutate Numba: 37.2 us
Here is the relative error of the approaches (based on the provided input function):
Numpy cumsum: 1e-13
Naive Numba: 5e-14
Accutate Numba: 2e-16
Perfect precision: 1e-16 (assuming 64-bit numbers are used)
If f can be easily computed using Numba (this is the case here), then vs[i] can be replaced by calls to f (inlined by Numba). This helps to reduce the memory consumption of the computation (N can be huge without saturating your RAM).
As for the interpolation, the splines often gives good numerical result but they are quite expensive to compute and AFAIK they require the whole array to be computed (each item of the array impact all the spline although some items may have a negligible impact alone). Regarding your needs, you could consider using Lagrange polynomials. You should be careful when using Lagrange polynomials on the edges. In your case, you can easily solve the numerical divergence issue on the edges by extending the array size with the border values (since you know the derivative on each edges of vs is 0). You can apply the interpolation on the fly with this method which can be good for both performance (typically if the computation is parallelized) and memory usage.

First, I created a version of the code I found more intuitive. Here I multiply cumulative sum values by bin widths. I believe there is a small error in the original version of the code related to the bin width issue.
import numpy as np
f = lambda x: np.exp(-x)*x
N = 1000
xs = np.linspace(0,100,N+1)
domainwidth = ( np.max(xs) - np.min(xs) )
binwidth = domainwidth / N
vs = f(xs)
avs = np.cumsum(vs)*binwidth
aavs = np.cumsum(avs)*binwidth
Next, for visualization here is some very simple plotting code:
import matplotlib
import matplotlib.pyplot as plt
plt.figure()
plt.scatter( xs, vs )
plt.figure()
plt.scatter( xs, avs )
plt.figure()
plt.scatter( xs, aavs )
plt.show()
The first integral matches the known result of the example expression and can be seen on wolfram
Below is a simple function that extracts an element from the second derivative. Note that int is a bad rounding function. I assume this is what you have implemented already.
def extract_double_antideriv_value(x):
return aavs[int(x/binwidth)]
singleresult = extract_double_antideriv_value(50.24)
print('singleresult', singleresult)
Whatever full computation steps are required, we need to know them before we can start optimizing. Do you have a million different functions to integrate? If you only need to query a single double anti-derivative many times, your original solution should be fairly ideal.
Symbolic Approximation:
Have you considered approximations to the original function f, which can have closed form integration solutions? You have a limited domain on which the function lives. Perhaps approximate f with a Taylor series (which can be constructed with known maximum error) then integrate exactly? (consider Pade, Taylor, Fourier, Cheby, Lagrange(as suggested by another answer), etc...)
Log Tricks:
Another alternative to dealing with spiky errors, would be to take the log of your original function. Is f always positive? Is the integration error caused because the neighborhood around the max is very small? If so, you can study ln(f) or even ln(ln(f)) instead. It would really help to understand what f looks like more.
Approximation Integration Tricks
There exist countless integration tricks in general, which can make approximate closed form solutions to undo-able integrals. A very common one when exponetnial functions are involved (I think yours is expoential?) is to use Laplace's Method. But which trick to pull out of the bag is highly dependent upon the conditions which f satisfies.

Calculating 1/r*d/dr(r*f) numerically in python when r=0. f is a function of r

Usually when you do this by hand there's no problem as the 1/r usually gets cancelled with another r. But doing this numerically with scipy.misc.derivative works like a charm for r different from zero. But of course, as soon as I ask for r = 0, I get division by zero, which I expected. So how else could you calculate this numerically. I insist on the fact that everything has to be done numerically as my function are now so complicated that I won't be able to find a derivative manually. Thank you!
My code:
rAtheta = lambda _r: _r*Atheta(_r,theta,z,t)
if r != 0:
return derivative(rAtheta,r,dx=1e-10,order=3)/r
else:
#What should go here so that it doesn't blow up when calculating the gradient?

tl;dr: use symbolic differentiation, or complex step differentiation if that fails
If you insist on using numerical methods, you really have to approximate the limit of the derivative as r->0 one way or the other.
I suggest trying complex step differentiation. The idea is to use complex arguments inside the function you're trying to differentiate, but it usually gets rid of the numerical instability that is imposed by standard finite difference schemes. The result is a procedure that needs complex arithmetic (hooray numpy, and python in general!) but in turn can be much more stable at small dx values.
Here's another point: complex step differentiation uses
F′(x0) = Im(F(x0+ih))/h + O(h^2)
Let's apply this to your r=0 case:
F′(0) = Im(F(ih))/h + O(h^2)
There are no singularities even for r=0! Choose a small h, possibly the same dx you're passing to your function, and use that:
def rAtheta(_r):
# note that named lambdas are usually frowned upon
return _r*Atheta(_r,theta,z,t)
tol = 1e-10
dr = 1e-12
if np.abs(r) > tol: # or math.abs or your favourite other abs
return derivative(rAtheta,r,dx=dr,order=3)/r
else:
return rAtheta(r + 1j*dr).imag/dr/r
Here is the above in action for f = r*ln(r):
The result is straightforwardly smooth, even though the points below r=1e-10 were computed with complex step differentiation.
Very important note: notice the separation between tol and dr in the code. The former is used to determine when to switch between methods, and the latter is used as a step in complex step differentiation. Look what happens when tol=dr=1e-10:
the result is a smoothly wrong function below r=1e-10! That's why you always have to be careful with numerical differentiation. And I wouldn't advise going too much below that in dr, as machine precision will bite you sooner or later.
But why stop here? I'm fairly certain that your functions could be written in a vectorized way, i.e. they could accept an array of radial points. Using complex step differentiation you don't have to loop over the radial points (which you would have to resort to using scipy.misc.derivative). Example:
import numpy as np
import matplotlib.pyplot as plt
def Atheta(r,*args):
return r*np.log(r) # <-- vectorized expression
def rAtheta(r):
return r*Atheta(r) #,theta,z,t) # <-- vectorized as much as Atheta is
def vectorized_difffun(rlist):
r = np.asarray(rlist)
dr = 1e-12
return (rAtheta(r + 1j*dr)).imag/dr/r
rarr = np.logspace(-12,-2,20)
darr = vectorized_difffun(rarr)
plt.figure()
plt.loglog(rarr,np.abs(darr),'.-')
plt.xlabel(r'$r$')
plt.ylabel(r'$|\frac{1}{r} \frac{d}{dr}(r^2 \ln r)|$')
plt.tight_layout()
plt.show()
The result should be familiar:
Having cleared the fun weirdness that is complex step differentiation, I should note that you should strongly consider using symbolic math. In cases like this when 1/r factors disappear exactly, it wouldn't hurt if you reached this conclusion exactly. After all double precision is still just double precision.
For this you'll need the sympy module, define your function symbolically once, differentiate it symbolically once, turn your simplified result into a numpy function using sympy.lambdify, and use this numerical function as much as you need (assuming that this whole process runs in finite time and the resulting function is not too slow to use). Example:
import sympy as sym
# only needed for the example:
import numpy as np
import matplotlib.pyplot as plt
r = sym.symbols('r')
f = r*sym.ln(r)
df = sym.diff(r*f,r)
res_sym = sym.simplify(df/r)
res_num = sym.lambdify(r,res_sym,'numpy')
rarr = np.logspace(-12,-2,20)
darr = res_num(rarr)
plt.figure()
plt.loglog(rarr,np.abs(darr),'.-')
plt.xlabel(r'$r$')
plt.ylabel(r'$|\frac{1}{r} \frac{d}{dr}(r^2 \ln r)|$')
plt.tight_layout()
plt.show()
resulting in
As you see, the resulting function was vectorized thanks to lambdify using numpy during the conversion from symbolic to numeric function. Obviously, the best solution is the symbolic one as long as the resulting function is not so complicated to make its practical use impossible. I urge you to first try the symbolic version, and if for some reason it's not applicable, switch to complex step differentiation, with due caution.

Improving runtime of weighted moving average filter function?

I have a weighted moving average function which smooths a curve by averaging 3*width values to the left and to the right of each point using a gaussian weighting mechanism. I am only worried about smoothing a region bounded by [start, end]. The following code works, but the problem is runtime with large arrays.
import numpy as np
def weighted_moving_average(x, y, start, end, width = 3):
def gaussian(x, a, m, s):
return a*exp(-(x-m)**2/(2*s**2))
cut = (x>=start-3*width)*(x<=end+3*width)
x, y = x[cut], y[cut]
x_avg = x[(x>=start)*(x<=end)]
y_avg = np.zeros(len(x_avg))
bin_vals = np.arange(-3*width,3*width+1)
weights = gaussian(bin_vals, 1, 0, width)
for i in range(len(x_avg)):
y_vals = y[i:i+6*width+1]
y_avg[i] = np.average(y_vals, weights = weights)
return x_avg, y_avg
From my understanding, it is generally inefficient to loop through a NumPy array. I was wondering if anyone had an idea to replace the for loop with something more runtime efficient.
Thanks

That slicing and summing/averaging on a weighted window basically corresponds to 1D convolution with the kernel being flipped. Now, for 1D convolution, NumPy has a very efficient implementation in np.convolve and that could be used to get rid of the loop and give us y_avg. Thus, we would have a vectorized implementation like so -
y_sums = np.convolve(y,weights[::-1],'valid')
y_avg = np.true_divide(y_sums,weights.sum())

The main concern with looping over a large array is that the memory allocation for the large array can be expensive, and the whole thing has to be initialized before the loop can start.
In this particular case I'd go with what Divakar is saying.
In general, if you find yourself in a circumstance where you really need to iterate over a large collection, use iterators instead of arrays. For a relatively simple case like this, just replace range with xrange (see https://docs.python.org/2/library/functions.html#xrange).

How to plot grad(f(x,y))?

I want to calculate and plot a gradient of any scalar function of two variables. If you really want a concrete example, lets say f=x^2+y^2 where x goes from -10 to 10 and same for y. How do I calculate and plot grad(f)? The solution should be vector and I should see vector lines. I am new to python so please use simple words.
EDIT:
#Andras Deak: thank you for your post, i tried what you suggested and instead of your test function (fun=3*x^2-5*y^2) I used function that i defined as V(x,y); this is how the code looks like but it reports an error
import numpy as np
import math
import sympy
import matplotlib.pyplot as plt
def V(x,y):
t=[]
for k in range (1,3):
for l in range (1,3):
t.append(0.000001*np.sin(2*math.pi*k*0.5)/((4*(math.pi)**2)* (k**2+l**2)))
term = t* np.sin(2 * math.pi * k * x/0.004) * np.cos(2 * math.pi * l * y/0.004)
return term
return term.sum()
x,y=sympy.symbols('x y')
fun=V(x,y)
gradfun=[sympy.diff(fun,var) for var in (x,y)]
numgradfun=sympy.lambdify([x,y],gradfun)
X,Y=np.meshgrid(np.arange(-10,11),np.arange(-10,11))
graddat=numgradfun(X,Y)
plt.figure()
plt.quiver(X,Y,graddat[0],graddat[1])
plt.show()
AttributeError: 'Mul' object has no attribute 'sin'
And lets say I remove sin, I get another error:
TypeError: can't multiply sequence by non-int of type 'Mul'
I read tutorial for sympy and it says "The real power of a symbolic computation system such as SymPy is the ability to do all sorts of computations symbolically". I get this, I just dont get why I cannot multiply x and y symbols with float numbers.
What is the way around this? :( Help please!
UPDATE
#Andras Deak: I wanted to make things shorter so I removed many constants from the original formulas for V(x,y) and Cn*Dm. As you pointed out, that caused the sin function to always return 0 (i just noticed). Apologies for that. I will update the post later today when i read your comment in details. Big thanks!
UPDATE 2
I changed coefficients in my expression for voltage and this is the result:
It looks good except that the arrows point in the opposite direction (they are supposed to go out of the reddish dot and into the blue one). Do you know how I could change that? And if possible, could you please tell me the way to increase the size of the arrows? I tried what was suggested in another topic (Computing and drawing vector fields):
skip = (slice(None, None, 3), slice(None, None, 3))
This plots only every third arrow and matplotlib does the autoscale but it doesnt work for me (nothing happens when i add this, for any number that i enter)
You were already of huge help , i cannot thank you enough!

Here's a solution using sympy and numpy. This is the first time I use sympy, so others will/could probably come up with much better and more elegant solutions.
import sympy
#define symbolic vars, function
x,y=sympy.symbols('x y')
fun=3*x**2-5*y**2
#take the gradient symbolically
gradfun=[sympy.diff(fun,var) for var in (x,y)]
#turn into a bivariate lambda for numpy
numgradfun=sympy.lambdify([x,y],gradfun)
now you can use numgradfun(1,3) to compute the gradient at (x,y)==(1,3). This function can then be used for plotting, which you said you can do.
For plotting, you can use, for instance, matplotlib's quiver, like so:
import numpy as np
import matplotlib.pyplot as plt
X,Y=np.meshgrid(np.arange(-10,11),np.arange(-10,11))
graddat=numgradfun(X,Y)
plt.figure()
plt.quiver(X,Y,graddat[0],graddat[1])
plt.show()
UPDATE
You added a specification for your function to be computed. It contains the product of terms depending on x and y, which seems to break my above solution. I managed to come up with a new one to suit your needs. However, your function seems to make little sense. From your edited question:
t.append(0.000001*np.sin(2*math.pi*k*0.5)/((4*(math.pi)**2)* (k**2+l**2)))
term = t* np.sin(2 * math.pi * k * x/0.004) * np.cos(2 * math.pi * l * y/0.004)
On the other hand, from your corresponding comment to this answer:
V(x,y) = Sum over n and m of [Cn * Dm * sin(2pinx) * cos(2pimy)]; sum goes from -10 to 10; Cn and Dm are coefficients, and i calculated
that CkDl = sin(2pik)/(k^2 +l^2) (i used here k and l as one of the
indices from the sum over n and m).
I have several problems with this: both sin(2*pi*k) and sin(2*pi*k/2) (the two competing versions in the prefactor are always zero for integer k, giving you a constant zero V at every (x,y). Furthermore, in your code you have magical frequency factors in the trigonometric functions, which are missing from the comment. If you multiply your x by 4e-3, you drastically change the spatial dependence of your function (by changing the wavelength by roughly a factor of a thousand). So you should really decide what your function is.
So here's a solution, where I assumed
V(x,y)=sum_{k,l = 1 to 10} C_{k,l} * sin(2*pi*k*x)*cos(2*pi*l*y), with
C_{k,l}=sin(2*pi*k/4)/((4*pi^2)*(k^2+l^2))*1e-6
This is a combination of your various versions of the function, with the modification of sin(2*pi*k/4) in the prefactor in order to have a non-zero function. I expect you to be able to fix the numerical factors to your actual needs, after you figure out the proper mathematical model.
So here's the full code:
import sympy as sp
import numpy as np
import matplotlib.pyplot as plt
def CD(k,l):
#return sp.sin(2*sp.pi*k/2)/((4*sp.pi**2)*(k**2+l**2))*1e-6
return sp.sin(2*sp.pi*k/4)/((4*sp.pi**2)*(k**2+l**2))*1e-6
def Vkl(x,y,k,l):
return CD(k,l)*sp.sin(2*sp.pi*k*x)*sp.cos(2*sp.pi*l*y)
def V(x,y,kmax,lmax):
k,l=sp.symbols('k l',integers=True)
return sp.summation(Vkl(x,y,k,l),(k,1,kmax),(l,1,lmax))
#define symbolic vars, function
kmax=10
lmax=10
x,y=sp.symbols('x y')
fun=V(x,y,kmax,lmax)
#take the gradient symbolically
gradfun=[sp.diff(fun,var) for var in (x,y)]
#turn into bivariate lambda for numpy
numgradfun=sp.lambdify([x,y],gradfun,'numpy')
numfun=sp.lambdify([x,y],fun,'numpy')
#plot
X,Y=np.meshgrid(np.linspace(-10,10,51),np.linspace(-10,10,51))
graddat=numgradfun(X,Y)
fundat=numfun(X,Y)
hf=plt.figure()
hc=plt.contourf(X,Y,fundat,np.linspace(fundat.min(),fundat.max(),25))
plt.quiver(X,Y,graddat[0],graddat[1])
plt.colorbar(hc)
plt.show()
I defined your V(x,y) function using some auxiliary functions for transparence. I left the summation cut-offs as literal parameters, kmax and lmax: in your code these were 3, in your comment they were said to be 10, and anyway they should be infinity.
The gradient is taken the same way as before, but when converting to a numpy function using lambdify you have to set an additional string parameter, 'numpy'. This will alow the resulting numpy lambda to accept array input (essentially it will use np.sin instead of math.sin and the same for cos).
I also changed the definition of the grid from array to np.linspace: this is usually more convenient. Since your function is almost constant at integer grid points, I created a denser mesh for plotting (51 points while keeping your original limits of (-10,10) fixed).
For clarity I included a few more plots: a contourf to show the value of the function (contour lines should always be orthogonal to the gradient vectors), and a colorbar to indicate the value of the function. Here's the result:
The composition is obviously not the best, but I didn't want to stray too much from your specifications. The arrows in this figure are actually hardly visible, but as you can see (and also evident from the definition of V) your function is periodic, so if you plot the same thing with smaller limits and less grid points, you'll see more features and larger arrows.