I have several test results of server response delay. According to our theory analysis, the delay distribution should have heavy-tail behavior. But how could I prove that the test result does follow heavy-tail distribution?
I think the most straightforward way is to make a best fit of half normal distribution to and look how well it describe the tail.
In Python you can make this with the help of scipy.stats.halfnorm.fit() or use a longtail module that is dedicated for plotting and analysing heavy tails (https://github.com/Mottl/longtail):
import numpy as np
import longtail
# generate random values from heavy tailed distribution (let's take Laplace)
X = np.random.laplace(size=10000)
X = X[X>0] # take only right half of the distribution
# get best fit of half normal distribution to our data:
params = longtail.fit_distributions(X, distributions=['halfnorm'])
# visualize X and best fit:
longtail.plot(X, params=params)
Since points at the tail are above the half normal approximation, the given distribution can be considered as having heavier tail than half normal:
I am not an expert, but I think that estimating the kurtosis of your delay distribution would be a good start.
If you know the theoretical delay distribution, you can also do a goodness of fit test.
Related
I have noticed that my gensim Doc2Vec (DBOW) model is sensitive to document tags. My understanding was that these tags are cosmetic and so they should not influence the learned embeddings. Am I misunderstanding something? Here is a minimal example:
from gensim.test.utils import common_texts
from gensim.models.doc2vec import Doc2Vec, TaggedDocument
import numpy as np
import os
os.environ['PYTHONHASHSEED'] = '0'
reps = []
for a in [0,500]:
documents = [TaggedDocument(doc, [i + a])
for i, doc in enumerate(common_texts)]
model = Doc2Vec(documents, vector_size=100, window=2, min_count=0,
workers=1, epochs=10, dm=0, seed=0)
reps.append(np.array([model.docvecs[k] for k in range(len(common_texts))])
reps[0].sum() == reps[1].sum()
This last line returns False. I am working with gensim 3.8.3 and Python 3.5.2. More generally, is there any role that the values of the tags play (assuming they are unique)? I ask because I have found that using different tags for documents in a classification task leads to widely varying performance.
Thanks in advance.
First & foremost, your test isn't even comparing vectors corresponding to the same texts!
In run #1, the vector for the 1st text in in model.docvecs[0]. In run #2, the vector for the 1st text is in model.docvecs[1].
And, in run #2, the vector at model.docvecs[0] is just a randomly-initialized, but never-trained, vector - because none of the training texts had a document tag of (int) 0. (If using pure ints as the doc-tags, Doc2Vec uses them as literal indexes - potentially leaving any unused slots less than your highest tag allocated-and-initialized, but never-trained.)
Since common_texts only has 11 entries, by the time you reach run #12, all the vectors in your reps array of the first 11 vectors are garbage uncorrelated with any of your texts/
However, even after correcting that:
As explained in the Gensim FAQ answer #11, determinism in this algorithm shouldn't generally be expected, given many sources of potential randomness, and the fuzzy/approximate nature of the whole approach. If you're relying on it, or testing for it, you're probably making some unwarranted assumptions.
In general, tests of these algorithms should be evaluating "roughly equivalent usefulness in comparative uses" rather than "identical (or even similar) specific vectors". For example, a test whether apple and orange are roughly at the same positions in each others' nearest-neighbor rankings makes more sense than checking their (somewhat arbitrary) exact vector positions or even cosine-similarity.
Additionally:
tiny toy datasets like common_texts won't show the algorithm's usual behavior/benefits
PYTHONHASHSEED is only consulted by the Python interpreter at startup; setting it from Python can't have any effect. But also, the kind of indeterminism it introduces only comes up with separate interpreter launches: a tight loop within a single interpreter run like this wouldn't be affected by that in any case.
Have you checked the magnitude of the differences?
Just running:
delta = reps[0].sum() - reps[1].sum()
for the aggregate differences results with -1.2598932e-05 when I run it.
Comparison dimension-wise:
eps = 10**-4
over = (np.abs(diff) <= eps).all()
Returns True on a vast majority of the runs which means that you are getting quite reproducible results given the complexity of the calculations.
I would blame numerical stability of the calculations or uncontrolled randomness. Even though you do try to control the random seed, there is a different random seed in NumPy and different in random standard library so you are not controlling for all of the sources of randomness. This can also have an influence on the results but I did not check the actual implementation in gensim and it's dependencies.
Change
import os
os.environ['PYTHONHASHSEED'] = '0'
to
import os
import sys
hashseed = os.getenv('PYTHONHASHSEED')
if not hashseed:
os.environ['PYTHONHASHSEED'] = '0'
os.execv(sys.executable, [sys.executable] + sys.argv)
I'm a physics PhD student solving lubrication equations (non-linear PDEs) related to the evaporation of binary droplets in a cylindrical pixel in order to simulate the shape evolution. So I split the system into N ODEs, each representing the height evolution (dh/dt) at point i, write as finite differences, and solve simultaneously using NDSolve. For single-liquid droplets this works perfectly, the droplet evaporates cleanly.
However for binary droplets I include N additional ODEs for the composition fraction evolution (dX/dt) at each point. This also adds a new term (marangoni stress) to the dh/dt equations. Immediately NDSolve says:
At t == 0.00029140763302667777`, step size is effectively zero;
singularity or stiff system suspected.
I plot the droplet height at this t-value and it shows a narrow spike at the origin (clearly exploding, hence the stiffness); plotting the composition shows a narrow plummet at the origin (also exploding, just negative. Also, the physics obviously shouldn't permit the composition fraction to be below 0).
Finally, both sets of equations have terms depending on dX/dr, the composition gradient. If this is set to zero and I also set the evaporation interaction to zero (meaning the two liquids evaporate at the same rate and there can be no X gradient) there should be no change in X anywhere and it should reduce to the single liquid case (dX/dt = 0 and dh/dt no longer depends on X). However, the procedure is introducing some small gradient in X nonetheless, which explodes and causes the same numerical instability.
My question is this: is there anything about NDSolve that might be causing this? I've been over the equations and discretisation a hundred times and am sure it's correct. I've also looked into the documentation of NDSolve, but didn't find anything that helps me. Could it be introducing a small numerical error in the composition gradient?
I can post an MRE code below, but it's pretty dense and obviously written in mathematica code (doesn't transfer to the real world well...) so I don't know how much it'd help. Anyway thank you for reading this!
I'm not sure StackOverflow is the right place to ask this question, because this question is half-programming and half-mathematics. And also really sorry if my question is stupid ^_^
I'm studying about Monte Carlo simulations via the "Monte Carlo Methods" book. One of the first thing I must learn is about Random Number Generator. The basic algorithm of RNG is:
1. Initialize: Draw the seed S0 from the distribution µ on S. Set t = 1.
2. Transition: Set St = f(St−1).
3. Output: Set Ut = g(St).
4. Repeat: Set t = t+ 1 and return to Step 2.
(µ is a probability distribution on the finite set of states S, the input is S0 and the random number we desire it the output Ut)
It is not hard to understand, but the problem here is I don't see the random factor which lie in the number of repeat. How can we decide when to stop the loop of the RNG? All examples I read which implement a RNG are loop for 100 times, and they returns the same value for a specific seed. It is not random at all >_<
Can someone explain what I'm missing here? Any help will be appreciated. Thanks everyone
You can't get a true sequence of random numbers on a computer, without specialized hardware. (Such specialized hardware performs the equivalent of an initial roll of the dice using physics to provide the randomness. Electronic ones often use the electronic noise of specialized diodes at constant temperatures; others use radioactive decay events.)
Without that specialized hardware, what you can generate are pseudorandom numbers which, as you've observed, always generate the same sequence of numbers for the same initial seed. For simple applications, you can often get away with generating an initial seed from the time of invocation, which is effectively random.
And when I say "simple applications," I am excluding cryptography. (Not just that, but especially that.)
Sometimes when you are trying to debug a simulation, you actually want to have a reproducible stream of "random" numbers so you might specifically sent a stream to start with a specific seed.
For instance in the answer Creating a facet_wrap plot with ggplot2 with different annotations in each plot rcs starts the answer by creating a reproducible set of data using the R code
set.seed(1)
df <- data.frame(x=rnorm(300), y=rnorm(300), cl=gl(3,100)) # create test data
before going on to demonstrate how to answer the actual question.
I have two audio recordings of a same signal by 2 different microphones (for example, in a WAV format), but one of them is recorded with delay, for example, several seconds.
It's easy to identify such a delay visually when viewing these signals in some kind of waveform viewer - i.e. just spotting first visible peak in every signal and ensuring that they're the same shape:
(source: greycat.ru)
But how do I do it programmatically - find out what this delay (t) is? Two digitized signals are slightly different (because microphones are different, were at different positions, due to ADC setups, etc).
I've digged around a bit and found out that this problem is usually called "time-delay estimation" and it has myriads of approaches to it - for example, one of them.
But are there any simple and ready-made solutions, such as command-line utility, library or straight-forward algorithm available?
Conclusion: I've found no simple implementation and done a simple command-line utility myself - available at https://bitbucket.org/GreyCat/calc-sound-delay (GPLv3-licensed). It implements a very simple search-for-maximum algorithm described at Wikipedia.
The technique you're looking for is called cross correlation. It's a very simple, if somewhat compute intensive technique which can be used for solving various problems, including measuring the time difference (aka lag) between two similar signals (the signals do not need to be identical).
If you have a reasonable idea of your lag value (or at least the range of lag values that are expected) then you can reduce the total amount of computation considerably. Ditto if you can put a definite limit on how much accuracy you need.
Having had the same problem and without success to find a tool to sync the start of video/audio recordings automatically,
I decided to make syncstart (github).
It is a command line tool. The basic code behind it is this:
import numpy as np
from scipy import fft
from scipy.io import wavfile
r1,s1 = wavfile.read(in1)
r2,s2 = wavfile.read(in2)
assert r1==r2, "syncstart normalizes using ffmpeg"
fs = r1
ls1 = len(s1)
ls2 = len(s2)
padsize = ls1+ls2+1
padsize = 2**(int(np.log(padsize)/np.log(2))+1)
s1pad = np.zeros(padsize)
s1pad[:ls1] = s1
s2pad = np.zeros(padsize)
s2pad[:ls2] = s2
corr = fft.ifft(fft.fft(s1pad)*np.conj(fft.fft(s2pad)))
ca = np.absolute(corr)
xmax = np.argmax(ca)
if xmax > padsize // 2:
file,offset = in2,(padsize-xmax)/fs
else:
file,offset = in1,xmax/fs
A very straight forward thing todo is just to check if the peaks exceed some threshold, the time between high-peak on line A and high-peak on line B is probably your delay. Just try tinkering a bit with the thresholds and if the graphs are usually as clear as the picture you posted, then you should be fine.
This picture from Wikipedia has a nice example of the sort of functions I'd ideally like to generate:
Right now I'm using the Irwin-Hall Distribution, which is more or less a polynomial approximation of the Gaussian distribution...basically, you use uniform random number generator and iterate it x times, and take the average. The more iterations, the more like a Gaussian Distribution it is.
It's pretty nice; however I'd like to be able to have one where I can vary the mean. For example, let's say I wanted a number between the range 0 and 10, but around 7. Like, the mean (if I repeated this function multiple times) would turn out to be 7, but the actual range is 0-10.
Is there one I should look up, or should I work on doing some fancy maths with standard Gaussian distributions?
I see a contradiction in your question. From one side you want normal distribution which is symmetrical by it's nature, from other side you want the range asymmetrically disposed to mean value.
I suspect you should try to look at other distributions density functions of which are like bell curve but asymmetrical. Like log distribution or beta distribution.
Look into generating normal random variates. You can generate pairs of normal random variates X = N(0,1) and tranform it into ANY normal random variate Y = N(m,s) (Y = m + s*X).
Sounds like the Truncated Normal distribution is just what the doctor ordered. It is not "computationally simple" per se, but easy to implement if you have an existing implementation of a normal distribution.
You can just generate the distribution with the mean you want, standard deviation you want, and the two ends wherever you want. You'll have to do some work beforehand to compute the mean and standard deviation of the underlying (non-truncated) normal distribution to get the mean for the TN that you want, but you can use the formulae in that article. Also note that you can adjust the variance as well using this method :)
I have Java code (based on the Commons Math framework) for both an accurate (slower) and quick (less accurate) implementation of this distribution, with PDF, CDF, and sampling.