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.
Related
I am testing RandomForestClassifier on simple dataset from sklearn. When I split the data with train_test_split, I get accuracy=0.89. If I use cross-validation with cross_val_score with same parameters of classifier, accuracy is smaller - about 0.83. Why?
Here is the code:
from sklearn.model_selection import cross_val_score, StratifiedKFold,GridSearchCV,train_test_split
from sklearn.metrics import accuracy_score,f1_score,make_scorer
from sklearn.ensemble import RandomForestClassifier
from sklearn.datasets import make_circles
np.random.seed(42)
#create dataset:
x, y = make_circles(n_samples=500, factor=0.1, noise=0.35, random_state=42)
#initialize stratified split:
skf = StratifiedKFold(n_splits=5, shuffle=True, random_state=42)
#create classifier:
clf = RandomForestClassifier(random_state=42, max_depth=12,n_jobs=-1,
oob_score=True,n_estimators=100,min_samples_leaf=10)
#average accuracy on cross-validation:
results = np.mean(cross_val_score(clf, x, y, cv=skf,scoring=make_scorer(accuracy_score)))
print("ACCURACY WITH CV = ",results)#prints 0.832
#use train_test_split
xtrain, xtest, ytrain, ytest = train_test_split(x, y, test_size=0.2)
clf=RandomForestClassifier(random_state=42, max_depth=12,n_jobs=-1, oob_score=True,n_estimators=100,min_samples_leaf=10)
clf.fit(xtrain,ytrain)
ypred=clf.predict(xtest)
print("ACCURACY WITHOUT CV = ",accuracy_score(ytest,ypred))#prints 0.89
what I got:
ACCURACY WITH CV = 0.83
ACCURACY WITHOUT CV = 0.89
Cross validation is used to run multiple experiments on different splits of data and then average their results. This is to ensure that the result of the experiment is not biased by one split, as it is in your case.
Your chosen seed along with some luck gave you a test train split which has higher accuracy than the average. The higher accuracy is an artifact of random sampling when making a split and not an indicator of better model performance.
Simply put:
Cross Validation makes multiple splits of data. Your model is trained
on all of these different splits and then the performance is
averaged.
If you pick one of these splits, you may get lucky and there might be
good overlap between the data points in your test and train set. Your
model will have high accuracy in this case.
Or you may get unlucky and there might not be a high overlap between
the data points in test and train set. Your model will have a lower
accuracy in this case.
Thus, cross validation is used to average the results of various such splits (5 in your case).
Here is your code run in a google colab notebook:
https://colab.research.google.com/drive/16-NotF-_WVLESmvGMONSGSZigxrT3KLx?usp=sharing
The last cell makes 5 different splits and then averages their accuracies. Notice how that is the same as the one you got from cross validation. Also notice how some splits have higher and some splits have a lower accuracy.
To further convince yourself, look at the output of:
cross_val_score(clf, x, y, cv=skf, scoring=make_scorer(accuracy_score))
The output is a list of scores (accuracies in your case) for the 5 different splits. You can see that they have varying values around 0.83
This is just up to chance for the split and the random state of the Random Forest Classifier. Try leaving the random_state=42 out and let it fit several times and you'll get a variance of different accuracies. By chance, I had one without CV of "just" 0.78! In contrast, the cv will give you and average (your calculated mean) PLUS an idea about how much your accuracy could vary around that.
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 writing a small program and I am training a random forest to predict a binary value. My dataset has around 20,000 entries and each entry has 25 features(continuous and categorical) with a binary target value to predict.
I am getting over 99% test accuracy which is surprisingly high. I tried to reduce the number of my features, even with two features I am still getting such high accuracy. I just want to make sure I am not doing anything wrong in my code, such as the training set leaking into my test set.
Here is the code snippet
data = pd.read_csv(r'test.csv')
data = data.drop_duplicates()
#spliting data
X = data.drop('label', axis=1)
y = data['label']
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=42)
#preproccessing the dataset by one hot encoding
l1 = OneHotEncoder(handle_unknown='ignore')
l1.fit(X_train)
X_train = l1.transform(X_train)
X_test = l1.transform(X_test)
from sklearn.ensemble import RandomForestRegressor, RandomForestClassifier
classifier = RandomForestClassifier(n_estimators=20, random_state=0)
classifier.fit(X_train, y_train.to_numpy())
#evaluation
y_pred = classifier.predict(X_test)
print(accuracy_score(y_test, y_pred))
additionally, I forgot to add that my dataset is balanced and precision and recall scores are 100% !
This is quite a big dataset. How balanced is your dataset? It might be the case your test split is filled mostly with the entries of one label and failed every time the entry was from the other label. Therefore, i would say accuracy is not a good measure to rely on in here.
Have a look at this:
Difference of model accuracy and performance
Have a look at your confusion matrix and inspect your splits.
I previously saw a post with code like this:
scalar = StandardScaler()
clf = svm.LinearSVC()
pipeline = Pipeline([('transformer', scalar), ('estimator', clf)])
cv = KFold(n_splits=4)
scores = cross_val_score(pipeline, X, y, cv = cv)
My understanding is that: when we apply scaler, we should use 3 out of the 4 folds to calculate mean and standard deviation, then we apply the mean and standard deviation to all 4 folds.
In the above code, how can I know that Sklearn is following the same strategy? On the other hand, if sklearn is not following the same strategy, which means sklearn would calculate the mean/std from all 4 folds. Would that mean I should not use the above codes?
I do like the above codes because it saves tons of time.
In the example you gave, I would add an additional step using sklearn.model_selection.train_test_split:
folds = 4
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=(1/folds), random_state=0, stratify=y)
scalar = StandardScaler()
clf = svm.LinearSVC()
pipeline = Pipeline([('transformer', scalar), ('estimator', clf)])
cv = KFold(n_splits=(folds - 1))
scores = cross_val_score(pipeline, X_train, y_train, cv = cv)
I think best practice is to only use the training data set (i.e., X_train, y_train) when tuning the hyperparameters of your model, and the test data set (i.e., X_test, y_test) should be used as a final check, to make sure your model isn't biased towards the validation folds. At that point you would apply the same scaler that you fit on your training data set to your testing data set.
Yes, this is done properly; this is one of the reasons for using pipelines: all the preprocessing is fitted only on training folds.
Some references.
Section 6.1.1 of the User Guide:
Safety
Pipelines help avoid leaking statistics from your test data into the trained model in cross-validation, by ensuring that the same samples are used to train the transformers and predictors.
The note at the end of section 3.1.1 of the User Guide:
Data transformation with held out data
Just as it is important to test a predictor on data held-out from training, preprocessing (such as standardization, feature selection, etc.) and similar data transformations similarly should be learnt from a training set and applied to held-out data for prediction:
...code sample...
A Pipeline makes it easier to compose estimators, providing this behavior under cross-validation:
...
Finally, you can look into the source for cross_val_score. It calls cross_validate, which clones and fits the estimator (in this case, the entire pipeline) on each training split. GitHub link.
I'm confused about using cross_val_predict in a test data set.
I created a simple Random Forest model and used cross_val_predict to make predictions:
from sklearn.ensemble import RandomForestClassifier
from sklearn.cross_validation import cross_val_predict, KFold
lr = RandomForestClassifier(random_state=1, class_weight="balanced", n_estimators=25, max_depth=6)
kf = KFold(train_df.shape[0], random_state=1)
predictions = cross_val_predict(lr,train_df[features_columns], train_df["target"], cv=kf)
predictions = pd.Series(predictions)
I'm confused on the next step here. How do I use what is learnt above to make predictions on the test data set?
I don't think cross_val_score or cross_val_predict uses fit before predicting. It does it on the fly. If you look at the documentation (section 3.1.1.1), you'll see that they never mention fit anywhere.
As #DmitryPolonskiy commented, the model has to be trained (with the fit method) before it can be used to predict.
# Train the model (a.k.a. `fit` training data to it).
lr.fit(train_df[features_columns], train_df["target"])
# Use the model to make predictions based on testing data.
y_pred = lr.predict(test_df[feature_columns])
# Compare the predicted y values to actual y values.
accuracy = (y_pred == test_df["target"]).mean()
cross_val_predict is a method of cross validation, which lets you determine the accuracy of your model. Take a look at sklearn's cross-validation page.
I am not sure the question was answered. I had a similar thought. I want compare the results (Accuracy for example) with the method that does not apply CV. The CV valiadte accuracy is on the X_train and y_train. The other method fit the model using X_trian and y_train, tested on the X_test and y_test. So the comparison is not fair since they are on different datasets.
What you can do is using the estimator returned by the cross_validate
lr_fit = cross_validate(lr, train_df[features_columns], train_df["target"], cv=kf, return_estimator=Ture)
y_pred = lr_fit.predict(test_df[feature_columns])
accuracy = (y_pred == test_df["target"]).mean()