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Considering the relative ease of implementation, classification accuracy with smaller datasets, and computational efficiency of Naive Bayes classifiers, I am su

Considering the relative ease of implementation, classification accuracy with smaller datasets, and computational efficiency of Naive Bayes classifiers, I am surprised that they are not mentioned as often as other machine learning competitors, such as random forest. Are there major drawbacks to Naive Bayes classifiers? Is it just that they aren't as accurate on large datasets?

data_scientist 9y ago
It doesn't have a good accuracy. I have yet to see a real-life dataset where it's better than to just call LogisticRegressionCV from scikit-learn. For bigger datasets, you may use vowpal wabbit or Fasttext. It may be a little bit slower for the training (but not so much), and as fast as LR for the training. What is the purpose of using an algorithm when another one is just better?
- boto3 9y ago
  There's a reason for that. NB needs to fit more params than LR (and other linear discriminant learners), e.g. for binary classification, NB needs to fit 2*N params while LR just N.
  - imh 9y ago
    Not true. Let's derive it. Here's NB for a binary classification: p(c|w)*p(w) = p(w|c) * p(c) p(c|w) = p(w|c) * p(c) / p(w) p(c|w) = p(w|c) * p(c) / [sum_i p(w|c_i) * p(c_i)] Let's look at the probability of class 1. p(c_1|w) = p(w|c_1) * p(c_1) / [sum_i p(w|c_i) * p(c_i)] Notice how the numerator is going to show up in the denominator. We can simplify that by bringing it into the denominator: p(c_1|w) = 1 / [sum_i p(w|c_i) * p(c_i) / p(w|c_1) / p(c_1] Then cancel it out: p(c_1|w) = 1 / [1 + p(w|c_0) / p(w|c_1) * p(c_0) / p(c_1)] Not let's apply the NB assumptions: p(w|c_0) / p(w|c_1) = prod_i p(w_i | c_0) / p(w_i | c_1) Now, if you take the log of the final p(c_1|w) I derived, the product in p(w|c_0) / p(w|c_1) turns into a exponentiated sum, giving you one parameter per word, plus an intercept for the log of p(c_0) / p(c_1). You end up with exactly the same 1/(1+exp(linear stuff)) with the same parametrization and form you have in logistic regression [0]. This is a broader thing that a given graphical model with a given fixed parametrization can be generatively or discriminatively trained. They will end up learning different models, but that's because they make different assumptions, not because of different parametrizations. [0] linear stuff = intercept + sum_i (coefficient of word_i)
    - abhgh 9y ago
      I may be missing something, but how do you take a log on the RHS of p(c_1|w)? The denominator has a sum i.e. "1+ something", which the log doesn't distribute over.
      - imh 9y ago
        Heh, nope you're right I was writing to quickly. I meant to say to take a log of of the probability term in p(c_1|w). So you take p(c_1|w) = 1/(1+stuff) and turn it into p(c_1|w) = 1/(1+exp(log(stuff))). stuff is a product, so log(stuff) is a sum. Good catch.
        abhgh 9y ago
        Oh, I see it now. Thanks for clarifying! In that case, you needn't have kept the denominator around at all. Since P(w) is not based on a class i.e. its not class conditioned, the classifier could have directly calculated P(C_1|w)/P(C_0|w). The P(w) term cancels out, and you end up with the product of ratios of feature probabilities conditioned on the classes. Note though that, for K-classes, K>2, the number of parameters you need to store would blow up. You would need have these N ratios: P(w_j|C_x)/P(w_j|C_y) for all possible classes x,y. Here N is the number of features. So, in all N*C(K,2) values. On the contrary, multiclass softmax regression (the discriminative analogue for NB) would need NK parameters.
        imh 9y ago
        In the multiclass problem, they have the same number of degrees of parameters. You can see this by choosing a reference class c_y. The NB parameter for class c_x and feature w_j is f_jxy = log(p(w_j|c_x)/p(w_j|c_y)). If you want a new reference class c_w, you can see that f_jxw = f_jxy - f_jwy. No new learned parameters needed. You need one of those parameters per N features and (K-1) classes that aren't c_y. So you get N(K-1) features. This is the same as for multiclass softmax regression: N(K-1) instead of the NK you wrote (which you can see by working it out with a reference class c_y). It really is the same parametrization. The analogue with NK parametrized softmax may be more straightforward if you just use NK naive bayes features of the form f_jx = log(p(w_j|c_x)) for each class and check the equivalence with softmax. It really is just a special case of a more general rule that a given PGM with a fixed parametrization can be trained discriminatively or generatively.
imh 9y ago
The (conditional) independence assumptions it makes are pretty strong and usually inaccurate. In the text example, consider "Steph Curry" and "Warriors." These are going to show up very frequently together, and both provide strong evidence that the topic is sports. However, they don't provide strongly independent evidence of sports. Supposing Curry announced he was making an investment in a local restaurant, the article would probably mention the Warriors just because it's Curry. The problem is that seeing "Warriors" when you see "Steph Curry" shouldn't make you that much more confident that it's sports than just seeing "Steph Curry" alone. Just like seeing "University" with "Stanford" shouldn't change any inferences much from seeing "Stanford" alone. In naive bayes models, these are treated as independent pieces of evidence and that can lead to overconfidence and errors. Naive bayes is a generative model, so in flipping bayes rule around to discriminate classes, you have to be sure your probability model is decent. In a discriminative model, you just go straight to learning the p(class | observations) and have no requirement for a decent model of p(observations). p(observations) is the kind of model that would have to know about "university" and "stanford" being likely to be observed together. Often, it's better to go straight to the discriminative model. In this case, logistic regression.
j2kun 9y ago
Boosted decision trees account for >50% of all Kaggle winners. That is the real surprise, since people rarely talk about boosted decision trees.
- abhgh 9y ago
  Deep Learning is in vogue so I think we see a lot around it written about in blogs, media etc. GBDTs are used in the industry though.
- nl 9y ago
  There is no greater discussion topic on Kaggle Slack than a comparison of XGB vs LightGBM vs CatGBM.
autokad 9y ago
many competition solutions use stacking, where you use the probabilities of several classifiers are used instead of their predictions. Since naive bayes makes an independence assumption, and most competitions have features that are highly correlated, its probability / confidence of its prediction is probably a lot more useless than other classifiers. besides that, nb doesn't do feature selection, regularization, and interactions like many tree based algorithms (such as XGBoost) do out of the box. interactions is a big one for me. With NB, you have to add an interaction feature by multiplying two together, but in creating the feature you must steal some probability away from the other stuff, and finding a good interaction among many features is hard. The tree structure of random forests naturally captures that edit: to your last point, nb does worse on larger data sets. when you dont have a lot of data, you have to make larger assumptions. NB makes a big independence assumption, thus it does well on very small data sets but falls short on large ones
abhgh 9y ago
Accuracy is a problem. NBs are linear classifiers, and more so, the linear boundary is found by assuming features to be independent. So, in cases: (1) data cannot be separated linearly - NBs don't do well for the same reason as any other linear classifier wouldn't work well (2) for linearly separable data - if the feature independence assumption is incompatible with the data, linear classifiers that don't make this assumption will beat it. For ex you can prove that logistic regression beats (or is at least as good as) Naive Bayes asymptotically. [1] [1] See proposition 1 here: "On Discriminative vs Generative classifiers ... Andrew ng, Michael Jordan. https://ai.stanford.edu/~ang/papers/nips01-discriminativegenerative.pdf https://ai.stanford.edu/~ang/papers/nips01-discriminativegen... [PDF]

data_scientist 9y ago

It doesn't have a good accuracy. I have yet to see a real-life dataset where it's better than to just call LogisticRegressionCV from scikit-learn. For bigger datasets, you may use vowpal wabbit or Fasttext. It may be a little bit slower for the training (but not so much), and as fast as LR for the training. What is the purpose of using an algorithm when another one is just better?

boto3 9y ago
There's a reason for that. NB needs to fit more params than LR (and other linear discriminant learners), e.g. for binary classification, NB needs to fit 2*N params while LR just N.
- imh 9y ago
  Not true. Let's derive it. Here's NB for a binary classification: p(c|w)*p(w) = p(w|c) * p(c) p(c|w) = p(w|c) * p(c) / p(w) p(c|w) = p(w|c) * p(c) / [sum_i p(w|c_i) * p(c_i)] Let's look at the probability of class 1. p(c_1|w) = p(w|c_1) * p(c_1) / [sum_i p(w|c_i) * p(c_i)] Notice how the numerator is going to show up in the denominator. We can simplify that by bringing it into the denominator: p(c_1|w) = 1 / [sum_i p(w|c_i) * p(c_i) / p(w|c_1) / p(c_1] Then cancel it out: p(c_1|w) = 1 / [1 + p(w|c_0) / p(w|c_1) * p(c_0) / p(c_1)] Not let's apply the NB assumptions: p(w|c_0) / p(w|c_1) = prod_i p(w_i | c_0) / p(w_i | c_1) Now, if you take the log of the final p(c_1|w) I derived, the product in p(w|c_0) / p(w|c_1) turns into a exponentiated sum, giving you one parameter per word, plus an intercept for the log of p(c_0) / p(c_1). You end up with exactly the same 1/(1+exp(linear stuff)) with the same parametrization and form you have in logistic regression [0]. This is a broader thing that a given graphical model with a given fixed parametrization can be generatively or discriminatively trained. They will end up learning different models, but that's because they make different assumptions, not because of different parametrizations. [0] linear stuff = intercept + sum_i (coefficient of word_i)
  - abhgh 9y ago
    I may be missing something, but how do you take a log on the RHS of p(c_1|w)? The denominator has a sum i.e. "1+ something", which the log doesn't distribute over.
    - imh 9y ago
      Heh, nope you're right I was writing to quickly. I meant to say to take a log of of the probability term in p(c_1|w). So you take p(c_1|w) = 1/(1+stuff) and turn it into p(c_1|w) = 1/(1+exp(log(stuff))). stuff is a product, so log(stuff) is a sum. Good catch.
      - abhgh 9y ago
        Oh, I see it now. Thanks for clarifying! In that case, you needn't have kept the denominator around at all. Since P(w) is not based on a class i.e. its not class conditioned, the classifier could have directly calculated P(C_1|w)/P(C_0|w). The P(w) term cancels out, and you end up with the product of ratios of feature probabilities conditioned on the classes. Note though that, for K-classes, K>2, the number of parameters you need to store would blow up. You would need have these N ratios: P(w_j|C_x)/P(w_j|C_y) for all possible classes x,y. Here N is the number of features. So, in all N*C(K,2) values. On the contrary, multiclass softmax regression (the discriminative analogue for NB) would need NK parameters.
        imh 9y ago
        In the multiclass problem, they have the same number of degrees of parameters. You can see this by choosing a reference class c_y. The NB parameter for class c_x and feature w_j is f_jxy = log(p(w_j|c_x)/p(w_j|c_y)). If you want a new reference class c_w, you can see that f_jxw = f_jxy - f_jwy. No new learned parameters needed. You need one of those parameters per N features and (K-1) classes that aren't c_y. So you get N(K-1) features. This is the same as for multiclass softmax regression: N(K-1) instead of the NK you wrote (which you can see by working it out with a reference class c_y). It really is the same parametrization. The analogue with NK parametrized softmax may be more straightforward if you just use NK naive bayes features of the form f_jx = log(p(w_j|c_x)) for each class and check the equivalence with softmax. It really is just a special case of a more general rule that a given PGM with a fixed parametrization can be trained discriminatively or generatively.

imh 9y ago

The (conditional) independence assumptions it makes are pretty strong and usually inaccurate. In the text example, consider "Steph Curry" and "Warriors." These are going to show up very frequently together, and both provide strong evidence that the topic is sports. However, they don't provide strongly independent evidence of sports. Supposing Curry announced he was making an investment in a local restaurant, the article would probably mention the Warriors just because it's Curry. The problem is that seeing "Warriors" when you see "Steph Curry" shouldn't make you that much more confident that it's sports than just seeing "Steph Curry" alone. Just like seeing "University" with "Stanford" shouldn't change any inferences much from seeing "Stanford" alone. In naive bayes models, these are treated as independent pieces of evidence and that can lead to overconfidence and errors. Naive bayes is a generative model, so in flipping bayes rule around to discriminate classes, you have to be sure your probability model is decent. In a discriminative model, you just go straight to learning the p(class | observations) and have no requirement for a decent model of p(observations). p(observations) is the kind of model that would have to know about "university" and "stanford" being likely to be observed together. Often, it's better to go straight to the discriminative model. In this case, logistic regression.

j2kun 9y ago

Boosted decision trees account for >50% of all Kaggle winners. That is the real surprise, since people rarely talk about boosted decision trees.

abhgh 9y ago
Deep Learning is in vogue so I think we see a lot around it written about in blogs, media etc. GBDTs are used in the industry though.
nl 9y ago
There is no greater discussion topic on Kaggle Slack than a comparison of XGB vs LightGBM vs CatGBM.

autokad 9y ago

many competition solutions use stacking, where you use the probabilities of several classifiers are used instead of their predictions. Since naive bayes makes an independence assumption, and most competitions have features that are highly correlated, its probability / confidence of its prediction is probably a lot more useless than other classifiers. besides that, nb doesn't do feature selection, regularization, and interactions like many tree based algorithms (such as XGBoost) do out of the box. interactions is a big one for me. With NB, you have to add an interaction feature by multiplying two together, but in creating the feature you must steal some probability away from the other stuff, and finding a good interaction among many features is hard. The tree structure of random forests naturally captures that edit: to your last point, nb does worse on larger data sets. when you dont have a lot of data, you have to make larger assumptions. NB makes a big independence assumption, thus it does well on very small data sets but falls short on large ones

abhgh 9y ago

Accuracy is a problem. NBs are linear classifiers, and more so, the linear boundary is found by assuming features to be independent. So, in cases: (1) data cannot be separated linearly - NBs don't do well for the same reason as any other linear classifier wouldn't work well (2) for linearly separable data - if the feature independence assumption is incompatible with the data, linear classifiers that don't make this assumption will beat it. For ex you can prove that logistic regression beats (or is at least as good as) Naive Bayes asymptotically. [1] [1] See proposition 1 here: "On Discriminative vs Generative classifiers ... Andrew ng, Michael Jordan. https://ai.stanford.edu/~ang/papers/nips01-discriminativegenerative.pdf https://ai.stanford.edu/~ang/papers/nips01-discriminativegen... [PDF]