knn.score(X_test, y_test)
Here X_test is a numpy array that contains test cases and y_test contains their correct labels.
This is the code that returns the reliability score of a model I made to differentiate between species of iris.
How does this function work, does it predict every value from X_test array and then compares it with y_test array and computes the mean?
The KNeighborsClassifier is a subclass of the sklearn.base.ClassifierMixin. From the documentation of the score method:
Returns the mean accuracy on the given test data and labels.
In multi-label classification, this is the subset accuracy which is a harsh metric since you require for each sample that each label set be correctly predicted.
The source code itself for the score method:
return accuracy_score(y, self.predict(X), sample_weight=sample_weight)
It's simply a shortcut for producing predictions on the test data and computing the accuracy score against the given labels.
Related
I am analyzing a dataset from kaggle and want to apply a logistic regression model to predict something. This is the data: https://www.kaggle.com/code/mohamedadelhosny/stroke-prediction-data-analysis-challenge/data
I split the data into train and test, and want to use cross validation to inssure highest accuracy possible. I did some pre-processing and used the dummy function over catigorical features, got to a certain point in the code, and and I don't know how to proceed. I cant figure out how to use the results of the cross validation, it's not so straight forward.
This is what I got so far:
from numpy import mean
from numpy import std
from sklearn.datasets import make_classification
from sklearn.model_selection import KFold
from sklearn.linear_model import LogisticRegression
X = data_Enco.iloc[:, data_Enco.columns != 'stroke'].values # features
Y = data_Enco.iloc[:, 6] # labels
X_train, X_test, Y_train, Y_test = train_test_split(X, Y, test_size=0.20)
scaler = MinMaxScaler()
scaled_X_train = scaler.fit_transform(X_train)
scaled_X_test = scaler.transform(X_test)
# prepare the cross-validation procedure
cv = KFold(n_splits=10, random_state=1, shuffle=True)
logisticModel = LogisticRegression(class_weight='balanced')
# evaluate model
scores = cross_val_score(logisticModel, scaled_X_train, Y_train, scoring='accuracy', cv=cv)
print('average score = ', np.mean(scores))
print('std of scores = ', np.std(scores))
average score = 0.7483538453549359
std of scores = 0.0190400919099899
So far so good.. I got the results of the model for each 10 splits. But now what? how do I build a confusion matrix? how do I calculate the recall, precesion..? I have the right code without performing cross validation, I just dont know how to adapt it.. how do I use the scores of the cross_val_score function ?
logisticModel = LogisticRegression(class_weight='balanced')
logisticModel.fit(scaled_X_train, Y_train) # Train the model
predictions_log = logisticModel.predict(scaled_X_test)
## Scoring the model
logisticModel.score(scaled_X_test,Y_test)
## Confusion Matrix
Y_pred = logisticModel.predict(scaled_X_test)
real_data = Y_test
print('Observe the difference between the real data and the data predicted by the knn classifier:\n')
print('Predictions: ',Y_pred,'\n\n')
print('Real Data:m', real_data,'\n')
cmtx = pd.DataFrame(
confusion_matrix(real_data, Y_pred, labels=[0, 1]),
index = ['real 0: ', 'real 1:'], columns = ['pred 0:', 'pred 1:']
)
print(cmtx)
print('Accuracy score is: ',accuracy_score(real_data, Y_pred))
print('Precision score is: ',precision_score(real_data, Y_pred))
print('Recall Score is: ',recall_score(real_data, Y_pred))
print('F1 Score is: ',f1_score(real_data, Y_pred))
The performance of a model on the training dataset is not a good estimator of the performance on new data because of overfitting.
Cross-validation is used to obtain an estimation of the performance of your model on new data, i.e. without overfitting. And you correctly applied it to compute the mean and variance of the accuracy of your model. This should be a much better approximation of the accuracy on your test dataset than the accuracy on your training dataset. And that is it.
However, cross-validation is usually used to do model selection. Say you have two logistic regression models that use different sets of independent variables. E.g., one is using only age and gender while the other one is using age, gender, and bmi. Or you want to compare logistic regression with an SVM model.
I.e. you have several possible models and you want to decide which one is best. Of course, you cannot just compare the training dataset accuracies of all the models because those are spoiled by overfitting. And if you use the performance on the test dataset for choosing the best model, the test dataset becomes part of the training, you will have leakage, and thus the performance on the test dataset cannot be used anymore for a final, untainted performance measure. That is why cross-validation is used which creates those splits that contain different versions of validation sets.
So the idea is to
apply cross-validation to each of your candidate models,
use the scores of those cross-validations to choose the best model,
retrain that best model on the complete training dataset to get a final version of your best model, and
to finally apply this final version to the test dataset to obtain some untainted evaluation.
But note, that those three steps are for model selection. However, you have only a single model, the logistic regression, so there is nothing to select from. If you fit your model, let's call it m(p) where p denotes the parameters, to e.g. five folds of CV, you get five different fitted versions m(p1), m(p2), ..., m(p5) of the same model.
So if you have only one model, you fit it to the complete training dataset, maybe use CV to have an additional estimate for the performance on new data, but that's it. But you have already done this. There is no "selection of best model", that is only for if you have several models as described above, like e.g. logistic regression and SVM.
I am using xgboost for a classification problem with an imbalanced dataset. I plan on using some combination of an f1-score or roc-auc as my primary criteria for judging the model.
Currently the default value returned from the score method is accuracy, but I would really like to have a specific evaluation metric returned instead. My big motivation for doing this is that I presume the feature_importances_ attribute from the model is determined from what's affecting the score method, and the columns that impact predictive accuracy might very well be different from the columns that impact roc-auc. Right now I am passing in values to eval_metric but it does not seem to be making a difference.
Here is some sample code:
from sklearn.model_selection import train_test_split
from xgboost import XGBClassifier
from sklearn.datasets import load_breast_cancer
from sklearn.metrics import roc_auc_score
data = load_breast_cancer()
X = data['data']
y = data['target']
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=42, test_size=0.2, stratify=y)
mod.fit(X_train, y_train)
Now at this point, mod.score(X_test, y_test) will return a value of ~ 0.96, and the roc_auc_score is ~ 0.99.
I was hoping the following snippet:
mod.fit(X_train, y_train, eval_metric='auc')
Would then allow mod.score(X_test, y_test) to return the roc_auc_score value, but it is still returning predictive accuracy, not roc_auc.
The purpose of this exercise is estimating the influence of different columns on the outcome, so if I could get feature_importances_ returned using f1 or roc_auc as the measure of impact this would be a huge boon, but I do not seem to be on the right path as of now.
Thank you.
There are two parts to your question, to use eval_metric, you need to provide data to evaluate using eval_set = :
mod = XGBClassifier()
mod.fit(X_train, y_train,eval_set=[(X_test,y_test)],eval_metric="auc")
You can check the auc using evals_result(), and it gives the auc for every iteration:
mod.evals_result()
{'validation_0': OrderedDict([('auc',
[0.965939,
0.9833,
0.984788,
[...]
0.991402,
0.991071,
0.991402,
0.991733])])}
The importance score is calculated based on the average gain across all splits the feature is used in see help page. From your question, I suppose you need the mdoel to maximize auc, like in cross-validation, but you cannot use the auc as an objective in xgboost. Gradient boosting methods require a differentiable loss function.
With imbalanced dataset, you can try to adjust the parameter scale_pos_weight, to adjust the balance of positive and negative weights. This is discussed in xgboost website
NOTE: I appreciate the massive quantity of comments suggesting that this is inappropriate to quantify model performance. However, this is irrelevant to my error, and this error occurs for a variety of other metrics. Also, see here for the appropriate way to respond when you think the OP is "asking the wrong question"
I have an sklearn logistic model for which I am attempting to get the RMSE. However, when I .predict_proba, I get a vector of probabilities. However, my y_test is in its categorical form, which sklearn.linear_model.LogisticRegression just sort of dealt with automagically.
How to I reconcile these two things to get the RMSE?
>>> sklearn.metrics.mean_squared_error(y_test, pred_proba, sample_weight=weights_test)
ValueError: y_true and y_pred have different number of output (1!=13)
predict_proba is predicting the probability that a sample belongs to a class. The arg max of those probabilities is the predicted class (categorical form). RMSE is not a metric for classification. If you want to evaluate your model, consider a different metric like accuracy_score:
from sklearn.metrics import accuracy_score
predictions = your_model.predict(X_test)
print("Accuracy: %.3f" % accuracy_score(y_test, predictions))
The brier score, basically the mean squared error, is a known and valid loss function for classification models that leverage probability scores; I would take a look at that as well.
To your particular issue, you want to compare the probabilities returned for your target class, i.e. for a binary class problem:
from sklearn.metrics import brier_score_loss
probs = your_model.predict_proba(X_test)
brier_score_loss(y_true, probs[:, 1])
I'm not sure brier is formally defined for multiclass problems. I would point to the idea of mean misclassification error, which averages the error across classes.
To leverage this within the sklearn API, encode your y_true categorically, i.e. each class gets its own column, and call
sklearn.metrics.mean_squared_error(y_true, probs, multioutput=’uniform_average’)
Here is how you can calculate RMSE:
import numpy as np
from sklearn.metrics import mean_squared_error
x = np.range(10)
y = x
rmse = np.sqrt(mean_squared_error(x, y))
One can transform the y_test into a format compatible with the predict_proba output as follows:
model = sklearn.linear_model.LogisticRegression().fit(X,y) # or whatever model
label_encoder = sklearn.preprocessing.LabelEncoder()
label_encoder.classes_ = model.classes_
y_test_onehot = sklearn.preprocessing.OneHotEncoder().fit_transform(label_encoder.transform(y_test).reshape((-1,1)))
You can now apply any of the metrics in sklearn.metric. This is essential for computing, say, the brier score.
I am trying to evaluate a relevance of features and I am using DecisionTreeRegressor()
The related part of the code is presented below:
# TODO: Make a copy of the DataFrame, using the 'drop' function to drop the given feature
new_data = data.drop(['Frozen'], axis = 1)
# TODO: Split the data into training and testing sets(0.25) using the given feature as the target
# TODO: Set a random state.
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(new_data, data['Frozen'], test_size = 0.25, random_state = 1)
# TODO: Create a decision tree regressor and fit it to the training set
from sklearn.tree import DecisionTreeRegressor
regressor = DecisionTreeRegressor(random_state=1)
regressor.fit(X_train, y_train)
# TODO: Report the score of the prediction using the testing set
from sklearn.model_selection import cross_val_score
#score = cross_val_score(regressor, X_test, y_test)
score = regressor.score(X_test, y_test)
print score # python 2.x
When I run the print function, it returns the given score:
-0.649574327334
You can find the score function implementatioin and some explanation below here and below:
Returns the coefficient of determination R^2 of the prediction.
...
The best possible score is 1.0 and it can be negative (because the
model can be arbitrarily worse).
I could not grasp the whole concept yet, so this explanation is not very helpful for me. For instance I could not understand why score could be negative and what exactly it indicates (if something is squared, I would expect it can only be positive).
What does this score indicates and why can it be negative?
If you know any article (for starters) it might be helpful as well!
R^2 can be negative from its definition (https://en.wikipedia.org/wiki/Coefficient_of_determination) if the model fits the data worse than a horizontal line. Basically
R^2 = 1 - SS_res/SS_tot
and SS_res and SS_tot are always positive. If SS_res >> SS_tot, you have a negative R^2. Look at this answer as well: https://stats.stackexchange.com/questions/12900/when-is-r-squared-negative
The article execute cross_val_score in which DecisionTreeRegressor is implemented. You may take a look at the documentation of scikitlearn DecisionTreeRegressor.
Basically, the score you see is R^2, or (1-u/v). U is the squared sum residual of your prediction, and v is the total square sum(sample sum of square).
u/v can be arbitrary large when you make really bad prediction, while it can only be as small as zero given u and v are sum of squared residual(>=0)
I am asking the question here, even though I hesitated to post it on CrossValidated (or DataScience) StackExchange. I have a dataset of 60 labeled objects (to be used for training) and 150 unlabeled objects (for test). The aim of the problem is to predict the labels of the 150 objects (this used to be given as a homework problem). For each object, I computed 258 features. Considering each object as a sample, I have X_train : (60,258), y_train : (60,) (labels of the objects used for training) and X_test : (150,258). Since the solution of the homework problem was given, I also have the true labels of the 150 objects, in y_test : (150,).
In order to predict the labels of the 150 objects, I choose to use a LogisticRegression (the Scikit-learn implementation). The classifier is trained on (X_train, y_train), after the data has been normalized, and used to make predictions for the 150 objects. Those predictions are compared to y_test to assess the performance of the model. For reproducibility, I copy the code I have used.
from sklearn import metrics
from sklearn.preprocessing import StandardScaler
from sklearn.linear_model import LogisticRegression
from sklearn.pipeline import make_pipeline
from sklearn.model_selection import cross_val_score, crosss_val_predict
# Fit classifier
LogReg = LogisticRegression(C=1, class_weight='balanced')
scaler = StandardScaler()
clf = make_pipeline(StandardScaler(), LogReg)
LogReg.fit(X_train, y_train)
# Performance on training data
CV_score = cross_val_score(clf, X_train, y_train, cv=10, scoring='roc_auc')
print(CV_score)
# Performance on test data
probas = LogReg.predict_proba(X_test)[:, 1]
AUC = metrics.roc_auc_score(y_test, probas)
print(AUC)
The matrices X_train,y_train,X_test and y_test are saved in a .mat file available at this link. My problem is the following :
Using this approach, I get a good performance on training data (CV_score = 0.8) but the performance on the test data is much worse : AUC = 0.54 for C=1 in LogReg and AUC = 0.40 for C=0.01. How can I get AUC<0.5 if a naive classifier should score AUC = 0.5 ? Is this due to the fact that I have a small number of samples for training ?
I have noticed that the performance on test data improves if I change the code for :
y_pred = cross_val_predict(clf, X_test, y_test, cv=5)
AUC = metrics.roc_auc_score(y_test, y_pred)
print(AUC)
Indeed, AUC=0.87 for C=1 and 0.9 for C=0.01. Why is the AUC score so much better using cross validation predictions ? Is it because cross validation allows to make predictions on subsets of the test data which do not contain objects/samples which decrease the AUC ?
Looks like you are encountering an overfitting problem, i.e. the classifier trained using the training data is overfitting to the training data. It has poor generalization ability. That is why the performance on the testing dataset isn't good.
cross_val_predict is actually training the classifier using part of your testing data and then predict on the rest. So the performance is much better.
Overall, there seems to be quite some difference between your training and testing datasets. So the classifier with the highest training accuracy doesn't work well on your testing set.
Another point not directly related with your question: since the number of your training samples is much smaller than the feature dimensions, it may be helpful to perform dimension reduction before feeding to classifier.
It looks like your training and test process are inconsistent. Although from your code you intend to standardize your data, you fail to do so during testing. What I mean:
clf = make_pipeline(StandardScaler(), LogReg)
LogReg.fit(X_train, y_train)
Although you define a pipeline, you do not fit the pipeline (clf.fit) but only the Logistic Regression. This matters, because your cross-validated score is calculated with the pipeline (CV_score = cross_val_score(clf, X_train, y_train, cv=10, scoring='roc_auc')) but during test instead of using the pipeline as expected to predict, you use only LogReg, hence the test data are not standardized.
The second point you raise is different. In y_pred = cross_val_predict(clf, X_test, y_test, cv=5)
you get predictions by doing cross-validation on the test data, while ignoring the train data. Here, you do data standardization since you use clf and thus your score is high; this is evidence that the standardization step is important.
To summarize, standardizing the test data, I believe will improve your test score.
Firstly it makes no sense to have 258 features for 60 training items. Secondly CV=10 for 60 items means you split the data into 10 train/test sets. Each of these has 6 items only in the test set. So whatever results you obtain will be useless. You need more training data and less features.