I am currently reading "Introduction to machine learning" by Ethem Alpaydin and I came across nearest centroid classifiers and tried to implement it. I guess I have correctly implemented the classifier but I am getting only 68% accuracy . So, is the nearest centroid classifier itself is inefficient or is there some error in my implementation (below) ?
The data set contains 1372 data points each having 4 features and there are 2 output classes
My MATLAB implementation :
DATA = load("-ascii", "data.txt");
#DATA is 1372x5 matrix with 762 data points of class 0 and 610 data points of class 1
#there are 4 features of each data point
X = DATA(:,1:4); #matrix to store all features
X0 = DATA(1:762,1:4); #matrix to store the features of class 0
X1 = DATA(763:1372,1:4); #matrix to store the features of class 1
X0 = X0(1:610,:); #to make sure both datasets have same size for prior probability to be equal
Y = DATA(:,5); # to store outputs
mean0 = sum(X0)/610; #mean of features of class 0
mean1 = sum(X1)/610; #mean of featurs of class 1
count = 0;
for i = 1:1372
pre = 0;
cost1 = X(i,:)*(mean0'); #calculates the dot product of dataset with mean of features of both classes
cost2 = X(i,:)*(mean1');
if (cost1<cost2)
pre = 1;
end
if pre == Y(i)
count = count+1; #counts the number of correctly predicted values
end
end
disp("accuracy"); #calculates the accuracy
disp((count/1372)*100);
There are at least a few things here:
You are using dot product to assign similarity in the input space, this is almost never valid. The only reason to use dot product would be the assumption that all your data points have the same norm, or that the norm does not matter (nearly never true). Try using Euclidean distance instead, as even though it is very naive - it should be significantly better
Is it an inefficient classifier? Depends on the definition of efficiency. It is an extremely simple and fast one, but in terms of predictive power it is extremely bad. In fact, it is worse than Naive Bayes, which is already considered "toy model".
There is something wrong with the code too
X0 = DATA(1:762,1:4); #matrix to store the features of class 0
X1 = DATA(763:1372,1:4); #matrix to store the features of class 1
X0 = X0(1:610,:); #to make sure both datasets have same size for prior probability to be equal
Once you subsamples X0, you have 1220 training samples, yet later during "testing" you test on both training and "missing elements of X0", this does not really make sense from probabilistic perspective. First of all you should never test accuracy on the training set (as it overestimates true accuracy), second of all by subsampling your training data your are not equalizing priors. Not in the method like this one, you are simply degrading quality of your centroid estimate, nothing else. These kind of techniques (sub/over- sampling) equalize priors for models that do model priors. Your method does not (as it is basically generative model with the assumed prior of 1/2), so nothing good can happen.
Related
I am tuning an SVM using a for loop to search in the range of hyperparameter's space. The svm model learned contains the following fields
SVMModel: [1×1 ClassificationSVM]
C: 2
FeaturesIdx: [4 6 8]
Score: 0.0142
Question1) What is the meaning of the field 'score' and its utility?
Question2) I am tuning the BoxConstraint, C value. Let, the number of features be denoted by the variable featsize. The variable gridC will contain the search space which can start from any value say 2^-5, 2^-3, to 2^15 etc. So, gridC = 2.^(-5:2:15). I cannot understand if there is a way to select the range?
1. score had been documented in here, which says:
Classification Score
The SVM classification score for classifying observation x is the signed distance from x to the decision boundary ranging from -∞ to +∞.
A positive score for a class indicates that x is predicted to be in
that class. A negative score indicates otherwise.
In two class cases, if there are six observations, and the predict function gave us some score value called TestScore, then we could determine which class does the specific observation ascribed by:
TestScore=[-0.4497 0.4497
-0.2602 0.2602;
-0.0746 0.0746;
0.1070 -0.1070;
0.2841 -0.2841;
0.4566 -0.4566;];
[~,Classes] = max(TestScore,[],2);
In the two-class classification, we can also use find(TestScore > 0) instead, and it is clear that the first three observations are belonging to the second class, and the 4th to 6th observations are belonging to the first class.
In multiclass cases, there could be several scores > 0, but the code max(scores,[],2) is still validate. For example, we could use the code (from here, an example called Find Multiple Class Boundaries Using Binary SVM) following to determine the classes of the predict Samples.
for j = 1:numel(classes);
[~,score] = predict(SVMModels{j},Samples);
Scores(:,j) = score(:,2); % Second column contains positive-class scores
end
[~,maxScore] = max(Scores,[],2);
Then the maxScore will denote the predicted classes of each sample.
2. The BoxConstraint denotes C in the SVM model, so we can train SVMs in different hyperparameters and select the best one by something like:
gridC = 2.^(-5:2:15);
for ii=1:length(gridC)
SVModel = fitcsvm(data3,theclass,'KernelFunction','rbf',...
'BoxConstraint',gridC(ii),'ClassNames',[-1,1]);
%if (%some constraints were meet)
% %save the current SVModel
%end
end
Note: Another way to implement this is using libsvm, a fast and easy-to-use SVM toolbox, which has the interface of MATLAB.
I have question please; concerning cross validation, for me the cross-validation is used to find the best parameters.
but I did not understand the role of this function "crossvalind":Generate cross-validation indices, it just takes a data set without model, like in this exemple :
load fisheriris
[g gn] = grp2idx(species);
[trainIdx testIdx] = crossvalind('HoldOut', species, 1/3);
crossvalind() function splits your data in two groups: the training set and the cross-validation set.
By your example:
[trainIdx testIdx] = crossvalind('HoldOut', size(species,1), 1/3); means split the data in species (2/3 in the training set and 1/3 in the cross-validation set).
Supposing that your data is like:
species=[datarow1;datarow2;datarow3;datarow4;datarow5;datarow6] then
trainIdx would be like [1;1;0;1;1;0] and testIdx would be like [0;0;1;0;0;1] meaning that from the 6 total elements in our set crossvalind function assigned 4 to the train set and 2 to the cross-validation set. Of course this is a random assignment meaning that the zero and ones indices will vary every time you call the function but the proportion between them will be fixed and trainIdx + testIdx will always be ones(size(species,1),1)
crossvalind('LeaveMout',size(species,1),2) would be exactly the same as crossvalind('HoldOut', size(species,1), 1/3) in this particular case. In the 'HoldOut' format you provide parameter P which takes values from 0 to 1 (like 1/3 in the example above) while with the option 'LeaveMout' you provide integer M like 2 samples from the 6 total or like 2000 samples from the 10000 total samples in your dataset. In case of 'Resubstitution': crossvalind('Resubstitution', size(species,1), [1/3,2/3]) would be yet the same but here you also have the option of let's say [1/3,3/4] meaning that some samples can be on both the train and cross-validation sets, or even [1,1] which means that all the samples are used in both sets (trainIdx=testIdx=[1;1;1;1;1;1] in the above example). I strongly suggest to type help crossvalind and take a look at the help file which is always a lot more detailed and helpful than i could ever be.
I am modelling the diffusion of movies through a contact network (based on telephone data) using a zero inflated negative binomial model (package: pscl)
m1 <- zeroinfl(LENGTH_OF_DIFF ~ ., data = trainData, type = "negbin")
(variables described below.)
The next step is to evaluate the performance of the model.
My attempt has been to do multiple out-of-sample predictions and calculate the MSE.
Using
predict(m1, newdata = testData)
I received a prediction for the mean length of a diffusion chain for each datapoint, and using
predict(m1, newdata = testData, type = "prob")
I received a matrix containing the probability of each datapoint being a certain length.
Problem with the evaluation: Since I have a 0 (and 1) inflated dataset, the model would be correct most of the time if it predicted 0 for all the values. The predictions I receive are good for chains of length zero (according to the MSE), but the deviation between the predicted and the true value for chains of length 1 or larger is substantial.
My question is:
How can we assess how well our model predicts chains of non-zero length?
Is this approach the correct way to make predictions from a zero inflated negative binomial model?
If yes: how do I interpret these results?
If no: what alternative can I use?
My variables are:
Dependent variable:
length of the diffusion chain (count [0,36])
Independent variables:
movie characteristics (both dummies and continuous variables).
Thanks!
It is straightforward to evaluate RMSPE (root mean square predictive error), but is probably best to transform your counts beforehand, to ensure that the really big counts do not dominate this sum.
You may find false negative and false positive error rates (FNR and FPR) to be useful here. FNR is the chance that a chain of actual non-zero length is predicted to have zero length (i.e. absence, also known as negative). FPR is the chance that a chain of actual zero length is falsely predicted to have non-zero (i.e. positive) length. I suggest doing a Google on these terms to find a paper in your favourite quantitative journals or a chapter in a book that helps explain these simply. For ecologists I tend to go back to Fielding & Bell (1997, Environmental Conservation).
First, let's define a repeatable example, that anyone can use (not sure where your trainData comes from). This is from help on zeroinfl function in the pscl library:
# an example from help on zeroinfl function in pscl library
library(pscl)
fm_zinb2 <- zeroinfl(art ~ . | ., data = bioChemists, dist = "negbin")
There are several packages in R that calculate these. But here's the by hand approach. First calculate observed and predicted values.
# store observed values, and determine how many are nonzero
obs <- bioChemists$art
obs.nonzero <- obs > 0
table(obs)
table(obs.nonzero)
# calculate predicted counts, and check their distribution
preds.count <- predict(fm_zinb2, type="response")
plot(density(preds.count))
# also the predicted probability that each item is nonzero
preds <- 1-predict(fm_zinb2, type = "prob")[,1]
preds.nonzero <- preds > 0.5
plot(density(preds))
table(preds.nonzero)
Then get the confusion matrix (basis of FNR, FPR)
# the confusion matrix is obtained by tabulating the dichotomized observations and predictions
confusion.matrix <- table(preds.nonzero, obs.nonzero)
FNR <- confusion.matrix[2,1] / sum(confusion.matrix[,1])
FNR
In terms of calibration we can do it visually or via calibration
# let's look at how well the counts are being predicted
library(ggplot2)
output <- as.data.frame(list(preds.count=preds.count, obs=obs))
ggplot(aes(x=obs, y=preds.count), data=output) + geom_point(alpha=0.3) + geom_smooth(col="aqua")
Transforming the counts to "see" what is going on:
output$log.obs <- log(output$obs)
output$log.preds.count <- log(output$preds.count)
ggplot(aes(x=log.obs, y=log.preds.count), data=output[!is.na(output$log.obs) & !is.na(output$log.preds.count),]) + geom_jitter(alpha=0.3, width=.15, size=2) + geom_smooth(col="blue") + labs(x="Observed count (non-zero, natural logarithm)", y="Predicted count (non-zero, natural logarithm)")
In your case you could also evaluate the correlations, between the predicted counts and the actual counts, either including or excluding the zeros.
So you could fit a regression as a kind of calibration to evaluate this!
However, since the predictions are not necessarily counts, we can't use a poisson
regression, so instead we can use a lognormal, by regressing the log
prediction against the log observed, assuming a Normal response.
calibrate <- lm(log(preds.count) ~ log(obs), data=output[output$obs!=0 & output$preds.count!=0,])
summary(calibrate)
sigma <- summary(calibrate)$sigma
sigma
There are more fancy ways of assessing calibration I suppose, as in any modelling exercise ... but this is a start.
For a more advanced assessment of zero-inflated models, check out the ways in which the log likelihood can be used, in the references provided for the zeroinfl function. This requires a bit of finesse.
This question already has answers here:
Evaluating K-means accuracy
(2 answers)
Closed 5 years ago.
How effectively evaluate the performance of the standard matlab k-means implementation.
For example I have a matrix X
X = [1 2;
3 4;
2 5;
83 76;
97 89]
For every point I have a gold standard clustering. Let's assume that (83,76), (97,89) is the first cluster and (1,2), (3,4), (2,5) is the second cluster. Then we run matlab
idx = kmeans(X,2)
And get the following results
idx = [1; 1; 2; 2; 2]
According the the NOMINAL values it's very bad clustering because only (2,5) is correct, but we don't care about nominal values, we care only about points that is clustered together. Therefore somehow we have to identify that only (2,5) gets to the incorrect cluster.
For me a newbie in matlab is not a trivial task to evaluate the performance of clustering. I would appreciate if you can share with us your ideas about how to evaluate the performance.
To evaluate the "best clustering" is somewhat ambiguous, especially if you have points in two different groups that may eventually cross over with respect to their features. When you get this case, how exactly do you define which cluster those points get merged to? Here's an example from the Fisher Iris dataset that you can get preloaded with MATLAB. Let's specifically take the sepal width and sepal length, which is the third and fourth columns of the data matrix, and plot the setosa and virginica classes:
load fisheriris;
plot(meas(101:150,3), meas(101:150,4), 'b.', meas(51:100,3), meas(51:100,4), 'r.', 'MarkerSize', 24)
This is what we get:
You can see that towards the middle, there is some overlap. You are lucky in that you knew what the clusters were before hand and so you can measure what the accuracy is, but if we were to get data such as the above and we didn't know what labels each point belonged to, how do you know which cluster the middle points belong to?
Instead, what you should do is try and minimize these classification errors by running kmeans more than once. Specifically, you can override the behaviour of kmeans by doing the following:
idx = kmeans(X, 2, 'Replicates', num);
The 'Replicates' flag tells kmeans to run for a total of num times. After running kmeans num times, the output memberships are those which the algorithm deemed to be the best over all of those times kmeans ran. I won't go into it, but they determine what the "best" average is out of all of the membership outputs and gives you those.
Not setting the Replicates flag obviously defaults to running one time. As such, try increasing the total number of times kmeans runs so that you have a higher probability of getting a higher quality of cluster memberships. By setting num = 10, this is what we get with your data:
X = [1 2;
3 4;
2 5;
83 76;
97 89];
num = 10;
idx = kmeans(X, 2, 'Replicates', num)
idx =
2
2
2
1
1
You'll see that the first three points belong to one cluster while the last two points belong to another. Even though the IDs are flipped, it doesn't matter as we want to be sure that there is a clear separation between the groups.
Minor note with regards to random algorithms
If you take a look at the comments above, you'll notice that several people tried running the kmeans algorithm on your data and they received different clustering results. The reason why is because when kmeans chooses the initial points for your cluster centres, these are chosen in a random fashion. As such, depending on what state their random number generator was in, it is not guaranteed that the initial points chosen for one person will be the same as another person.
Therefore, if you want reproducible results, you should set the random seed of your random seed generator to be the same before running kmeans. On that note, try using rng with an integer that is known before hand, like 123. If we did this before the code above, everyone who runs the code will be able to reproduce the same results.
As such:
rng(123);
X = [1 2;
3 4;
2 5;
83 76;
97 89];
num = 10;
idx = kmeans(X, 2, 'Replicates', num)
idx =
1
1
1
2
2
Here the labels are reversed, but I guarantee that if any else runs the above code, they will get the same labelling as what was produced above each time.
I have a full dataset of lets say 50000 observations which are assigned to 16 classes.
I now want to draw a Sample of let's say 70% of the full data, but I want MATLAB to take the same number of samples from each class (if possible of course, because some classes have less numbers than needed)
Is there a MATLAB function that can do this, or do I have to program a new one for myself? I'm just trying to save time here.
I found cvpartition, but as far as I know this can be used only to take a sample with the same distribution over the classes as the original dataset and not a uniformly distributed sample.
Thank you for your help!
It shouldn't be too hard. Let's say that the observations are in a vector observations. Then you can do
fraction = 0.7;
classes = unique(observations);
nObs = length(observations);
nClasses = length(classes);
nSamples = round(nObs * fraction / nClasses);
for ii = 1:nClasses
idx = observations == classes(ii);
samples((ii-1)*nSamples+1:ii*nSamples) = randsample(observations(idx), nSamples);
end
Now samples is a vector of length nClasses * nsamples that contains your sampled observations, with an equal number from each class.
At the moment it will fail if one of the classes doesn't contain at least nSamples observations. The simplest fix is to add the additional arguments 'replace','true' to the call to randsample, which will tell it to replace each observation after being picked.