I looked at the documentation of scikit-learn but it is not clear to me what sort of classification method is used under the hood of the VotingClassifier? Is it logistic regression, SVM or some sort of a tree method?
I'm interested in ways to vary the classifier method used under the hood. If Scikit-learn is not offering such an option is there a python package which can be integrated easily with scikit-learn which would offer such functionality?
EDIT:
I meant the classifier method used for the second level model. I'm perfectly aware that the first level classifiers can be any type of classifier supported by scikit-learn.
The second level classifier uses the predictions of the first level classifiers as inputs. So my question is - what method does this second level classifier use? Is it logistic regression? Or something else? Can I change it?
General
The VotingClassifier is not limited to one specific method/algorithm. You can choose multiple different algorithms and combine them to one VotingClassifier. See example below:
iris = datasets.load_iris()
X, y = iris.data[:, 1:3], iris.target
clf1 = LogisticRegression(...)
clf2 = RandomForestClassifier(...)
clf3 = SVC(...)
eclf = VotingClassifier(estimators=[('lr', clf1), ('rf', clf2), ('svm', clf3)], voting='hard')
Read more about the usage here: VotingClassifier-Usage.
When it comes down to how the VotingClassifier "votes" you can either specify voting='hard' or voting='soft'. See the paragraph below for more detail.
Voting
Majority Class Labels (Majority/Hard Voting)
In majority voting, the predicted class label for a particular sample
is the class label that represents the majority (mode) of the class
labels predicted by each individual classifier.
E.g., if the prediction for a given sample is
classifier 1 -> class 1 classifier 2 -> class 1 classifier 3 -> class
2 the VotingClassifier (with voting='hard') would classify the sample
as “class 1” based on the majority class label.
Source: scikit-learn-majority-class-labels-majority-hard-voting
Weighted Average Probabilities (Soft Voting)
In contrast to majority voting (hard voting), soft voting returns the
class label as argmax of the sum of predicted probabilities.
Specific weights can be assigned to each classifier via the weights
parameter. When weights are provided, the predicted class
probabilities for each classifier are collected, multiplied by the
classifier weight, and averaged. The final class label is then derived
from the class label with the highest average probability.
Source/Read more here: scikit-learn-weighted-average-probabilities-soft-voting
The VotingClassifier does not fit any meta model on the first level of classifiers output.
It just aggregates the output of each classifier in the first level by the mode (if voting is hard) or averaging the probabilities (if the voting is soft).
In simple terms, VotingClassifier does not learn anything from the first level of classifiers. It only consolidates the output of individual classifiers.
If you want your meta model to be more intelligent, try using the adaboost, gradientBoosting models.
Related
I'm using OneVsRestClassifier on a multiclass problem with svm.SVC as the base estimator.
The argmax from the predict_proba() does not match the predicted class:
Is there some normalization going on in the background? How do I get predict_proba() and predict()
to match?
According to the scikit learn's SVC documentation on multi-class classification, there can be discrepancies between the output of predict and the argmax of predict_proba (emphasis mine):
The decision_function method of SVC and NuSVC gives per-class scores for each sample (or a single score per sample in the binary case). When the constructor option probability is set to True, class membership probability estimates (from the methods predict_proba and predict_log_proba) are enabled. In the binary case, the probabilities are calibrated using Platt scaling: logistic regression on the SVM’s scores, fit by an additional cross-validation on the training data. In the multiclass case, this is extended as per Wu et al. (2004).
Needless to say, the cross-validation involved in Platt scaling is an expensive operation for large datasets. In addition, the probability estimates may be inconsistent with the scores, in the sense that the “argmax” of the scores may not be the argmax of the probabilities. (E.g., in binary classification, a sample may be labeled by predict as belonging to a class that has probability <½ according to predict_proba.) Platt’s method is also known to have theoretical issues. If confidence scores are required, but these do not have to be probabilities, then it is advisable to set probability=False and use decision_function instead of predict_proba.
You cannot get them to match using a SVC. You can try another model if you need the probabilities. If you do not need probabilities, as stated in the documentation, you can use decision_function (see here for more details.)
How does the RandomForestClassifier of sklearn handle a multilabel problem (under the hood)?
For example, does it brake the problem in distinct one-label problems?
Just to be clear, I have not really tested it yet but I see y : array-like, shape = [n_samples] or [n_samples, n_outputs] at the .fit() function of the RandomForestClassifier.
Let me cite scikit-learn. The user guide of random forest:
Like decision trees, forests of trees also extend to multi-output problems (if Y is an array of size [n_samples, n_outputs]).
The section multi-output problems of the user guide of decision trees:
… to support multi-output problems. This requires the following changes:
Store n output values in leaves, instead of 1;
Use splitting criteria that compute the average reduction across all n outputs.
And I hope this will answer your question. If not, you can look at the section's reference:
M. Dumont et al., Fast multi-class image annotation with random subwindows and multiple output randomized trees, International Conference on Computer Vision Theory and Applications, 2009.
I was a bit confused when I started using trees. If you refer to the sklearn doc:
https://scikit-learn.org/stable/modules/generated/sklearn.tree.DecisionTreeClassifier.html#sklearn.tree.DecisionTreeClassifier
If you go down on the methods to predict_proba, you can see:
"The predicted class probability is the fraction of samples of the same class in a leaf."
So in predict, the class is the mode of the classes on that node. This can change if you use weighted classes
"class_weight : dict, list of dicts, “balanced” or None, default=None
Weights associated with classes in the form {class_label: weight}. If not given, all classes are supposed to have weight one."
Hope this helps! :)
I do this:
from sklearn.linear_model import SGDClassifier
sgclass = SGDClassifier(random_state=10)
sgclass.fit(X_train,y_train)
pred = sgclass.predict(X_test)
from sklearn.metrics import classification_report,accuracy_score
print(classification_report(y_test, pred))
print(accuracy_score(y_test, pred))
These are useful reports on the recall and precision of the model.
However how to acquire the most influential independent variables that predict the dependent variable? I started with about 12 candidates and want to see their rank order in terms of influence in the model.
As the documentation specifies, you can use the coef_ attribute to get feature weights. The greater the absolute value of the feature is, the greater is its importance.
You can see that in action in the feature selection class from scikit, SelectFromModel. The best features are selected from any classifier that has attributes feature_importances_ or coef_.
In the documentation of scikit-learn in section 1.9.2.1 (excerpt is posted below), why does the implementation of random forest differ from the original paper by Breiman? As far as I'm aware, Breiman opted for a majority vote (mode) for classification and an average for regression (paper written by Liaw and Wiener, the maintainers of the original R code with citation below) when aggregating the ensembles of classifiers.
Why does scikit-learn use probabilistic prediction instead of a majority vote?
Is there any advantage in using probabilistic prediction?
The section in question:
In contrast to the original publication [B2001], the scikit-learn
implementation combines classifiers by averaging their probabilistic
prediction, instead of letting each classifier vote for a single
class.
Source: Liaw, A., & Wiener, M. (2002). Classification and Regression by randomForest. R news, 2(3), 18-22.
This question has now been answered on Cross Validated. Included here for reference:
Such questions are always best answered by looking at the code, if
you're fluent in Python.
RandomForestClassifier.predict, at least in the current version
0.16.1, predicts the class with highest probability estimate, as given by predict_proba. (this
line)
The documentation for predict_proba says:
The predicted class probabilities of an input sample is computed as the mean predicted class probabilities of the trees in the forest. The
class probability of a single tree is the fraction of samples of the
same class in a leaf.
The difference from the original method is probably just so that
predict gives predictions consistent with predict_proba. The
result is sometimes called "soft voting", rather than the "hard"
majority vote used in the original Breiman paper. I couldn't in quick
searching find an appropriate comparison of the performance of the two
methods, but they both seem fairly reasonable in this situation.
The predict documentation is at best quite misleading; I've
submitted a pull
request to
fix it.
If you want to do majority vote prediction instead, here's a function
to do it. Call it like predict_majvote(clf, X) rather than
clf.predict(X). (Based on predict_proba; only lightly tested, but
I think it should work.)
from scipy.stats import mode
from sklearn.ensemble.forest import _partition_estimators, _parallel_helper
from sklearn.tree._tree import DTYPE
from sklearn.externals.joblib import Parallel, delayed
from sklearn.utils import check_array
from sklearn.utils.validation import check_is_fitted
def predict_majvote(forest, X):
"""Predict class for X.
Uses majority voting, rather than the soft voting scheme
used by RandomForestClassifier.predict.
Parameters
----------
X : array-like or sparse matrix of shape = [n_samples, n_features]
The input samples. Internally, it will be converted to
``dtype=np.float32`` and if a sparse matrix is provided
to a sparse ``csr_matrix``.
Returns
-------
y : array of shape = [n_samples] or [n_samples, n_outputs]
The predicted classes.
"""
check_is_fitted(forest, 'n_outputs_')
# Check data
X = check_array(X, dtype=DTYPE, accept_sparse="csr")
# Assign chunk of trees to jobs
n_jobs, n_trees, starts = _partition_estimators(forest.n_estimators,
forest.n_jobs)
# Parallel loop
all_preds = Parallel(n_jobs=n_jobs, verbose=forest.verbose,
backend="threading")(
delayed(_parallel_helper)(e, 'predict', X, check_input=False)
for e in forest.estimators_)
# Reduce
modes, counts = mode(all_preds, axis=0)
if forest.n_outputs_ == 1:
return forest.classes_.take(modes[0], axis=0)
else:
n_samples = all_preds[0].shape[0]
preds = np.zeros((n_samples, forest.n_outputs_),
dtype=forest.classes_.dtype)
for k in range(forest.n_outputs_):
preds[:, k] = forest.classes_[k].take(modes[:, k], axis=0)
return preds
On the dumb synthetic case I tried, predictions agreed with the
predict method every time.
This was studied by Breiman in Bagging predictor (http://statistics.berkeley.edu/sites/default/files/tech-reports/421.pdf).
This gives nearly identical results, but using soft voting gives smoother probabilities. Note that if you are using completely developed tree, you won't have any difference.
is there any possibility to configure an svm classifier from sci-kit such that:
1.) the svm classifier is trained with examples from 0,...,n - 1
2.) If none of the single classifiers (one-vs-rest) delivers a positive result (class membership), then the output is a designated label n which means "none of them"
Thanks!
By construction, the OvR multiclass wrapper sklearn.multiclass.OneVsRestClassifier selects the maximum decision_function output or the maximum predict_proba to be decisive of predicted class. This means that there will always be a predicted class.
If you wanted e.g. to predict "None of these" when decision_function / predict_proba all stay under a certain threshold (for all OvR problems), then you would have to write this estimator yourself, but could get inspiration from the code of sklearn.multiclass.OneVsRestClassifier and just modify the decision logic.