In a confusion matrix, the diagonals tell us the number of correct predictions for each class and the off-diagonals are the errors. Consider a toy problem where out of 28 examples belonging to the negative class (label 0), 28 were predicted correctly (TN) and out of 8 positive examples (label 1), 6 were predicted correctly (TP). So, there were 2 examples from the negative classes that were not captured by the classifier.
Accuracy is the overall class performance. But if I want to find out the individual class performance, I should use the diagonals for each class and divide by the total examples in that class. I applied the formula and got opposite results for the individual class performance. Can somebody please help what is going on: why I got 100% for the positive class when two examples were missed? Here is what I did
cmMatrix =
28 2
0 6
acc_0 = 100*(cmMatrix(1,1))/sum(cmMatrix(1,:)) = 93.33
acc_1 = 100*(cmMatrix(2,2))/sum(cmMatrix(2,:)) = 100
The above values seem counter-intuitive!! I should be getting 100% for acc_0 since the classifier did not miss any examples (diagonal = 28) and for acc_1 for class 1 I should be getting 93.33
confusionmat arguments are:
cmMatrix = confusionmat(yrtue, ypredicted)
you most likely swapped the arguments.
Related
I have built a pretty basic naive bayes over apache spark and using mllib of course. But I have a few clarifications on what exactly neutrality means.
From what I understand, in a given dataset there are pre-labeled sentences which comprise of the necessary classes, let's take 3 for example below.
0-> Negative sentiment
1-> Positive sentiment
2-> Neutral sentiment
This neutral is pre-labeled in the training set itself.
Is there any other form of neutrality handling. Suppose if there are no neutral sentences available in the dataset then is it possible that I can calculate it from the scale of probability like
0.0 - 0.4 => Negative
0.4- - 0.6 => Neutral
0.6 - 1.0 => Positive
Is such kind of mapping possible in spark. I searched around but could not find any. The NaiveBayesModel class in the RDD API has a predict method which just returns a double that is mapped according to the training set i.e if only 0,1 is there it will return only 0,1 and not in a scaled manner such as 0.0 - 1.0 as above.
Any pointers/advice on this would be incredibly helpful.
Edit - 1
Sample code
//Performs tokenization,pos tagging and then lemmatization
//Returns a array of string
val tokenizedString = Util.tokenizeData(text)
val hashingTF = new HashingTF()
//Returns a double
//According to the training set 1.0 => Positive, 0.0 => Negative
val status = model.predict(hashingTF.transform(tokenizedString.toSeq))
if(status == 1.0) "Positive" else "Negative"
Sample dataset content
1,Awesome movie
0,This movie sucks
Of course the original dataset contains more longer sentences, but this should be enough for explanations I guess
Using the above code I am calculating. My question is the same
1) Neutrality handling in dataset
In the above dataset if I am adding another category such as
2,This movie can be enjoyed by kids
For arguments sake, lets assume that it is a neutral review, then the model.predict method will give either 1.0,0.0,2.0 based on the passed in sentence.
2) Using the model.predictProbabilities it gives an array of doubles, but I am not sure in what order it gives the result i.e index 0 is for negative or for positive? With three features i.e Negative,Positive,Neutral then in what order will that method return the predictions?
It would have been helpful to have the code that builds the model (for your example to work, the 0.0 from the dataset must be converted to 0.0 as a Double in the model, either after indexing it with a StringIndexer stage, or if you converted that from the file), but assuming that this code works:
val status = model.predict(hashingTF.transform(tokenizedString.toSeq))
if(status == 1.0) "Positive" else "Negative"
Then yes, it means the probabilities at index 0 is that of negative and at 1 that of positive (it's a bit strange and there must be a reason, but everything is a double in ML, even feature and category indexes). If you have something like this in your code:
val labelIndexer = new StringIndexer()
.setInputCol("sentiment")
.setOutputCol("indexedsentiment")
.fit(trainingData)
Then you can use labelIndexer.labels to identify the labels (probability at index 0 is for labelIndexer.labels at index 0.
Now regarding your other questions.
Neutrality can mean two different things. Type 1: a review contains as much positive and negative words Type 2: there is (almost) no sentiment expressed.
A Neutral category can be very helpful if you want to manage Type 2. If that is the case, you need neutral examples in your dataset. Naive Bayes is not a good classifier to apply thresholding on the probabilities in order to determine Type 2 neutrality.
Option 1: Build a dataset (if you think you will have to deal with a lot of Type 2 neutral texts). The good news is, building a neutral dataset is not too difficult. For instance you can pick random texts that are not movie reviews and assume they are neutral. It would be even better if you could pick content that is closely related to movies (but neutral), like a dataset of movie synopsis. You could then create a multi-class Naive Bayes classifier (between neutral, positive and negative) or a hierarchical classifier (first step is a binary classifier that determines whether a text is a movie review or not, second step to determine the overall sentiment).
Option 2 (can be used to deal with both Type 1 and 2). As I said, Naive Bayes is not very great to deal with thresholds on the probabilities, but you can try that. Without a dataset though, it will be difficult to determine the thresholds to use. Another approach is to identify the number of words or stems that have a significant polarity. One quick and dirty way to achieve that is to query your classifier with each individual word and count the number of times it returns "positive" with a probability significantly higher than the negative class (discard if the probabilities are too close to each other, for instance within 25% - a bit of experimentations will be needed here). At the end, you may end up with say 20 positive words vs 15 negative ones and determine it is neutral because it is balanced or if you have 0 positive and 1 negative, return neutral because the count of polarized words is too low.
Good luck and hope this helped.
I am not sure if I understand the problem but:
prior in Naive Bayes is computed from the data and cannot be set manually.
in MLLib you can use predictProbabilities to obtain class probabilities.
in ML you can use setThresholds to set prediction threshold for each class.
I have training set of size 54 * 65536 and a testing set of 18 * 65536.
I want to use a knn classifier, but I have some questions:
1) How should I define trainlabel?
Class = knnclassify(TestVec,TrainVec, TrainLabel,k);
Is it a vector of size 54 * 1 that defines to which group each row in training set belongs? Here the group is numbered as 1 ,2,..
2) To find the accuracy I used this:
cp = classperf(TrainLabel);
Class = knnclassify(TestVec,TrainVec, TrainLabel);
cp = classperf(TestLabel,Class);
cp.CorrectRate*100
Is this right? Is there another method to calculate it?
3) How can I enhance the accuracy?
4) How do I choose the best value of k?
I do not know matlab nor the implementation of the knn you are providing, so I can answer only a few of your questions.
1) You assumption is correct. trainlabel is a 54*1 vector or an array of size 54 or something equivalent that defines which group each datapoint (row) in training set belongs to.
2) ... MATLAB / implementation related, sorry.
3) That is a very big discussion. Possible ways are:
Choose a better value of K.
Preprocess the data (or make preprocessing better if already applied).
Get a better / bigger trainset.
to name a few...
4) You can use different values while measuring the accuracy for each one and keep the best. (Note: If you do that, make sure you do not measure the accuracy of the classifier per value of k only once, but rather you use some technique like 10-Folding or something).
There is more than a fair chance that the library you are using for the K-NNclassifier provides such utilities.
I want to fit a data sets with Gaussian mixture model, the data sets contains about 120k samples and each sample has about 130 dimensions. When I use matlab to do it, so I run scripts (with cluster number 1000):
gm = fitgmdist(data, 1000, 'Options', statset('Display', 'iter'), 'RegularizationValue', 0.01);
I get the following outputs:
iter log-likelihood
1 -6.66298e+07
2 -1.87763e+07
3 -5.00384e+06
4 -1.11863e+06
5 299767
6 985834
7 1.39525e+06
8 1.70956e+06
9 1.94637e+06
The log likelihood is bigger than 0! I think it's unreasonable, and don't know why.
Could somebody help me?
First of all, it is not a problem of how large your dataset is.
Here is some code that produces similar results with a quite small dataset:
options = statset('Display', 'iter');
x = ones(5,2) + (rand(5,2)-0.5)/1000;
fitgmdist(x,1,'Options',options);
this produces
iter log-likelihood
1 64.4731
2 73.4987
3 73.4987
Of course you know that the log function (the natural logarithm) has a range from -inf to +inf. I guess your problem is that you think the input to the log (i.e. the aposteriori function) should be bounded by [0,1]. Well, the aposteriori function is a pdf function, which means that its value can be very large for very dense dataset.
PDFs must be positive (which is why we can use the log on them) and must integrate to 1. But they are not bounded by [0,1].
You can verify this by reducing the density in the above code
x = ones(5,2) + (rand(5,2)-0.5)/1;
fitgmdist(x,1,'Options',options);
this produces
iter log-likelihood
1 -8.99083
2 -3.06465
3 -3.06465
So, I would rather assume that your dataset contains several duplicate (or very close) values.
Problem: 3 class classification with labels 1,2,3.
Tool: LibSVM for MATLAB
svmModel = svmtrain(<Trainfeatures>, <TrainclassLabels>, '-b 1 -c <someCValue> -g <someGammaValue>');
[predLabels, classAccuracy, **probEstimates**] = svmpredict(<TestFeatures>, <TestClassLabels>, '-b 1');
AFter this step, I get the first ten rows of probEstimates to be,
0.9129 0.0749 0.0122
0.9059 0.0552 0.0389
0.8231 0.0183 0.1586
0.9077 0.0098 0.0825
0.9074 0.0668 0.0257
0.8685 0.0146 0.1169
0.8962 0.0664 0.0374
0.9074 0.0548 0.0377
0.9474 0.0054 0.0472
0.9178 0.0642 0.0180
but the first ten predicted labels to be:
2
2
2
2
2
2
2
2
2
2
Questions:
My understanding was that the probability estimate was the probability that a particular item would belong to a particular class, given its feature vector. However, if that were true, then these items should belong to class 1 and not class 2. Does the libsvm change the order of classes or am I missing something here? If I am wrong, can someone please explain what the real interpretation of probability estimate is?
If I have to move the decision boundary to increase the precision of class 1 (have less items to be predicted to be class 1 and hence be more conservative in the decision boundary), which of these class probabilities should I have to deal with and how?
I came across the same problem recently.
The reason is related to the order of training data.
If you want the index of post-probability vector to correspond to the label of training data, the training data should be sorted according to the label.
For example, if the label of the the first data point is 4, then the first entry of post-probability vector is related to data points labeled 4.
The order of the the labels stored in the model may different from what we thought it should be. You can check using svmModel.Label. And the probability estimates are outputted according to this order.
I have production (q) values from 4 different methods stored in the 4 matrices. Each of the 4 matrices contains q values from a different method as:
Matrix_1 = 1 row x 20 column
Matrix_2 = 100 rows x 20 columns
Matrix_3 = 100 rows x 20 columns
Matrix_4 = 100 rows x 20 columns
The number of columns indicate the number of years. 1 row would contain the production values corresponding to the 20 years. Other 99 rows for matrix 2, 3 and 4 are just the different realizations (or simulation runs). So basically the other 99 rows for matrix 2,3 and 4 are repeat cases (but not with exact values because of random numbers).
Consider Matrix_1 as the reference truth (or base case ). Now I want to compare the other 3 matrices with Matrix_1 to see which one among those three matrices (each with 100 repeats) compares best, or closely imitates, with Matrix_1.
How can this be done in Matlab?
I know, manually, that we use confidence interval (CI) by plotting the mean of Matrix_1, and drawing each distribution of mean of Matrix_2, mean of Matrix_3 and mean of Matrix_4. The largest CI among matrix 2, 3 and 4 which contains the reference truth (or mean of Matrix_1) will be the answer.
mean of Matrix_1 = (1 row x 1 column)
mean of Matrix_2 = (100 rows x 1 column)
mean of Matrix_3 = (100 rows x 1 column)
mean of Matrix_4 = (100 rows x 1 column)
I hope the question is clear and relevant to SO. Otherwise please feel free to edit/suggest anything in question. Thanks!
EDIT: My three methods I talked about are a1, a2 and a3 respectively. Here's my result:
ci_a1 =
1.0e+008 *
4.084733001497999
4.097677503988565
ci_a2 =
1.0e+008 *
5.424396063219890
5.586301025525149
ci_a3 =
1.0e+008 *
2.429145282593182
2.838897116739112
p_a1 =
8.094614835195452e-130
p_a2 =
2.824626709966993e-072
p_a3 =
3.054667629953656e-012
h_a1 = 1; h_a2 = 1; h_a3 = 1
None of my CI, from the three methods, includes the mean ( = 3.454992884900722e+008) inside it. So do we still consider p-value to choose the best result?
If I understand correctly the calculation in MATLAB is pretty strait-forward.
Steps 1-2 (mean calculation):
k1_mean = mean(k1);
k2_mean = mean(k2);
k3_mean = mean(k3);
k4_mean = mean(k4);
Step 3, use HIST to plot distribution histograms:
hist([k2_mean; k3_mean; k4_mean]')
Step 4. You can do t-test comparing your vectors 2, 3 and 4 against normal distribution with mean k1_mean and unknown variance. See TTEST for details.
[h,p,ci] = ttest(k2_mean,k1_mean);
EDIT : I misinterpreted your question. See the answer of Yuk and following comments. My answer is what you need if you want to compare distributions of two vectors instead of a vector against a single value. Apparently, the latter is the case here.
Regarding your t-tests, you should keep in mind that they test against a "true" mean. Given the number of values for each matrix and the confidence intervals it's not too difficult to guess the standard deviation on your results. This is a measure of the "spread" of your results. Now the error on your mean is calculated as the standard deviation of your results divided by the number of observations. And the confidence interval is calculated by multiplying that standard error with appx. 2.
This confidence interval contains the true mean in 95% of the cases. So if the true mean is exactly at the border of that interval, the p-value is 0.05 the further away the mean, the lower the p-value. This can be interpreted as the chance that the values you have in matrix 2, 3 or 4 come from a population with a mean as in matrix 1. If you see your p-values, these chances can be said to be non-existent.
So you see that when the number of values get high, the confidence interval becomes smaller and the t-test becomes very sensitive. What this tells you, is nothing more that the three matrices differ significantly from the mean. If you have to choose one, I'd take a look at the distributions anyway. Otherwise the one with the closest mean seems a good guess. If you want to get deeper into this, you could also ask on stats.stackexchange.com
Your question and your method aren't really clear :
Is the distribution equal in all columns? This is important, as two distributions can have the same mean, but differ significantly :
is there a reason why you don't use the Central Limit Theorem? This seems to me like a very complex way of obtaining a result that can easily be found using the fact that the distribution of a mean approaches a normal distribution where sd(mean) = sd(observations)/number of observations. Saves you quite some work -if the distributions are alike! -
Now if the question is really the comparison of distributions, you should consider looking at a qqplot for a general idea, and at a 2-sample kolmogorov-smirnov test for formal testing. But please read in on this test, as you have to understand what it does in order to interprete the results correctly.
On a sidenote : if you do this test on multiple cases, make sure you understand the problem of multiple comparisons and use the appropriate correction, eg. Bonferroni or Dunn-Sidak.