# Tag Info

## Hot answers tagged classification

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Based on the question and your comments, you can consider every 1-vs-all classification problem as a so-called PU learning problem (*P*ositive and *U*nlabeled). PU learning is a branch of semi-supervised learning in which you only have labels of the positive class and (typically a lot of) unlabeled instances. Every tag is a class in your case. Instances ...

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Another possibility are neural networks, if you use the cross-entropy as the cost functional with sigmoidal output units. That will provide you with the estimates you are looking for. Neural networks, as well as logistic regression, are discriminative classifiers, meaning that they attempt to maximize the conditional distribution on the training data. ...

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SVM is closely related to logistic regression, and can be used to predict the probabilities as well based on the distance to the hyperplane (the score of each point). You do this by making score -> probability mapping some way, which is relatively easy as the problem is one-dimensional. One way is to fit an S-curve (e.g. the logistic curve, or its slope) to ...

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Another question: When two or more classifiers assign a test instance as positive, which one should I go with. I read somewhere that you should use "output function of the SVM". Now what's that? When used as a classifier, the label for instance $\mathbf{x}$ is decided based on the sign of the decision value $f(\mathbf{x})$. The decision value of an SVM ...

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If by technique you mean classification method (logistic regression, classification tree, ...), you can use any of these methods to obtain the result you want. Each method usually has a build in cost-function that you can adjust to obtain your desired results. All of these methods end up as being equivalent to building an ROC curve and choosing which point ...

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Let's normalize the confusion matrix, i.e. $TP + FP + FN + TN = 1$. We have: $F_1 = 2 \cdot \frac{\mathrm{precision} \cdot \mathrm{recall}}{\mathrm{precision} + \mathrm{recall}} = 2 \cdot \frac{\frac{tp}{tp+fp} \cdot \frac{tp}{tp+fn}}{\frac{tp}{tp+fp} + \frac{tp}{tp+fn}} = 2 \frac{TP} {2 TP + FP + FN} = 2 \frac{TP} {TP + 1 - TN}$ Therefore: $\text{F-1 ... 2 I am responding to a request for alternative graphical techniques that show how well simulated failure events match observed failure events. The question arose in "Probabilistic Programming and Bayesian Methods for Hackers " found here. Here's my graphical approach: Code found here. 2 Yes, it is true that the training data should be the as close as possible to the data the model will be applied on eventually. But as you said, cleaning the input data automatically by fixing typos via spellcorrector or normalizing synonyms to one term will improve make the data more "precise" for the classification task. However, to make the model ... 2 I don't know if you split your dataset randomly (each sample receives a random subset of observations) or not. I assume that your split is random. why does the training error start so high, then suddenly drop, then start to rise again as training set size increases? This is merely a noise caused by small size of your training and test sets, as well as the ... 2 In the bioinformatics literature, this is known as sequence coevolution. Disclaimer: I'm an author. See this article. I'm familiar with one branch of the literature. The early literature on coevolution formed positional covariance using$\chi^2$measures. Someone got the idea to compare positional covariance to structural contact maps derived from pdbs. ... 1 Yes, ridge regression can be used as a classifier, just code the response labels as -1 and +1 and fit the regression model as normal. Allen's PRESS statistic (i.e. the leave-one-out estimate of the squared error) works fine as a model selection criterion (e.g. for selecting the ridge parameter). In my experience it works about as well as a linear support ... 1 Outliers in small samples can always be very tricky to detect. In most cases actually I would advocate that if you feel that your data are not bluntly corrupted, an "outlierish" value might not be problematic and its exclusion might be unreasonable. Probably using robust statistical techniques will be more sensible and closer to a middle-ground solution. You ... 1 You can change the loss function that is used for fitting the model parameters and evaluating the forecasts. For binary models, there is a large class of loss functions called "proper scoring rules", including e.g. negative likelihood and squared error. These loss functions are sensible from a stat theory perspective since they set the incentive to identify ... 1 From the linked paper (Paper #1) Each chromosome is defined as a floating-point vector, whose length corresponds to the number of variable in a certain problem. Each element in a vector is called as a gene and each chromosome consists of$N(N-1)$genes, which are floating point numbers in the range$[-1, 1]$. When computing the FCM classifier's ... 1 Choose 1 when you don't have access to large unlabeled data Choose 2 when you do have access to large unlabeled data 3 is incorrect Ideally, your test and training data should have the same distribution, which implies that mean and variance should have matched. That is, 2] is simply a better estimate of mean than 1]. However, your improved result for 3] ... 1 For the imbalanced data sets classification, downsampling is mostly used in the case when there is huge difference between the size of two groups, while for your case, maybe upsampling (aka duplicate your set1 to alter the balance) is a better strategy. You can also design your own strategy of resampling (bootstrap for example). Besides, studies have shown ... 1 I assume that to compute$AUC(c_i,c_j)$, you use the classifier obtained by comparing the probabilities of pertaining to these two classes. Then, for the random classifier which draws uniformly a point$(p_1, \dots, p_n)$in the$n$-simplex (I denote$n = |C|$), you have$AUC(c_i, c_j) = 0.5$. As there are$n( n - 1)$pairs$(c_i, c_j)$, if each$AUC(c_i, ...

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There are many - and what works best depends on the data. There are also many ways to cheat - for example, you can perform probability calibration on the outputs of any classifier that gives some semblance of a score (i.e.: a dot product between the weight vector and the input). The most common example of this is called Platt's scaling. There is also the ...

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This is a really deep question. I am going to try to answer it for your specific case and make broader points at the same time. Has anyone an idea on how to proceed here? How to proceed from here is really a question of which method to use. The answer for your particular case seems to be CART (Classification and Regression Trees). CARTs will allow ...

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The terminal nodes are mutually exclusive in that observations cannot be classified into more than one node. They are not mutually exclusive in the sense that they can use the same characteristics/variables to classify observations. The issue here may be that often in statistical packages like R the node number has little meaning (at least to me). So that ...

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The transform from a classification to regression of SVM is explained pretty will in this new svm paper. A margin-based loss is used for regression with the loss function max(0, |x - f(x)| - epsilon). libsvm implemented this idea as well.

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