Develop Reference

Does numpy.fft2 results produce results that follow standard ordering as described in numpy.fft documentation?

I've read the documentation and searched Stack Overflow for the answer to this question, but can't find it. Sorry if it has already been answered.
I'm working with the results of an np.fft.fft2(Z) where Z is some 2d NumPy array. I would expect positive frequencies to be stored in values less than the Nyquist wavenumber in both x and y directions. From my tests, it seems this is the approach Matlab takes. In NumPy documentation they write positive frequencies are stored below the Nyquist number and negative frequencies above; this does not seem to be the case for fft2.
Some positive frequencies terms are stored at locations greater than the Nyquist wavenumber. For example, a mode at location (127,1) with associated amplitude stored at (1,127), will produce a 2D sinusoid with 4 peaks indicating that the wavenumber should be around 4, not 127.
I can't tell which is the positive and negative frequency in my example above because they are not following standard ordering.
So the main question I have is what kind of order does the fft2 follow for storing positive and negative frequencies?
I didn't post any examples because my question is a universal one and shouldn't be problem specific.
import numpy as np
from heapq import nlargest
## Setting up a simple example
lx = 4.0
ly = 4.0;
lz = 1.5;
nx = 128;
ny = 128;
L = 1.0
H = .4
x = np.linspace(0, lx, nx);
y = np.linspace(0, ly, ny);
x0 = 2.0;
y0 = 2.0;
z1 = np.zeros([ny,nx])
zm= np.zeros([ny,nx])
for j in range(1,ny):
for i in range(1,nx):
if np.sqrt(abs(x[i] - x0)** 2 + abs(y[j] - y0) ** 2) < L:
if np.abs(x[i] - x0) < L:
z1[j, i] = H * np.cos(np.pi * abs(x[i] - x0) / (2 *L))**2;
z1 = z1+np.transpose(z1)/2.0
## Here I take the fft
nf = np.shape(z1)[0]/2
fz1 = np.fft.fft2(z1)
spec_fz1 = np.abs(fz1)**2
valmax = nlargest(1000, spec_fz1.flatten())
## Here I search for amplitude pairs above nyquist number
for i in range(1,len(valmax),2):
xy = return_xy(valmax[i], spec_fz1)
if len(xy) >2:
if ((xy[0] > nf or xy[1]> nf) and (xy[2] > nf or xy[3]> nf) ):
print('both index locations above nyquist frequency')
else:
xy2 = return_xy(valmax[i+1], spec_fz1)
if ((xy[0] > nf or xy[1]> nf) and (xy2[0] > nf or xy2[1]> nf) ):
print('both index locations above nyquist frequency')
def return_xy(mode,spec_topo):
kxky = np.array([])
for i in range(np.shape(spec_topo)[0]):
for j in range(np.shape(spec_topo)[1]):
if spec_topo[i,j] == mode:
kxky= np.append(kxky,[i,j])
if len(kxky)> 1:
return kxky
else:
return kxky[0]
After sorting by the largest amplitude at the 21st index two amplitude pairs are stored at (127,1) and (1,127) which is above the Nyquist number. How should I interpret this wavenumber? note return_xy does same thing as np.where

I think this bit of code demonstrates how the 2D DFT output of np.fft.fft2 is organized:
import numpy as np
import matplotlib.pyplot as plt
n = 16
x = np.arange(n) / n * 2 * np.pi
y = np.arange(n) / n * 2 * np.pi
for kx in range(4):
for ky in range(4):
f = np.cos(kx * x[None,:] + ky * y[:,None])
F = np.fft.fft2(f)
plt.subplot(4, 4, 1 + ky * 4 + kx)
plt.imshow(np.abs(F))
plt.axis('off')
plt.title(f'kx = {kx}, ky = {ky}', fontsize=10)
plt.tight_layout()
plt.show()
We can see that the origin, kx=0 and ky=0 is at the top-left of the array. For a horizontal wave with exactly one period in the input, we see we have a pair of peaks at kx=1 and kx=N-1 (which is equivalent to kx=-1). With two periods in the input, kx=2 and kx=-2, etc. Vertical waves produce the same result but along the vertical axis, and diagonal waves at 45 degrees have the peaks at 45 degrees.
This is the exact same ordering as the 1D DFT (np.fft.fft) produces. The 2D DFT is simply the 1D DFT applied along the columns, and then along the rows of the result (or the other way around, it doesn't matter).
As for the test shown in the question, it is the superposition of two sine waves (one horizontal and one vertical) multiplied by a round window (a "pillbox" function). In the Fourier domain (continuous world), this corresponds to four impulse functions (two along the horizontal axis for the one sine wave, two along the vertical axis for the other sine wave), convolved with the Bessel function of the first kind of order 1 (J1). Because the sine waves have a low frequency, the four impulse functions are close together, and after the convolution appear as a somewhat wider Bessel function, centered around the origin:
plt.imshow(np.log(np.abs(fz1) + 1e-6))
plt.show()
What we see is the peak centered on the origin (at the top-left corner), with things to the left of the origin wrapped around to the right edge, and things to the top of the origin wrapped around to the bottom edge. Applying np.fft.fftshift moves the origin to the middle of the array, yielding a more recognizable shape.

Which loss function calculates the distance between two contours

In my contour generation network I am using the nn.L1Loss() to caculate how many pixels are wrong. This works for training, but the 2D-Distance between the real contour and the fake would be way better. My aim is to measure the length of generated contours afterwards. This code example of two binary images shows where the nn.L1Loss() fails.
import cv2
import torch
from torch import nn
p1 = [(15, 15),(45,45)]
p2 = [(16, 15),(46,45)]
real = cv2.rectangle(np.ones((60,60)), p1[0], p1[1], color=0, thickness=1)
fake = cv2.rectangle(np.ones((60,60)), p2[0], p2[1], color=0, thickness=1)
cv2.imshow('image',np.hstack((real,fake)))
cv2.waitKey(0)
real = torch.tensor(real)
fake = torch.tensor(fake)
losss = [nn.L1Loss(), nn.MSELoss(), nn.BCELoss(), nn.HingeEmbeddingLoss(), nn.SmoothL1Loss()]
print(my_loss(real, fake))
for k, loss in enumerate(losss):
err = loss(real, fake)
print(err*60)
If I move the rectangle 1 pixel to the right:
-> L1 loss is 0.0333 * 60 = 2
If I move the rectangle 1 pixel to the right and 1 to the left:
-> L1 loss is 0.0656 * 60 = 3.933
If I move the rectangle 10 pixel to the right and 10 to the left:
-> L1 loss is 0.0656 * 60 = 3.933 still the same! Which is no suprise, the amount of wrong pixels is the same. But the distance to them changed by 10 * 2**1/2.
I also thought about the distance between both centers with this:
M = cv2.moments(c)
cX = int(M['m10'] /M['m00'])
cY = int(M['m01'] /M['m00'])
centers.append([cX,cY])
The problem here is that the generated contours are not identical to the real and thus have diffrent centers.
This answer is close to what I am looking for, but is computional very expensive?!
https://stackoverflow.com/a/36505073/12337147
Is there a custom loss functions to determine the distance like I described?

Is this the equation you want
As opposed to the area between the curves that is given by the next equation if the contours are sufficiently similar to each other
It means the accumulated squared distances from points in one contour to points
to the closest point in the other contour?
Given two arrays with the points on the contours, I can compute this with complexity O(M * N) where C1 has M points and C2 has N points, directly on GPU. Alternatively it could be computed in a O(W * H) where W * H is the dimension of the images.
If this is exactly what you want I may can post a solution.
The solution
First Let's create some example data.
import torch
import math
from torch import nn
import matplotlib.pyplot as plt;
# Number of points in each contour
M, N = 1000, 1500
t1 = torch.linspace(0, 2*math.pi, M).view(1, -1)
t2 = torch.linspace(0, 2*math.pi, N).view(1, -1)
c1 = torch.stack([torch.sin(t1),torch.cos(t1)], dim=2) # (1 x M x 2)
c2 = 1 - 2* torch.sigmoid(torch.stack([torch.sin(t2)*3 + 1, torch.cos(t2)*3 + 2], dim=2)) # (1 x N x 2)
With this we can compute the distance between every pair of points using torch.cdist. Here I used torch.argmin to find the position of the minimum of each column in the array. For computing the loss function what is important is the distance itself, and that can be computed with torch.amin.
distances = torch.cdist(c1, c2); # (1 x M x N)
plt.imshow(distances[0]);
plt.xlabel('index in countor 1');
plt.ylabel('index in countor 2');
plt.plot(torch.argmin(distances[0], axis=0), '.r')
Now basically however, what you to accumulate is not the distance, is a function of the distance. this could be easily obtained with torch.min(f(distances)), and assuming f(.) is monotonic can be simplified to f(torch.min(distances)).
To approximate the integral we can use the trapezoidal rule, that integrates a linear interpolation of a sampled function, in our case the contour sampled at the points you give.
This gives you a loss function
def contour_divergence(c1, c2, func = lambda x: x**2):
c1 = torch.atleast_3d(c1);
c2 = torch.atleast_3d(c2);
f = func(torch.amin(torch.cdist(c1, c2), dim=2));
# this computes the length of each segment connecting two consecutive points
df = torch.sum((c1[:, 1:, :] - c1[:, :-1, :])**2, axis=2)**0.5;
# here is the trapesoid rule
return torch.sum((f[:, :-1] + f[:, 1:]) * df[:, :], axis=1) / 4.0;
def contour_dist(c1, c2, func = lambda x: x**2):
return contour_divergence(c1, c2, func) + contour_divergence(c2, c1, func)
For the case where the line connecting the closest point is always perpendicular to the trajectory contour_dist(c1, c2, lambda x: x) gives the area.
This will give the area of the circle of radius 1 (all the points of the second circle is on the origin).
print(contour_dist(c1, c1*0, lambda x: x) / math.pi) # should print 1
Consider now the distance between a circle of radius 1 and a circle of radius 1 (it will be pi * (1 - 1/4) = 0.75*pi)
print(contour_dist(c1, c1*0.5, lambda x: x) / math.pi) # should print 0.75
And if you want any loss that accumulates the square distance you just use contour_dist(c1, c2), and you can pass an arbitrary function as parameter to the function. You can backpropagate the loss as long as you can backpropagate the passed func.

Calculating MSE between numpy arrays

The Scientific Question:
I have lots of 3D volumes all with a cylinder in them orientated with the cylinder 'upright' on the z axis. The volumes containing the cylinder are incredibly noisy, like super noisy you can't see the cylinder in them as a human. If I average together 1000s of these volumes I can see the cylinder. Each volume contains a copy of the cylinder but in a few cases the cylinder may not be orientated correctly so I want a way of figuring this out.
The Solution I have come up with:
I have taken the averaged volume and projected it down the z and x axis (just projecting the numpy array) so that I get a nice circle in one direction and a rectangle in the other. I then take each 3D volume and project every single one down the Z axis. The SNR is still so bad that I cannot see a circle but if I average the 2D slices I can begin to see a circle after averaging a few hundred and it is easy to see after the first 1000 are averaged. To calculate a score of how each volume I figured calculating the MSE of the 3D volumes projected down z against three other arrays, the first would be the average projected down Z, then the average projected down y or x, and finally an array with a normal distribution of noise in it.
Currently I have the following where RawParticle is the 3D data and Ave is the average:
def normalise(array):
min = np.amin(array)
max = np.amax(array)
normarray = (array - min) / (max - min)
return normarray
def Noise(mag):
NoiseArray = np.random.normal(0, mag, size=(200,200,200))
return NoiseArray
#3D volume (normally use a for loop to iterate through al particles but for this example just showing one)
RawParticleProjected = np.sum(RawParticle, 0)
RawParticleProjectedNorm = normalise(RawParticleProjected)
#Average
AveProjected = np.sum(Ave, 0)
AveProjectedNorm = normalise(AveProjected)
#Noise Array
NoiseArray = Noise(0.5)
NoiseNorm = normalise(NoiseArray)
#Mean squared error
MSE = (np.square(np.subtract(RawParticleProjectedNorm, AveProjectedNorm))).mean()
I then repeat this with the Ave summed down axis 1 and then again compared the Raw particle to the Noise array.
However my output from this gives highest MSE when I am comparing the projections that should both be circles as shown below:
My understanding of MSE is that the other two populations should have high MSE and my populations that agree should have low MSE. Perhaps my data is too noisy for this type of analysis? but if that is true then I don't really know how to do what I am doing.
If anyone could glance at my code or enlighten my understanding of MSE I would be super appreciative.
Thank you for taking the time to look and read.

If I understood your question correctly you want to figure how close your different samples are to the average.
And by comparing the samples you want to find the outliers which contain a disoriented cylinder.
This fits pretty well to the definition of the L2 norm, so MSE should work here.
I would calculate the average 3D image of all samples and than compute the distance of each sample to this average. Then I would just compare those values.
The idea of comparing the samples to an image of artificial noise is not bad, but I am not sure if a normal distribution and your normalization work out as you planned. I could be apple and oranges.
And I don't think it is a good idea to look at projections along different axis,
just compare the 3D images.
I made some small tests with a circle in 2D with a parameter alpha which indicates how much noise and how much circle there is in a picture.
(alpha=0 means only noise, alpha=1 means only circle`)
import numpy as np
import matplotlib.pyplot as plt
grid_size = 20
radius = 5
mag = 1
def get_circle_stencil(radius):
xx, yy = np.meshgrid(np.linspace(-grid_size/2+1/2, grid_size/2-1/2, grid_size),
np.linspace(-grid_size/2+1/2, grid_size/2-1/2, grid_size))
dist = np.sqrt(xx**2 + yy**2)
inner = dist < (radius - 1/2)
return inner.astype(float)
def create_noise(mag, n_dim=2):
# return np.random.normal(0, mag, size=(grid_size,)*n_dim)
return np.random.uniform(0, mag, size=(grid_size,)*n_dim)
def create_noisy_sample(alpha, n_dim=2):
return (np.random.uniform(0, 1-alpha, size=(grid_size,)*n_dim) +
alpha*get_circle_stencil(radius))
fig = plt.figure()
ax = fig.subplots(nrows=3, ncols=3)
np.unravel_index(3, shape=(3, 3))
alpha_list = np.arange(9) / 10
for i, alpha in enumerate(alpha_list):
r, c = np.unravel_index(i, shape=(3, 3))
ax[r][c].imshow(*norm(create_noisy_sample(alpha=alpha)), cmap='Greys')
ax[r][c].set_title(f"alpha={alpha}")
ax[r][c].xaxis.set_ticklabels([])
ax[r][c].yaxis.set_ticklabels([])
Than I tried some metrics (mse, cosine similarity and binary cross entropy
and looked how they behave for different values of alpha.
def normalize(*args):
return [a / np.linalg.norm(a) for a in args]
def cosim(a, b):
return np.sum(a * b)
def mse(a, b):
return np.sqrt(np.sum((a-b)**2))
def bce(a, b):
# binary cross entropy implemented from tensorflow / keras
eps = 1e-7
res = a * np.log(b + eps)
res += (1 - a) * np.log(1 - b + eps)
return np.mean(-res)
I compared NoiseA-NoiseB, Circle-Circle, Circle-Noise, Noise-Sample, Circle-Sample
alpha = 0.1
noise = create_noise(mag=1, grid_size=grid_size)
noise_b = create_noise(mag=1, grid_size=grid_size)
circle_reference = get_circle_stencil(radius=radius, grid_size=grid_size)
sample = create_noise(mag=1, grid_size=grid_size) + alpha * circle_reference
print('NoiseA-NoiseB:', mse(*norm(noise, noise_b))) # 0.718
print('Circle-Circle:', mse(*norm(circle, circle))) # 0.000
print('Circle-Noise:', mse(*norm(circle, noise))) # 1.168
print('Noise-Sample:', mse(*norm(noise, sample))) # 0.697
print('Circle-Sample:', mse(*norm(circle, sample))) # 1.100
print('NoiseA-NoiseB:', cosim(*norm(noise, noise_b))) # 0.741
print('Circle-Circle:', cosim(*norm(circle, circle))) # 1.000
print('Circle-Noise:', cosim(*norm(circle, noise))) # 0.317
print('Noise-Sample:', cosim(*norm(noise, sample))) # 0.757
print('Circle-Sample:', cosim(*norm(circle, sample))) # 0.393
print('NoiseA-NoiseB:', bce(*norm(noise, noise_b))) # 0.194
print('Circle-Circle:', bce(*norm(circle, circle))) # 0.057
print('Circle-Noise:', bce(*norm(circle, noise))) # 0.111
print('Noise-Circle:', bce(*norm(noise, circle))) # 0.636
print('Noise-Sample:', bce(*norm(noise, sample))) # 0.192
print('Circle-Sample:', bce(*norm(circle, sample))) # 0.104
n = 1000
ns = np.zeros(n)
cs = np.zeros(n)
for i, alpha in enumerate(np.linspace(0, 1, n)):
sample = create_noisy_sample(alpha=alpha)
ns[i] = mse(*norm(noise, sample))
cs[i] = mse(*norm(circle, sample))
fig, ax = plt.subplots()
ax.plot(np.linspace(0, 1, n), ns, c='b', label='noise-sample')
ax.plot(np.linspace(0, 1, n), cs, c='r', label='circle-sample')
ax.set_xlabel('alpha')
ax.set_ylabel('mse')
ax.legend()
For your problem I would just look at the comparison circle-sample (red).
Different samples will behave as if they have different alpha values and you can group them accordingly. And you should be able to detect outliers because they should have an higher mse.
You said you have to combine 100-1000 pictures to see the cylinder, which indictas you have an really small alpha value in your problem, but in average mse
should work.

Euler Angles and Rotation Matrix from two 3D points

I am trying to find the Euler angles that allow the transformation from point A to point B in 3D space.
Consider the normalized vectors A = [1, 0, 0] and B = [0.32 0.88 -0.34].
I understand that by computing the cross product A × B I get the rotation axis. The angle between A and B is given by tan⁻¹(||cross||, A·B), where A·B is the dot product between A and B.
This gives me the rotation vector rotvec = [0 0.36 0.93 1.24359531111], which is rotvec = [A × B; angle] (the cross product is normalized).
Now my question is: How do I move from here to get the Euler angles that correspond to the transformation from A to B?
In MATLAB the function vrrotvec2mat receives as input a rotation vector and outputs a rotation matrix. Then the function rotm2eul should return the corresponding Euler angles. I get the following result (in radians): [0.2456 0.3490 1.2216], according to the XYZ convention. Yet, this is not the expected result.
The correct answer is [0 0.3490 1.2216] that corresponds to a rotation of 20° and 70° in Y and Z, respectively.
When I use eul2rot([0 0.3490 1.2216]) (with eul2rot taken from here) to verify the resulting rotation matrix, this one is different from the one I obtain when using vrrotvec2mat(rotvec).
I also have a Python spinet that yields the exactly same results as described above.
--- Python (2.7) using transform3d ---
import numpy as np
import transforms3d
cross = np.cross(A, B)
dot = np.dot(A, B.transpose())
angle = math.atan2(np.linalg.norm(cross), dot)
rotation_axes = sklearn.preprocessing.normalize(cross)
rotation_m = transforms3d.axangles.axangle2mat(rotation_axes[0], angle, True)
rotation_angles = transforms3d.euler.mat2euler(rotation_m, 'sxyz')
What I am missing here? What should I be doing instead?
Thank you

A rotation matrix has 3 degrees of freedom but the constraints of your problem only constrain 2 of those degrees.
This can be made more concrete by considering the case where we have a rotation matrix R which rotates from A to B so R*A == B. If we then construct another rotation matrix RB which rotates about vector B then applying this rotation to R*A won't have any effect, i.e. B == R*A == RB*R*A. It will, however, produce a different rotation matrix RB*R with different Euler angles.
Here's an example in MATLAB:
A = [1; 0; 0];
B = [0.32; 0.88; -0.34];
A = A / norm(A);
B = B / norm(B);
ax = cross(A, B);
ang = atan2(norm(ax), dot(A, B)); % ang = acos(dot(A, B)) works too
R = axang2rotm([ax; ang].');
ang_arbitrary = rand()*2*pi;
RB = axang2rotm([B; ang_arbitrary].');
R*A - B
RB*R*A - B
rotm2eul(R)
rotm2eul(RB*R)
Result
ans =
1.0e-15 *
-0.0555
0.1110
0
ans =
1.0e-15 *
0.2220
0.7772
-0.2776
ans =
1.2220 0.3483 0.2452
ans =
1.2220 0.3483 0.7549

I will give you a solution based on Euler's rotation theorem.
This solution gives you only the one angle, but the other angles can be derived.
import numpy as np
a_vec = np.array([1, 0, 0])/np.linalg.norm(np.array([1, 0, 0]))
b_vec = np.array([0.32, 0.88, -0.34])/np.linalg.norm(np.array([0.32, 0.88, -0.34]))
cross = np.cross(a_vec, b_vec)
ab_angle = np.arccos(np.dot(a_vec,b_vec))
vx = np.array([[0,-cross[2],cross[1]],[cross[2],0,-cross[0]],[-cross[1],cross[0],0]])
R = np.identity(3)*np.cos(ab_angle) + (1-np.cos(ab_angle))*np.outer(cross,cross) + np.sin(ab_angle)*vx
validation=np.matmul(R,a_vec)
This uses the common axis of rotation (eigenvector in this case), as the cross product.
The matrix R is then the rotation matrix.
This is a general way of doing it, and very simple.

Evenly distributing n points on a sphere

I need an algorithm that can give me positions around a sphere for N points (less than 20, probably) that vaguely spreads them out. There's no need for "perfection", but I just need it so none of them are bunched together.
This question provided good code, but I couldn't find a way to make this uniform, as this seemed 100% randomized.
This blog post recommended had two ways allowing input of number of points on the sphere, but the Saff and Kuijlaars algorithm is exactly in psuedocode I could transcribe, and the code example I found contained "node[k]", which I couldn't see explained and ruined that possibility. The second blog example was the Golden Section Spiral, which gave me strange, bunched up results, with no clear way to define a constant radius.
This algorithm from this question seems like it could possibly work, but I can't piece together what's on that page into psuedocode or anything.
A few other question threads I came across spoke of randomized uniform distribution, which adds a level of complexity I'm not concerned about. I apologize that this is such a silly question, but I wanted to show that I've truly looked hard and still come up short.
So, what I'm looking for is simple pseudocode to evenly distribute N points around a unit sphere, that either returns in spherical or Cartesian coordinates. Even better if it can even distribute with a bit of randomization (think planets around a star, decently spread out, but with room for leeway).

The Fibonacci sphere algorithm is great for this. It is fast and gives results that at a glance will easily fool the human eye. You can see an example done with processing which will show the result over time as points are added. Here's another great interactive example made by #gman. And here's a simple implementation in python.
import math
def fibonacci_sphere(samples=1000):
points = []
phi = math.pi * (3. - math.sqrt(5.)) # golden angle in radians
for i in range(samples):
y = 1 - (i / float(samples - 1)) * 2 # y goes from 1 to -1
radius = math.sqrt(1 - y * y) # radius at y
theta = phi * i # golden angle increment
x = math.cos(theta) * radius
z = math.sin(theta) * radius
points.append((x, y, z))
return points
1000 samples gives you this:

The golden spiral method
You said you couldn’t get the golden spiral method to work and that’s a shame because it’s really, really good. I would like to give you a complete understanding of it so that maybe you can understand how to keep this away from being “bunched up.”
So here’s a fast, non-random way to create a lattice that is approximately correct; as discussed above, no lattice will be perfect, but this may be good enough. It is compared to other methods e.g. at BendWavy.org but it just has a nice and pretty look as well as a guarantee about even spacing in the limit.
Primer: sunflower spirals on the unit disk
To understand this algorithm, I first invite you to look at the 2D sunflower spiral algorithm. This is based on the fact that the most irrational number is the golden ratio (1 + sqrt(5))/2 and if one emits points by the approach “stand at the center, turn a golden ratio of whole turns, then emit another point in that direction,” one naturally constructs a spiral which, as you get to higher and higher numbers of points, nevertheless refuses to have well-defined ‘bars’ that the points line up on.(Note 1.)
The algorithm for even spacing on a disk is,
from numpy import pi, cos, sin, sqrt, arange
import matplotlib.pyplot as pp
num_pts = 100
indices = arange(0, num_pts, dtype=float) + 0.5
r = sqrt(indices/num_pts)
theta = pi * (1 + 5**0.5) * indices
pp.scatter(r*cos(theta), r*sin(theta))
pp.show()
and it produces results that look like (n=100 and n=1000):
Spacing the points radially
The key strange thing is the formula r = sqrt(indices / num_pts); how did I come to that one? (Note 2.)
Well, I am using the square root here because I want these to have even-area spacing around the disk. That is the same as saying that in the limit of large N I want a little region R ∈ (r, r + dr), Θ ∈ (θ, θ + dθ) to contain a number of points proportional to its area, which is r dr dθ. Now if we pretend that we are talking about a random variable here, this has a straightforward interpretation as saying that the joint probability density for (R, Θ) is just c r for some constant c. Normalization on the unit disk would then force c = 1/π.
Now let me introduce a trick. It comes from probability theory where it’s known as sampling the inverse CDF: suppose you wanted to generate a random variable with a probability density f(z) and you have a random variable U ~ Uniform(0, 1), just like comes out of random() in most programming languages. How do you do this?
First, turn your density into a cumulative distribution function or CDF, which we will call F(z). A CDF, remember, increases monotonically from 0 to 1 with derivative f(z).
Then calculate the CDF’s inverse function F-1(z).
You will find that Z = F-1(U) is distributed according to the target density. (Note 3).
Now the golden-ratio spiral trick spaces the points out in a nicely even pattern for θ so let’s integrate that out; for the unit disk we are left with F(r) = r2. So the inverse function is F-1(u) = u1/2, and therefore we would generate random points on the disk in polar coordinates with r = sqrt(random()); theta = 2 * pi * random().
Now instead of randomly sampling this inverse function we’re uniformly sampling it, and the nice thing about uniform sampling is that our results about how points are spread out in the limit of large N will behave as if we had randomly sampled it. This combination is the trick. Instead of random() we use (arange(0, num_pts, dtype=float) + 0.5)/num_pts, so that, say, if we want to sample 10 points they are r = 0.05, 0.15, 0.25, ... 0.95. We uniformly sample r to get equal-area spacing, and we use the sunflower increment to avoid awful “bars” of points in the output.
Now doing the sunflower on a sphere
The changes that we need to make to dot the sphere with points merely involve switching out the polar coordinates for spherical coordinates. The radial coordinate of course doesn't enter into this because we're on a unit sphere. To keep things a little more consistent here, even though I was trained as a physicist I'll use mathematicians' coordinates where 0 ≤ φ ≤ π is latitude coming down from the pole and 0 ≤ θ ≤ 2π is longitude. So the difference from above is that we are basically replacing the variable r with φ.
Our area element, which was r dr dθ, now becomes the not-much-more-complicated sin(φ) dφ dθ. So our joint density for uniform spacing is sin(φ)/4π. Integrating out θ, we find f(φ) = sin(φ)/2, thus F(φ) = (1 − cos(φ))/2. Inverting this we can see that a uniform random variable would look like acos(1 - 2 u), but we sample uniformly instead of randomly, so we instead use φk = acos(1 − 2 (k + 0.5)/N). And the rest of the algorithm is just projecting this onto the x, y, and z coordinates:
from numpy import pi, cos, sin, arccos, arange
import mpl_toolkits.mplot3d
import matplotlib.pyplot as pp
num_pts = 1000
indices = arange(0, num_pts, dtype=float) + 0.5
phi = arccos(1 - 2*indices/num_pts)
theta = pi * (1 + 5**0.5) * indices
x, y, z = cos(theta) * sin(phi), sin(theta) * sin(phi), cos(phi);
pp.figure().add_subplot(111, projection='3d').scatter(x, y, z);
pp.show()
Again for n=100 and n=1000 the results look like:
Further research
I wanted to give a shout out to Martin Roberts’s blog. Note that above I created an offset of my indices by adding 0.5 to each index. This was just visually appealing to me, but it turns out that the choice of offset matters a lot and is not constant over the interval and can mean getting as much as 8% better accuracy in packing if chosen correctly. There should also be a way to get his R2 sequence to cover a sphere and it would be interesting to see if this also produced a nice even covering, perhaps as-is but perhaps needing to be, say, taken from only a half of the unit square cut diagonally or so and stretched around to get a circle.
Notes
Those “bars” are formed by rational approximations to a number, and the best rational approximations to a number come from its continued fraction expression, z + 1/(n_1 + 1/(n_2 + 1/(n_3 + ...))) where z is an integer and n_1, n_2, n_3, ... is either a finite or infinite sequence of positive integers:
def continued_fraction(r):
while r != 0:
n = floor(r)
yield n
r = 1/(r - n)
Since the fraction part 1/(...) is always between zero and one, a large integer in the continued fraction allows for a particularly good rational approximation: “one divided by something between 100 and 101” is better than “one divided by something between 1 and 2.” The most irrational number is therefore the one which is 1 + 1/(1 + 1/(1 + ...)) and has no particularly good rational approximations; one can solve φ = 1 + 1/φ by multiplying through by φ to get the formula for the golden ratio.
For folks who are not so familiar with NumPy -- all of the functions are “vectorized,” so that sqrt(array) is the same as what other languages might write map(sqrt, array). So this is a component-by-component sqrt application. The same also holds for division by a scalar or addition with scalars -- those apply to all components in parallel.
The proof is simple once you know that this is the result. If you ask what's the probability that z < Z < z + dz, this is the same as asking what's the probability that z < F-1(U) < z + dz, apply F to all three expressions noting that it is a monotonically increasing function, hence F(z) < U < F(z + dz), expand the right hand side out to find F(z) + f(z) dz, and since U is uniform this probability is just f(z) dz as promised.

This is known as packing points on a sphere, and there is no (known) general, perfect solution. However, there are plenty of imperfect solutions. The three most popular seem to be:
Create a simulation. Treat each point as an electron constrained to a sphere, then run a simulation for a certain number of steps. The electrons' repulsion will naturally tend the system to a more stable state, where the points are about as far away from each other as they can get.
Hypercube rejection. This fancy-sounding method is actually really simple: you uniformly choose points (much more than n of them) inside of the cube surrounding the sphere, then reject the points outside of the sphere. Treat the remaining points as vectors, and normalize them. These are your "samples" - choose n of them using some method (randomly, greedy, etc).
Spiral approximations. You trace a spiral around a sphere, and evenly-distribute the points around the spiral. Because of the mathematics involved, these are more complicated to understand than the simulation, but much faster (and probably involving less code). The most popular seems to be by Saff, et al.
A lot more information about this problem can be found here

In this example code node[k] is just the kth node. You are generating an array N points and node[k] is the kth (from 0 to N-1). If that is all that is confusing you, hopefully you can use that now.
(in other words, k is an array of size N that is defined before the code fragment starts, and which contains a list of the points).
Alternatively, building on the other answer here (and using Python):
> cat ll.py
from math import asin
nx = 4; ny = 5
for x in range(nx):
lon = 360 * ((x+0.5) / nx)
for y in range(ny):
midpt = (y+0.5) / ny
lat = 180 * asin(2*((y+0.5)/ny-0.5))
print lon,lat
> python2.7 ll.py
45.0 -166.91313924
45.0 -74.0730322921
45.0 0.0
45.0 74.0730322921
45.0 166.91313924
135.0 -166.91313924
135.0 -74.0730322921
135.0 0.0
135.0 74.0730322921
135.0 166.91313924
225.0 -166.91313924
225.0 -74.0730322921
225.0 0.0
225.0 74.0730322921
225.0 166.91313924
315.0 -166.91313924
315.0 -74.0730322921
315.0 0.0
315.0 74.0730322921
315.0 166.91313924
If you plot that, you'll see that the vertical spacing is larger near the poles so that each point is situated in about the same total area of space (near the poles there's less space "horizontally", so it gives more "vertically").
This isn't the same as all points having about the same distance to their neighbours (which is what I think your links are talking about), but it may be sufficient for what you want and improves on simply making a uniform lat/lon grid.

What you are looking for is called a spherical covering. The spherical covering problem is very hard and solutions are unknown except for small numbers of points. One thing that is known for sure is that given n points on a sphere, there always exist two points of distance d = (4-csc^2(\pi n/6(n-2)))^(1/2) or closer.
If you want a probabilistic method for generating points uniformly distributed on a sphere, it's easy: generate points in space uniformly by Gaussian distribution (it's built into Java, not hard to find the code for other languages). So in 3-dimensional space, you need something like
Random r = new Random();
double[] p = { r.nextGaussian(), r.nextGaussian(), r.nextGaussian() };
Then project the point onto the sphere by normalizing its distance from the origin
double norm = Math.sqrt( (p[0])^2 + (p[1])^2 + (p[2])^2 );
double[] sphereRandomPoint = { p[0]/norm, p[1]/norm, p[2]/norm };
The Gaussian distribution in n dimensions is spherically symmetric so the projection onto the sphere is uniform.
Of course, there's no guarantee that the distance between any two points in a collection of uniformly generated points will be bounded below, so you can use rejection to enforce any such conditions that you might have: probably it's best to generate the whole collection and then reject the whole collection if necessary. (Or use "early rejection" to reject the whole collection you've generated so far; just don't keep some points and drop others.) You can use the formula for d given above, minus some slack, to determine the min distance between points below which you will reject a set of points. You'll have to calculate n choose 2 distances, and the probability of rejection will depend on the slack; it's hard to say how, so run a simulation to get a feel for the relevant statistics.

This answer is based on the same 'theory' that is outlined well by this answer
I'm adding this answer as:
-- None of the other options fit the 'uniformity' need 'spot-on' (or not obviously-clearly so). (Noting to get the planet like distribution looking behavior particurally wanted in the original ask, you just reject from the finite list of the k uniformly created points at random (random wrt the index count in the k items back).)
--The closest other impl forced you to decide the 'N' by 'angular axis', vs. just 'one value of N' across both angular axis values ( which at low counts of N is very tricky to know what may, or may not matter (e.g. you want '5' points -- have fun ) )
--Furthermore, it's very hard to 'grok' how to differentiate between the other options without any imagery, so here's what this option looks like (below), and the ready-to-run implementation that goes with it.
with N at 20:
and then N at 80:
here's the ready-to-run python3 code, where the emulation is that same source: " http://web.archive.org/web/20120421191837/http://www.cgafaq.info/wiki/Evenly_distributed_points_on_sphere " found by others. ( The plotting I've included, that fires when run as 'main,' is taken from: http://www.scipy.org/Cookbook/Matplotlib/mplot3D )
from math import cos, sin, pi, sqrt
def GetPointsEquiAngularlyDistancedOnSphere(numberOfPoints=45):
""" each point you get will be of form 'x, y, z'; in cartesian coordinates
eg. the 'l2 distance' from the origion [0., 0., 0.] for each point will be 1.0
------------
converted from: http://web.archive.org/web/20120421191837/http://www.cgafaq.info/wiki/Evenly_distributed_points_on_sphere )
"""
dlong = pi*(3.0-sqrt(5.0)) # ~2.39996323
dz = 2.0/numberOfPoints
long = 0.0
z = 1.0 - dz/2.0
ptsOnSphere =[]
for k in range( 0, numberOfPoints):
r = sqrt(1.0-z*z)
ptNew = (cos(long)*r, sin(long)*r, z)
ptsOnSphere.append( ptNew )
z = z - dz
long = long + dlong
return ptsOnSphere
if __name__ == '__main__':
ptsOnSphere = GetPointsEquiAngularlyDistancedOnSphere( 80)
#toggle True/False to print them
if( True ):
for pt in ptsOnSphere: print( pt)
#toggle True/False to plot them
if(True):
from numpy import *
import pylab as p
import mpl_toolkits.mplot3d.axes3d as p3
fig=p.figure()
ax = p3.Axes3D(fig)
x_s=[];y_s=[]; z_s=[]
for pt in ptsOnSphere:
x_s.append( pt[0]); y_s.append( pt[1]); z_s.append( pt[2])
ax.scatter3D( array( x_s), array( y_s), array( z_s) )
ax.set_xlabel('X'); ax.set_ylabel('Y'); ax.set_zlabel('Z')
p.show()
#end
tested at low counts (N in 2, 5, 7, 13, etc) and seems to work 'nice'

Try:
function sphere ( N:float,k:int):Vector3 {
var inc = Mathf.PI * (3 - Mathf.Sqrt(5));
var off = 2 / N;
var y = k * off - 1 + (off / 2);
var r = Mathf.Sqrt(1 - y*y);
var phi = k * inc;
return Vector3((Mathf.Cos(phi)*r), y, Mathf.Sin(phi)*r);
};
The above function should run in loop with N loop total and k loop current iteration.
It is based on a sunflower seeds pattern, except the sunflower seeds are curved around into a half dome, and again into a sphere.
Here is a picture, except I put the camera half way inside the sphere so it looks 2d instead of 3d because the camera is same distance from all points.
http://3.bp.blogspot.com/-9lbPHLccQHA/USXf88_bvVI/AAAAAAAAADY/j7qhQsSZsA8/s640/sphere.jpg

Healpix solves a closely related problem (pixelating the sphere with equal area pixels):
http://healpix.sourceforge.net/
It's probably overkill, but maybe after looking at it you'll realize some of it's other nice properties are interesting to you. It's way more than just a function that outputs a point cloud.
I landed here trying to find it again; the name "healpix" doesn't exactly evoke spheres...

edit: This does not answer the question the OP meant to ask, leaving it here in case people find it useful somehow.
We use the multiplication rule of probability, combined with infinitessimals. This results in 2 lines of code to achieve your desired result:
longitude: φ = uniform([0,2pi))
azimuth: θ = -arcsin(1 - 2*uniform([0,1]))
(defined in the following coordinate system:)
Your language typically has a uniform random number primitive. For example in python you can use random.random() to return a number in the range [0,1). You can multiply this number by k to get a random number in the range [0,k). Thus in python, uniform([0,2pi)) would mean random.random()*2*math.pi.
Proof
Now we can't assign θ uniformly, otherwise we'd get clumping at the poles. We wish to assign probabilities proportional to the surface area of the spherical wedge (the θ in this diagram is actually φ):
An angular displacement dφ at the equator will result in a displacement of dφ*r. What will that displacement be at an arbitrary azimuth θ? Well, the radius from the z-axis is r*sin(θ), so the arclength of that "latitude" intersecting the wedge is dφ * r*sin(θ). Thus we calculate the cumulative distribution of the area to sample from it, by integrating the area of the slice from the south pole to the north pole.
(where stuff=dφ*r)
We will now attempt to get the inverse of the CDF to sample from it: http://en.wikipedia.org/wiki/Inverse_transform_sampling
First we normalize by dividing our almost-CDF by its maximum value. This has the side-effect of cancelling out the dφ and r.
azimuthalCDF: cumProb = (sin(θ)+1)/2 from -pi/2 to pi/2
inverseCDF: θ = -sin^(-1)(1 - 2*cumProb)
Thus:
let x by a random float in range [0,1]
θ = -arcsin(1-2*x)

with small numbers of points you could run a simulation:
from random import random,randint
r = 10
n = 20
best_closest_d = 0
best_points = []
points = [(r,0,0) for i in range(n)]
for simulation in range(10000):
x = random()*r
y = random()*r
z = r-(x**2+y**2)**0.5
if randint(0,1):
x = -x
if randint(0,1):
y = -y
if randint(0,1):
z = -z
closest_dist = (2*r)**2
closest_index = None
for i in range(n):
for j in range(n):
if i==j:
continue
p1,p2 = points[i],points[j]
x1,y1,z1 = p1
x2,y2,z2 = p2
d = (x1-x2)**2+(y1-y2)**2+(z1-z2)**2
if d < closest_dist:
closest_dist = d
closest_index = i
if simulation % 100 == 0:
print simulation,closest_dist
if closest_dist > best_closest_d:
best_closest_d = closest_dist
best_points = points[:]
points[closest_index]=(x,y,z)
print best_points
>>> best_points
[(9.921692138442777, -9.930808529773849, 4.037839326088124),
(5.141893371460546, 1.7274947332807744, -4.575674650522637),
(-4.917695758662436, -1.090127967097737, -4.9629263893193745),
(3.6164803265540666, 7.004158551438312, -2.1172868271109184),
(-9.550655088997003, -9.580386054762917, 3.5277052594769422),
(-0.062238110294250415, 6.803105171979587, 3.1966101417463655),
(-9.600996012203195, 9.488067284474834, -3.498242301168819),
(-8.601522086624803, 4.519484132245867, -0.2834204048792728),
(-1.1198210500791472, -2.2916581379035694, 7.44937337008726),
(7.981831370440529, 8.539378431788634, 1.6889099589074377),
(0.513546008372332, -2.974333486904779, -6.981657873262494),
(-4.13615438946178, -6.707488383678717, 2.1197605651446807),
(2.2859494919024326, -8.14336582650039, 1.5418694699275672),
(-7.241410895247996, 9.907335206038226, 2.271647103735541),
(-9.433349952523232, -7.999106443463781, -2.3682575660694347),
(3.704772125650199, 1.0526567864085812, 6.148581714099761),
(-3.5710511242327048, 5.512552040316693, -3.4318468250897647),
(-7.483466337225052, -1.506434920354559, 2.36641535124918),
(7.73363824231576, -8.460241422163824, -1.4623228616326003),
(10, 0, 0)]

Take the two largest factors of your N, if N==20 then the two largest factors are {5,4}, or, more generally {a,b}. Calculate
dlat = 180/(a+1)
dlong = 360/(b+1})
Put your first point at {90-dlat/2,(dlong/2)-180}, your second at {90-dlat/2,(3*dlong/2)-180}, your 3rd at {90-dlat/2,(5*dlong/2)-180}, until you've tripped round the world once, by which time you've got to about {75,150} when you go next to {90-3*dlat/2,(dlong/2)-180}.
Obviously I'm working this in degrees on the surface of the spherical earth, with the usual conventions for translating +/- to N/S or E/W. And obviously this gives you a completely non-random distribution, but it is uniform and the points are not bunched together.
To add some degree of randomness, you could generate 2 normally-distributed (with mean 0 and std dev of {dlat/3, dlong/3} as appropriate) and add them to your uniformly distributed points.

OR... to place 20 points, compute the centers of the icosahedronal faces. For 12 points, find the vertices of the icosahedron. For 30 points, the mid point of the edges of the icosahedron. you can do the same thing with the tetrahedron, cube, dodecahedron and octahedrons: one set of points is on the vertices, another on the center of the face and another on the center of the edges. They cannot be mixed, however.

Based on fnord's answer, here is a Unity3D version with added ranges :
Code :
// golden angle in radians
static float Phi = Mathf.PI * ( 3f - Mathf.Sqrt( 5f ) );
static float Pi2 = Mathf.PI * 2;
public static Vector3 Point( float radius , int index , int total , float min = 0f, float max = 1f , float angleStartDeg = 0f, float angleRangeDeg = 360 )
{
// y goes from min (-) to max (+)
var y = ( ( index / ( total - 1f ) ) * ( max - min ) + min ) * 2f - 1f;
// golden angle increment
var theta = Phi * index ;
if( angleStartDeg != 0 || angleRangeDeg != 360 )
{
theta = ( theta % ( Pi2 ) ) ;
theta = theta < 0 ? theta + Pi2 : theta ;
var a1 = angleStartDeg * Mathf.Deg2Rad;
var a2 = angleRangeDeg * Mathf.Deg2Rad;
theta = theta * a2 / Pi2 + a1;
}
// https://stackoverflow.com/a/26127012/2496170
// radius at y
var rY = Mathf.Sqrt( 1 - y * y );
var x = Mathf.Cos( theta ) * rY;
var z = Mathf.Sin( theta ) * rY;
return new Vector3( x, y, z ) * radius;
}
Gist : https://gist.github.com/nukadelic/7449f0872f708065bc1afeb19df666f7/edit
Preview:

# create uniform spiral grid
numOfPoints = varargin[0]
vxyz = zeros((numOfPoints,3),dtype=float)
sq0 = 0.00033333333**2
sq2 = 0.9999998**2
sumsq = 2*sq0 + sq2
vxyz[numOfPoints -1] = array([(sqrt(sq0/sumsq)),
(sqrt(sq0/sumsq)),
(-sqrt(sq2/sumsq))])
vxyz[0] = -vxyz[numOfPoints -1]
phi2 = sqrt(5)*0.5 + 2.5
rootCnt = sqrt(numOfPoints)
prevLongitude = 0
for index in arange(1, (numOfPoints -1), 1, dtype=float):
zInc = (2*index)/(numOfPoints) -1
radius = sqrt(1-zInc**2)
longitude = phi2/(rootCnt*radius)
longitude = longitude + prevLongitude
while (longitude > 2*pi):
longitude = longitude - 2*pi
prevLongitude = longitude
if (longitude > pi):
longitude = longitude - 2*pi
latitude = arccos(zInc) - pi/2
vxyz[index] = array([ (cos(latitude) * cos(longitude)) ,
(cos(latitude) * sin(longitude)),
sin(latitude)])

#robert king It's a really nice solution but has some sloppy bugs in it. I know it helped me a lot though, so never mind the sloppiness. :)
Here is a cleaned up version....
from math import pi, asin, sin, degrees
halfpi, twopi = .5 * pi, 2 * pi
sphere_area = lambda R=1.0: 4 * pi * R ** 2
lat_dist = lambda lat, R=1.0: R*(1-sin(lat))
#A = 2*pi*R^2(1-sin(lat))
def sphere_latarea(lat, R=1.0):
if -halfpi > lat or lat > halfpi:
raise ValueError("lat must be between -halfpi and halfpi")
return 2 * pi * R ** 2 * (1-sin(lat))
sphere_lonarea = lambda lon, R=1.0: \
4 * pi * R ** 2 * lon / twopi
#A = 2*pi*R^2 |sin(lat1)-sin(lat2)| |lon1-lon2|/360
# = (pi/180)R^2 |sin(lat1)-sin(lat2)| |lon1-lon2|
sphere_rectarea = lambda lat0, lat1, lon0, lon1, R=1.0: \
(sphere_latarea(lat0, R)-sphere_latarea(lat1, R)) * (lon1-lon0) / twopi
def test_sphere(n_lats=10, n_lons=19, radius=540.0):
total_area = 0.0
for i_lons in range(n_lons):
lon0 = twopi * float(i_lons) / n_lons
lon1 = twopi * float(i_lons+1) / n_lons
for i_lats in range(n_lats):
lat0 = asin(2 * float(i_lats) / n_lats - 1)
lat1 = asin(2 * float(i_lats+1)/n_lats - 1)
area = sphere_rectarea(lat0, lat1, lon0, lon1, radius)
print("{:} {:}: {:9.4f} to {:9.4f}, {:9.4f} to {:9.4f} => area {:10.4f}"
.format(i_lats, i_lons
, degrees(lat0), degrees(lat1)
, degrees(lon0), degrees(lon1)
, area))
total_area += area
print("total_area = {:10.4f} (difference of {:10.4f})"
.format(total_area, abs(total_area) - sphere_area(radius)))
test_sphere()

This works and it's deadly simple. As many points as you want:
private function moveTweets():void {
var newScale:Number=Scale(meshes.length,50,500,6,2);
trace("new scale:"+newScale);
var l:Number=this.meshes.length;
var tweetMeshInstance:TweetMesh;
var destx:Number;
var desty:Number;
var destz:Number;
for (var i:Number=0;i<this.meshes.length;i++){
tweetMeshInstance=meshes[i];
var phi:Number = Math.acos( -1 + ( 2 * i ) / l );
var theta:Number = Math.sqrt( l * Math.PI ) * phi;
tweetMeshInstance.origX = (sphereRadius+5) * Math.cos( theta ) * Math.sin( phi );
tweetMeshInstance.origY= (sphereRadius+5) * Math.sin( theta ) * Math.sin( phi );
tweetMeshInstance.origZ = (sphereRadius+5) * Math.cos( phi );
destx=sphereRadius * Math.cos( theta ) * Math.sin( phi );
desty=sphereRadius * Math.sin( theta ) * Math.sin( phi );
destz=sphereRadius * Math.cos( phi );
tweetMeshInstance.lookAt(new Vector3D());
TweenMax.to(tweetMeshInstance, 1, {scaleX:newScale,scaleY:newScale,x:destx,y:desty,z:destz,onUpdate:onLookAtTween, onUpdateParams:[tweetMeshInstance]});
}
}
private function onLookAtTween(theMesh:TweetMesh):void {
theMesh.lookAt(new Vector3D());
}