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# Rayleigh test, getting only one answer from function (Python3)

By : user2175952
Date : October 14 2020, 02:24 PM
it fixes the issue Your main() function contains list1 = list(map(rayleightest,t,f)). This is how you get one answer per pair of values in t and f and got a bunch of values as a result. In other words the code computes rayleightest(t[0], f[0]), then rayleightest(t[1], f[1]), and so on.
When you use the code z = rayleightest(t, f) you pass 2 numpy arrays instead of 2 numbers. The result is a single value because the formula in rayleightest() contains np.sum() - you sum up all the values in arrays in a single value.
code :

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## ALLFITDIST function in MATLAB return 'rayleigh' as a best fit for normally distributed data

By : user2305787
Date : March 29 2020, 07:55 AM
Any of those help Following user3278640's answer, I tried out several versions of matlab. 2012b showed the normal distribution, while 2013b and 2014b showed the rayleigh as the best approximation.
While inspecting the values, it seems that the negative log-likelihood of the rayleigh in 2013b and 2014b is a complex number. I don't think that makes much sense, so I changed the code around line 209 from:
code :
``````%If NLL is non-finite number, produce error to ignore distribution
if ~isfinite(NLL)
error('non-finite NLL');
end
``````
``````%If NLL is non-finite or complex number, produce error to ignore
%distribution
if ~isfinite(NLL) || ~isreal(NLL)
error('non-finite or non-real NLL');
end
``````

## Generates rayleigh distribution using Jmeter

By : Danil Kudabaev
Date : March 29 2020, 07:55 AM
around this issue You would need to implement the algorithm using for example:
JSR223Timer Groovy as a language in the timer And one of those algorithm implementations: How to generate a random number with a specific probability density function?
code :
``````import org.apache.commons.math3.distribution.WeibullDistribution;
alpha = Integer.parseInt(vars.get("alpha"));
beta = Integer.parseInt(vars.get("beta"));
rng= Integer.parseInt(bsh.args[0]);
try {
WeibullDistribution g= new WeibullDistribution(rng, alpha,beta);
return g.sample();
} catch (Throwable ex) {
log.error("Something wrong", ex);
throw ex;
}
``````

## MATLAB pdf of filtered Rayleigh distribution

By : ye zihui
Date : March 29 2020, 07:55 AM
I wish this help you To get the pdf of a filtered Rayleigh Distribution, you have to take the original pdf equation and substitute any instance of sigma^2 with the mean square value of the filtered rayleigh distribution. So, the equation becomes
2x/MSV * exp(-x^2 / MSV)
code :
``````            x1 = randn(N, 1);
y1 = randn(N, 1);
x1_LPF = filter(LPF, 1, x1);
y1_LPF = filter(LPF, 1, y1);
ray1_f = abs(x1_LPF + 1i*y1_LPF);
range = [0:0.01:4];
subplot(1, 2, 2);
histogram(ray1_f, 'Normalization', 'pdf');
title('Normalized Histogram of Raleigh Distribution (filtered)')
xlabel('Random Variable')
ylabel('Probability')

mean_square = mean(ray1_f .^ 2);
filter_theory = (range) * 2 / mean_square .* exp( - (range.^2) ./ mean_square);
hold on
plot(range, filter_theory, 'Linewidth', 1.5)
xlim([0 .6])
legend('Simulation', 'Theoretical')
``````

By : invinci
Date : March 29 2020, 07:55 AM
hop of those help? Your tests failed because you are using the non-breaking space. The error code on Codewars is showing:
code :
``````let space = " ";
``````

## rayleigh distribution curve on histogram

Date : March 29 2020, 07:55 AM
will be helpful for those in need You might need to first follow the link to rv_continuous, from which rayleigh is subclassed. And from there to the ppf to find out that ppf is the 'Percent point function'. x0 = ppf(0.01) tells at which spot everything less than x0 has accumulated 1% of its total 'weight' and similarly x1 = ppf(0.99) is where 99% of the 'weight' is accumulated. np.linspace(x0, x1, 100) divides the space from x0 to x1 in 100 short intervals. As a continuous distribution can be infinite, these x0 and x1 limits are needed to only show the interesting interval.
rayleigh.pdf(x) gives the pdf at x. So, an indication of how probable each x is.
code :
``````estimated_rayleigh_scale = samples.mean() / np.sqrt(np.pi / 2)
``````
``````from matplotlib import pyplot as plt
import numpy as np
from scipy.stats import rayleigh

N = 1000
# V = np.random.uniform(0, 0.1, 2*N).reshape((N,2))
# V_norm = (np.linalg.norm(V, axis=1))
scale = 0.08
V_norm_hist = scale * np.sqrt( -2* np.log (np.random.uniform(0, 1, N)))
fig, ax = plt.subplots(1, 1)

num_bins = 60
_binvalues, bins, _patches = plt.hist(V_norm_hist, bins=num_bins, density=False, rwidth=1, ec='white', label='Histogram')
x = np.linspace(bins[0], bins[-1], 100)
binwidth = (bins[-1] - bins[0]) / num_bins

scale = V_norm_hist.mean() / np.sqrt(np.pi / 2)

plt.plot(x, rayleigh(loc=0, scale=scale).pdf(x)*len(V_norm_hist)*binwidth, lw=5, alpha=0.6, label=f'Rayleigh pdf (s={scale:.3f})')
plt.legend()
plt.show()
``````