Develop Reference

cross validation for split test and train datasets

Unlike standart data, I have dataset contain separetly as train, test1 and test2. I implemented ML algorithms and got performance metrics. But when i apply cross validation, it's getting complicated.. May be someone help me.. Thank you..
It's my code..
train = pd.read_csv('train-alldata.csv',sep=";")
test = pd.read_csv('test1-alldata.csv',sep=";")
test2 = pd.read_csv('test2-alldata.csv',sep=";")
X_train = train_pca_son.drop('churn_yn',axis=1)
y_train = train_pca_son['churn_yn']
X_test = test_pca_son.drop('churn_yn',axis=1)
y_test = test_pca_son['churn_yn']
X_test_2 = test2_pca_son.drop('churn_yn',axis=1)
y_test_2 = test2_pca_son['churn_yn']
For example, KNN Classifier.
knn_classifier = KNeighborsClassifier(n_neighbors =7,metric='euclidean')
knn_classifier.fit(X_train, y_train)
For K-Fold.
from sklearn import datasets
from sklearn.tree import DecisionTreeClassifier
from sklearn.model_selection import KFold, cross_val_score
dtc = DecisionTreeClassifier(random_state=42)
k_folds = KFold(n_splits = 5)
scores = cross_val_score(dtc, X, y, cv = k_folds)
print("Cross Validation Scores: ", scores)
print("Average CV Score: ", scores.mean())
print("Number of CV Scores used in Average: ", len(scores))

This is a variation on the "holdout test data" pattern (see also: Wikipedia: Training, Validation, Test / Confusion in terminology). For churn prediction: this may arise if you have two types of customers, or are evaluating on two time frames.
X_train, y_train ← perform training and hyperparameter tuning with this
X_test1, y_test1 ← test on this
X_test2, y_test2 ← test on this as well
Cross validation estimates holdout error using the training data—it may come up if you estimate hyperparameters with GridSearchCV. Final evaluation involves estimating performance on two test sets, separately or averaged over the two:
from sklearn.datasets import make_classification
from sklearn.model_selection import train_test_split
from sklearn.neighbors import KNeighborsClassifier
from sklearn.metrics import f1_score
X, y = make_classification(n_samples=1000, random_state=42)
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=.4)
X_test1, X_test2, y_test1, y_test2 = train_test_split(X_test, y_test, test_size=.5)
print(y_train.shape, y_test1.shape, y_test2.shape)
# (600,) (200,) (200,)
clf = KNeighborsClassifier(n_neighbors=7).fit(X_train, y_train)
print(f1_score(y_test1, clf.predict(X_test1)))
print(f1_score(y_test2, clf.predict(X_test2)))
# 0.819
# 0.805

Performance metric when using XGboost regressor with sklearn learning_curve

I've created xgboost regressor model and want to see how training and test performance changes as number of training set increases.
xgbm_reg = XGBRegressor()
tr_sizes, tr_scs, test_scs = learning_curve(estimator=xgbm_reg,
X=ori_X,y=y,
train_sizes=np.linspace(0.1, 1, 5),
cv=5)
What is performance is it using for tr_scs, and test_scs?
Sklearn doc tells me that
scoring : str or callable, default=None
A str (see model evaluation documentation) or a scorer callable object / function
with signature scorer(estimator, X, y)
So I've looked at XGboost documentation which says objective is default = reg:squarederror does this mean results of tr_scs, and test_scs are in terms of squared error?
I want to check by using cross_val_score
scoring = "neg_mean_squared_error"
cv_results = cross_val_score(xgbm_reg, ori_X, y, cv=5, scoring=scoring)
however not quite sure how to get squared_error from cross_val_score

The XGBRegressor's built-in scorer is the R-squared and this is the default scorer used in learning_curve and cross_val_score, see the code below.
from xgboost import XGBRegressor
from sklearn.datasets import make_regression
from sklearn.model_selection import learning_curve, cross_val_score, KFold
from sklearn.metrics import r2_score
# generate the data
X, y = make_regression(n_features=10, random_state=100)
# generate 5 CV splits
kf = KFold(n_splits=5, shuffle=False)
# calculate the CV scores using `learning_curve`, use 100% train size for comparison purposes
_, _, lc_scores = learning_curve(estimator=XGBRegressor(), X=X, y=y, train_sizes=[1.0], cv=kf)
print(lc_scores)
# [[0.51444244 0.70020972 0.64521668 0.36608259 0.81670165]]
# calculate the CV scores using `cross_val_score`
cv_scores = cross_val_score(estimator=XGBRegressor(), X=X, y=y, cv=kf)
print(cv_scores)
# [0.51444244 0.70020972 0.64521668 0.36608259 0.81670165]
# calculate the CV scores manually
xgb_scores = []
r2_scores = []
# iterate across the CV splits
for train_index, test_index in kf.split(X):
# extract the training and test data
X_train, X_test = X[train_index], X[test_index]
y_train, y_test = y[train_index], y[test_index]
# fit the model to the training data
estimator = XGBRegressor()
estimator.fit(X_train, y_train)
# score the test data using the XGBRegressor built-in scorer
xgb_scores.append(estimator.score(X_test, y_test))
# score the test data using the R-squared
y_pred = estimator.predict(X_test)
r2_scores.append(r2_score(y_test, y_pred))
print(xgb_scores)
# [0.5144424362721487, 0.7002097211679331, 0.645216683969211, 0.3660825936288453, 0.8167016490227281]
print(r2_scores)
# [0.5144424362721487, 0.7002097211679331, 0.645216683969211, 0.3660825936288453, 0.8167016490227281]

Repeated holdout method

How can I make "Repeated" holdout method, I made holdout method and get accuracy but need to repeat holdout method for 30 times
There is my code for holdout method
[IN]
X_train, X_test, Y_train, Y_test = train_test_split(X, Y.values.ravel(), random_state=100)
model = LogisticRegression()
model.fit(X_train, Y_train)
result = model.score(X_test, Y_test)
print("Accuracy: %.2f%%" % (result*100.0))
[OUT]
Accuracy: 49.62%
I see many codes for repeated method but only for K fold cross, nothing for holdout method

So to use a repeated holdout you could use the ShuffleSplit method from sklearn. A minimum working example (following the name conventions that you used) might be as follows:
from sklearn.modelselection import ShuffleSplit
from sklearn.linear_model import LogisticRegression
from sklearn.datasets import make_classification
# Create some artificial data to train on, can be replace by your own data
X, Y = make_classification()
rs = ShuffleSplit(n_splits=30, test_size=0.25, random_state=100)
model = LogisticRegression()
for train_index, test_index in rs.split(X):
X_train, Y_train = X[train_index], Y[train_index]
X_test, Y_test = X[test_index], Y[test_index]
model.fit(X_train,Y_train)
result = model.score(X_test, Y_test)
print("Accuracy: %.2f%%" % (result*100.0))
n_splits determines how many time you would like to repeat the holdout. test_size deterimines the fraction of samples that is sampled as a test set. In this case 75% is sampled as train set, whereas 25% is sampled to your test set. For reproducible results you can set the random_state (any number suffices, as long as you use the same number consistently).

Python: I want to perform 5 fold cross validation for logistic regression and report scores. Do I use LogisticRegressionCV() or cross_val_score()?

cross_val_scores gives different results than LogisticRegressionCV, and I can't figure out why.
Here is my code:
seed = 42
test_size = .33
X_train, X_test, Y_train, Y_test = train_test_split(scale(X),Y, test_size=test_size, random_state=seed)
#Below is my model that I use throughout the program.
model = LogisticRegressionCV(random_state=42)
print('Logistic Regression results:')
#For cross_val_score below, I just call LogisticRegression (and not LogRegCV) with the same parameters.
scores = cross_val_score(LogisticRegression(random_state=42), X_train, Y_train, scoring='accuracy', cv=5)
print(np.amax(scores)*100)
print("%.2f%% average accuracy with a standard deviation of %0.2f" % (scores.mean() * 100, scores.std() * 100))
model.fit(X_train, Y_train)
y_pred = model.predict(X_test)
predictions = [round(value) for value in y_pred]
accuracy = accuracy_score(Y_test, predictions)
coef=np.round(model.coef_,2)
print("Accuracy: %.2f%%" % (accuracy * 100.0))
The output is this.
Logistic Regression results:
79.90483019359885
79.69% average accuracy with a standard deviation of 0.14
Accuracy: 79.81%
Why is the maximum accuracy from cross_val_score higher than the accuracy used by LogisticRegressionCV?
And, I recognize that cross_val_scores does not return a model, which is why I want to use LogisticRegressionCV, but I am struggling to understand why it is not performing as well. Likewise, I am not sure how to get the standard deviations of the predictors from LogisticRegressionCV.

For me, there might be some points to take into consideration:
Cross validation is generally used whenever you should simulate a validation set (for instance when the training set is not that big to be divided into training, validation and test sets) and only uses training data. In your case you're computing accuracy of model on test data, making it impossible to exactly compare results.
According to the docs:
Cross-validation estimators are named EstimatorCV and tend to be roughly equivalent to GridSearchCV(Estimator(), ...). The advantage of using a cross-validation estimator over the canonical estimator class along with grid search is that they can take advantage of warm-starting by reusing precomputed results in the previous steps of the cross-validation process. This generally leads to speed improvements.
If you look at this snippet, you'll see that's what happens indeed:
import numpy as np
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression, LogisticRegressionCV
from sklearn.model_selection import cross_val_score, GridSearchCV, train_test_split
data = load_breast_cancer()
X, y = data['data'], data['target']
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
estimator = LogisticRegression(random_state=42, solver='liblinear')
grid = {
'C': np.power(10.0, np.arange(-10, 10)),
}
gs = GridSearchCV(estimator, param_grid=grid, scoring='accuracy', cv=5)
gs.fit(X_train, y_train)
print(gs.best_score_) # 0.953846153846154
lrcv = LogisticRegressionCV(Cs=list(np.power(10.0, np.arange(-10, 10))),
cv=5, scoring='accuracy', solver='liblinear', random_state=42)
lrcv.fit(X_train, y_train)
print(lrcv.scores_[1].mean(axis=0).max()) # 0.953846153846154
I would suggest to have a look here, too, so as to get the details of lrcv.scores_[1].mean(axis=0).max().
Eventually, to get the same results with cross_val_score you should better write:
score = cross_val_score(gs.best_estimator_, X_train, y_train, cv=5, scoring='accuracy')
score.mean() # 0.953846153846154

How to run SVC classifier after running 10-fold cross validation in sklearn?

I'm relatively new to machine learning and would like some help in the following:
I ran a Support Vector Machine Classifier (SVC) on my data with 10-fold cross validation and calculated the accuracy score (which was around 89%). I'm using Python and scikit-learn to perform the task. Here's a code snippet:
def get_scores(features,target,classifier):
X_train, X_test, y_train, y_test =train_test_split(features, target ,
test_size=0.3)
scores = cross_val_score(
classifier,
X_train,
y_train,
cv=10,
scoring='accuracy',
n_jobs=-1)
return(scores)
get_scores(features_from_df,target_from_df,svm.SVC())
Now, how can I use my classifier (after running the 10-folds cv) to test it on X_test and compare the predicted results to y_test? As you may have noticed, I only used X_train and y_train in the cross validation process.
I noticed that sklearn have cross_val_predict:
http://scikit-learn.org/stable/modules/generated/sklearn.model_selection.cross_val_predict.html should I replace my cross_val_score by cross_val_predict? just FYI: my target data column is binarized (have values of 0s and 1s).
If my approach is wrong, please advise me with the best way to proceed with.
Thanks!

You only need to split your X and y. Do not split the train and test.
Then you can pass your classifier in your case svm to the cross_val_score function to get the accuracy for each experiment.
In just 3 lines of code:
clf = svm.SVC(kernel='linear', C=1)
scores = cross_val_score(clf, X, y, cv=10)
print scores

from sklearn.metrics import classification_report
classifier = svm.SVC()
classifier.fit(X_train, y_train)
y_pred = classifier.predict(X_test)
print(classification_report(y_test , y_pred)

You're almost there:
# Build your classifier
classifier = svm.SVC()
# Train it on the entire training data set
classifier.fit(X_train, y_train)
# Get predictions on the test set
y_pred = classifier.predict(X_test)
At this point, you can use any metric from the sklearn.metrics module to determine how well you did. For example:
from sklearn.metrics import accuracy_score
print(accuracy_score(y_test, y_pred))