I'm working with a evolutionary algorithm and I'm trying to generate new population using probability density function. We have many classical individual(Xij) and his fitness( f(Xij) ), to have probabilities for each of them, normalize fitness for instance I have Xij with probability 0.10 and I can get pdf.
During iterations I will have many pdf's, I want to save information of last pdf in the actual pdf. I have tried adding but I think it's not the best choice. What would you recommend?
Example of pdf's
PD: you can watch peaks of the pdf's in the picture.
To answer this question I will have to make some assumptions:
You have individuals Xij
You have individual fitness or scores f(Xij)
You normalize the score/fitness to a probability density function by applying a function to the fitness. In other words: pdf(Xij) = g(f(Xij)
You require persistence in the pdf to allow selection of specimen that performed well over several iterations.
I would either use a running average of the fitness score and then normalize the averaged score; or use a running average of the normalized score for a new pdf.
Related
My input data $X$ has 100 samples and 1724 features. I want to classify my data into two classes. But explained various ratio value is too low like 0.05,0.04, No matter how many principal components I fix while performing PCA. I have seen in the literature that usually explained variance ratio is close to 1. I'm unable to understand my mistake. I tried to perform PCA by reducing number of features and increasing number of samples. But it doesn't make any significant effect on my result.
`pca = PCA(n_components=10).fit(X)
Xtrain = pca.transform(X)
explained_ratio=pca.explained_variance_ratio_
EX=explained_ratio
fig,ax1=plt.subplots(ncols=1,nrows=1)
ax1.plot(np.arange(len(EX)),EX,marker='o',color='blue')
ax1.set_ylabel(r"$\lambda$")
ax1.set_xlabel("l")
I have a bundle of high-dimensional data and the instances are labeled as outliers or not. I am looking to get some insights around where these outliers reside within the data. I seek to answer questions like:
Are the outliers spread far apart from each other? Or are they clustered together?
Are the outliers lying 'in-between' clusters of good data? Or are they on the 'edge' boundaries of the data?
If outliers are clustered together, how do these cluster densities compare with clusters of good data?
'Where' are the outliers?
What kind of techniques will let me find these insights? If the data was 2 or 3-dimensional, I can easily plot the data and just look at it. But I can't do it high-dimensional data.
Analyzing the Statistical Properties of Outliers
First of all, if you can choose to focus on specific features. For
example, if you know a featues is subject to high variation, you can
draw a box plot. You can also draw a 2D graph if you want to focus on
2 features. THis shows how much the labelled outliers vary.
Next, there's a metric called a Z-score, which basically says how
many standard devations a point varies compared to the mean. The
Z-score is signed, meaning if a point is below the mean, the Z-score
will be negative. This can be used to analyze all the features of the
dataset. You can find the threshold value in your labelled dataset for which all the points above that threshold are labelled outliers
Lastly, we can find the interquartile range and similarly filter
based on it. The IQR is simply the difference between the 75
percentile and 25 percentile. You can also use this similarly to Z-score.
Using these techniques, we can analyze some of the statistical properties of the outliers.
If you also want to analyze the clusters, you can adapt the DBSCAN algorithm to your problem. This algorithm clusters data based on densities, so it will be easy to apply the techniques to outliers.
I use the python implementation of XGBoost. One of the objectives is rank:pairwise and it minimizes the pairwise loss (Documentation). However, it does not say anything about the scope of the output. I see numbers between -10 and 10, but can it be in principle -inf to inf?
good question. you may have a look in kaggle competition:
Actually, in Learning to Rank field, we are trying to predict the relative score for each document to a specific query. That is, this is not a regression problem or classification problem. Hence, if a document, attached to a query, gets a negative predict score, it means and only means that it's relatively less relative to the query, when comparing to other document(s), with positive scores.
It gives predicted score for ranking.
However, the scores are valid for ranking only in their own groups.
So we must set the groups for input data.
For esay ranking, refer to my project xgboostExtension
If I understand your questions correctly, you mean the output of the predict function on a model fitted using rank:pairwise.
Predict gives the predicted variable (y_hat).
This is the same for reg:linear / binary:logistic etc. The only difference is that reg:linear builds trees to Min(RMSE(y, y_hat)), while rank:pairwise build trees to Max(Map(Rank(y), Rank(y_hat))). However, output is always y_hat.
Depending on the values of your dependent variables, output can be anything. But I typically expect output to be much smaller in variance vs the dependent variable. This is usually the case as it is not necessary to fit extreme data values, the tree just needs to produce predictors that are large/small enough to be ranked first/last in the group.
I am now implementing an email filtering application using the Naive Bayes algorithm. My application uses the Spambase Data Set from the UCI Machine Learning Repository. Since the attributes are continuous, I calculate the probability using the Probability Density Function (PDF). However, when I evaluate the data using the k-fold cross validation, a training set may contain only 0 for one of its attributes. For this reason, I got a 0 standard deviation and the PDF returns NaN and it leads to a huge number of spams are not correctly classified with that training set. What should I do to fix the problem?
You could use a discrete PDF, which will always be bounded.
Alternatively, simply ignore any attribute with zero variance. There is no point in including distributions with zero variance, because they won't actually do anything. For example, you want to know how old I am, and then I tell you that I live on planet Earth. That shouldn't change your estimate, because every single piece of data you have is for people on planet Earth.
I searched the site but did not find exactly what I was looking for... I wanted to generate a discrete random number from normal distribution.
For example, if I have a range from a minimum of 4 and a maximum of 10 and an average of 7. What code or function call ( Objective C preferred ) would I need to return a number in that range. Naturally, due to normal distribution more numbers returned would center round the average of 7.
As a second example, can the bell curve/distribution be skewed toward one end of the other? Lets say I need to generate a random number with a range of minimum of 4 and maximum of 10, and I want the majority of the numbers returned to center around the number 8 with a natural fall of based on a skewed bell curve.
Any help is greatly appreciated....
Anthony
What do you need this for? Can you do it the craps player's way?
Generate two random integers in the range of 2 to 5 (inclusive, of course) and add them together. Or flip a coin (0,1) six times and add 4 to the result.
Summing multiple dice produces a normal distribution (a "bell curve"), while eliminating high or low throws can be used to skew the distribution in various ways.
The key is you are going for discrete numbers (and I hope you mean integers by that). Multiple dice throws famously generate a normal distribution. In fact, I think that's how we were first introduced to the Gaussian curve in school.
Of course the more throws, the more closely you approximate the bell curve. Rolling a single die gives a flat line. Rolling two dice just creates a ramp up and down that isn't terribly close to a bell. Six coin flips gets you closer.
So consider this...
If I understand your question correctly, you only have seven possible outcomes--the integers (4,5,6,7,8,9,10). You can set up an array of seven probabilities to approximate any distribution you like.
Many frameworks and libraries have this built-in.
Also, just like TokenMacGuy said a normal distribution isn't characterized by the interval it's defined on, but rather by two parameters: Mean μ and standard deviation σ. With both these parameters you can confine a certain quantile of the distribution to a certain interval, so that 95 % of all points fall in that interval. But resticting it completely to any interval other than (−∞, ∞) is impossible.
There are several methods to generate normal-distributed values from uniform random values (which is what most random or pseudorandom number generators are generating:
The Box-Muller transform is probably the easiest although not exactly fast to compute. Depending on the number of numbers you need, it should be sufficient, though and definitely very easy to write.
Another option is Marsaglia's Polar method which is usually faster1.
A third method is the Ziggurat algorithm which is considerably faster to compute but much more complex to program. In applications that really use a lot of random numbers it may be the best choice, though.
As a general advice, though: Don't write it yourself if you have access to a library that generates normal-distributed random numbers for you already.
For skewing your distribution I'd just use a regular normal distribution, choosing μ and σ appropriately for one side of your curve and then determine on which side of your wanted mean a point fell, stretching it appropriately to fit your desired distribution.
For generating only integers I'd suggest you just round towards the nearest integer when the random number happens to fall within your desired interval and reject it if it doesn't (drawing a new random number then). This way you won't artificially skew the distribution (such as you would if you were clamping the values at 4 or 10, respectively).
1 In testing with deliberately bad random number generators (yes, worse than RANDU) I've noticed that the polar method results in an endless loop, rejecting every sample. Won't happen with random numbers that fulfill the usual statistic expectations to them, though.
Yes, there are sophisticated mathematical solutions, but for "simple but practical" I'd go with Nosredna's comment. For a simple Java solution:
Random random=new Random();
public int bell7()
{
int n=4;
for (int x=0;x<6;++x)
n+=random.nextInt(2);
return n;
}
If you're not a Java person, Random.nextInt(n) returns a random integer between 0 and n-1. I think the rest should be similar to what you'd see in any programming language.
If the range was large, then instead of nextInt(2)'s I'd use a bigger number in there so there would be fewer iterations through the loop, depending on frequency of call and performance requirements.
Dan Dyer and Jay are exactly right. What you really want is a binomial distribution, not a normal distribution. The shape of a binomial distribution looks a lot like a normal distribution, but it is discrete and bounded whereas a normal distribution is continuous and unbounded.
Jay's code generates a binomial distribution with 6 trials and a 50% probability of success on each trial. If you want to "skew" your distribution, simply change the line that decides whether to add 1 to n so that the probability is something other than 50%.
The normal distribution is not described by its endpoints. Normally it's described by it's mean (which you have given to be 7) and its standard deviation. An important feature of this is that it is possible to get a value far outside the expected range from this distribution, although that will be vanishingly rare, the further you get from the mean.
The usual means for getting a value from a distribution is to generate a random value from a uniform distribution, which is quite easily done with, for example, rand(), and then use that as an argument to a cumulative distribution function, which maps probabilities to upper bounds. For the standard distribution, this function is
F(x) = 0.5 - 0.5*erf( (x-μ)/(σ * sqrt(2.0)))
where erf() is the error function which may be described by a taylor series:
erf(z) = 2.0/sqrt(2.0) * Σ∞n=0 ((-1)nz2n + 1)/(n!(2n + 1))
I'll leave it as an excercise to translate this into C.
If you prefer not to engage in the exercise, you might consider using the Gnu Scientific Library, which among many other features, has a technique to generate random numbers in one of many common distributions, of which the Gaussian Distribution (hint) is one.
Obviously, all of these functions return floating point values. You will have to use some rounding strategy to convert to a discrete value. A useful (but naive) approach is to simply downcast to integer.