07_cross_validation


Cross-validation for parameter tuning, model selection, and feature selection

From the video series: Introduction to machine learning with scikit-learn

Agenda

  • What is the drawback of using the train/test split procedure for model evaluation?
  • How does K-fold cross-validation overcome this limitation?
  • How can cross-validation be used for selecting tuning parameters, choosing between models, and selecting features?
  • What are some possible improvements to cross-validation?

Review of model evaluation procedures

Motivation: Need a way to choose between machine learning models

  • Goal is to estimate likely performance of a model on out-of-sample data

Initial idea: Train and test on the same data

  • But, maximizing training accuracy rewards overly complex models which overfit the training data

Alternative idea: Train/test split

  • Split the dataset into two pieces, so that the model can be trained and tested on different data
  • Testing accuracy is a better estimate than training accuracy of out-of-sample performance
  • But, it provides a high variance estimate since changing which observations happen to be in the testing set can significantly change testing accuracy

In [2]:
from sklearn.datasets import load_iris
from sklearn.cross_validation import train_test_split
from sklearn.neighbors import KNeighborsClassifier
from sklearn import metrics

In [3]:
# read in the iris data
iris = load_iris()

# create X (features) and y (response)
X = iris.data
y = iris.target

In [4]:
# use train/test split with different random_state values
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=4)

# check classification accuracy of KNN with K=5
knn = KNeighborsClassifier(n_neighbors=5)
knn.fit(X_train, y_train)
y_pred = knn.predict(X_test)
print metrics.accuracy_score(y_test, y_pred)


0.973684210526

Question: What if we created a bunch of train/test splits, calculated the testing accuracy for each, and averaged the results together?

Answer: That's the essense of cross-validation!

Steps for K-fold cross-validation

  1. Split the dataset into K equal partitions (or "folds").
  2. Use fold 1 as the testing set and the union of the other folds as the training set.
  3. Calculate testing accuracy.
  4. Repeat steps 2 and 3 K times, using a different fold as the testing set each time.
  5. Use the average testing accuracy as the estimate of out-of-sample accuracy.

Diagram of 5-fold cross-validation:


In [5]:
# simulate splitting a dataset of 25 observations into 5 folds
from sklearn.cross_validation import KFold
kf = KFold(25, n_folds=5, shuffle=False)

# print the contents of each training and testing set
print '{} {:^61} {}'.format('Iteration', 'Training set observations', 'Testing set observations')
for iteration, data in enumerate(kf, start=1):
    print '{:^9} {} {:^25}'.format(iteration, data[0], data[1])


Iteration                   Training set observations                   Testing set observations
    1     [ 5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24]        [0 1 2 3 4]       
    2     [ 0  1  2  3  4 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24]        [5 6 7 8 9]       
    3     [ 0  1  2  3  4  5  6  7  8  9 15 16 17 18 19 20 21 22 23 24]     [10 11 12 13 14]     
    4     [ 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 20 21 22 23 24]     [15 16 17 18 19]     
    5     [ 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19]     [20 21 22 23 24]     
  • Dataset contains 25 observations (numbered 0 through 24)
  • 5-fold cross-validation, thus it runs for 5 iterations
  • For each iteration, every observation is either in the training set or the testing set, but not both
  • Every observation is in the testing set exactly once

Comparing cross-validation to train/test split

Advantages of cross-validation:

  • More accurate estimate of out-of-sample accuracy
  • More "efficient" use of data (every observation is used for both training and testing)

Advantages of train/test split:

  • Runs K times faster than K-fold cross-validation
  • Simpler to examine the detailed results of the testing process

Cross-validation recommendations

  1. K can be any number, but K=10 is generally recommended
  2. For classification problems, stratified sampling is recommended for creating the folds
    • Each response class should be represented with equal proportions in each of the K folds
    • scikit-learn's cross_val_score function does this by default

Cross-validation example: parameter tuning

Goal: Select the best tuning parameters (aka "hyperparameters") for KNN on the iris dataset


In [6]:
from sklearn.cross_validation import cross_val_score

In [7]:
# 10-fold cross-validation with K=5 for KNN (the n_neighbors parameter)
knn = KNeighborsClassifier(n_neighbors=5)
scores = cross_val_score(knn, X, y, cv=10, scoring='accuracy')
print scores


[ 1.          0.93333333  1.          1.          0.86666667  0.93333333
  0.93333333  1.          1.          1.        ]

In [8]:
# use average accuracy as an estimate of out-of-sample accuracy
print scores.mean()


0.966666666667

In [9]:
# search for an optimal value of K for KNN
k_range = range(1, 31)
k_scores = []
for k in k_range:
    knn = KNeighborsClassifier(n_neighbors=k)
    scores = cross_val_score(knn, X, y, cv=10, scoring='accuracy')
    k_scores.append(scores.mean())
print k_scores


[0.95999999999999996, 0.95333333333333337, 0.96666666666666656, 0.96666666666666656, 0.96666666666666679, 0.96666666666666679, 0.96666666666666679, 0.96666666666666679, 0.97333333333333338, 0.96666666666666679, 0.96666666666666679, 0.97333333333333338, 0.98000000000000009, 0.97333333333333338, 0.97333333333333338, 0.97333333333333338, 0.97333333333333338, 0.98000000000000009, 0.97333333333333338, 0.98000000000000009, 0.96666666666666656, 0.96666666666666656, 0.97333333333333338, 0.95999999999999996, 0.96666666666666656, 0.95999999999999996, 0.96666666666666656, 0.95333333333333337, 0.95333333333333337, 0.95333333333333337]

In [10]:
import matplotlib.pyplot as plt
%matplotlib inline

# plot the value of K for KNN (x-axis) versus the cross-validated accuracy (y-axis)
plt.plot(k_range, k_scores)
plt.xlabel('Value of K for KNN')
plt.ylabel('Cross-Validated Accuracy')


Out[10]:
<matplotlib.text.Text at 0x168cc3c8>

Cross-validation example: model selection

Goal: Compare the best KNN model with logistic regression on the iris dataset


In [11]:
# 10-fold cross-validation with the best KNN model
knn = KNeighborsClassifier(n_neighbors=20)
print cross_val_score(knn, X, y, cv=10, scoring='accuracy').mean()


0.98

In [12]:
# 10-fold cross-validation with logistic regression
from sklearn.linear_model import LogisticRegression
logreg = LogisticRegression()
print cross_val_score(logreg, X, y, cv=10, scoring='accuracy').mean()


0.953333333333

Cross-validation example: feature selection

Goal: Select whether the Newspaper feature should be included in the linear regression model on the advertising dataset


In [13]:
import pandas as pd
import numpy as np
from sklearn.linear_model import LinearRegression

In [14]:
# read in the advertising dataset
data = pd.read_csv('http://www-bcf.usc.edu/~gareth/ISL/Advertising.csv', index_col=0)

In [15]:
# create a Python list of three feature names
feature_cols = ['TV', 'Radio', 'Newspaper']

# use the list to select a subset of the DataFrame (X)
X = data[feature_cols]

# select the Sales column as the response (y)
y = data.Sales

In [16]:
# 10-fold cross-validation with all three features
lm = LinearRegression()
scores = cross_val_score(lm, X, y, cv=10, scoring='mean_squared_error')
print scores


[-3.56038438 -3.29767522 -2.08943356 -2.82474283 -1.3027754  -1.74163618
 -8.17338214 -2.11409746 -3.04273109 -2.45281793]

In [17]:
# fix the sign of MSE scores
mse_scores = -scores
print mse_scores


[ 3.56038438  3.29767522  2.08943356  2.82474283  1.3027754   1.74163618
  8.17338214  2.11409746  3.04273109  2.45281793]

In [18]:
# convert from MSE to RMSE
rmse_scores = np.sqrt(mse_scores)
print rmse_scores


[ 1.88689808  1.81595022  1.44548731  1.68069713  1.14139187  1.31971064
  2.85891276  1.45399362  1.7443426   1.56614748]

In [19]:
# calculate the average RMSE
print rmse_scores.mean()


1.69135317081

In [20]:
# 10-fold cross-validation with two features (excluding Newspaper)
feature_cols = ['TV', 'Radio']
X = data[feature_cols]
print np.sqrt(-cross_val_score(lm, X, y, cv=10, scoring='mean_squared_error')).mean()


1.67967484191

Improvements to cross-validation

Repeated cross-validation

  • Repeat cross-validation multiple times (with different random splits of the data) and average the results
  • More reliable estimate of out-of-sample performance by reducing the variance associated with a single trial of cross-validation

Creating a hold-out set

  • "Hold out" a portion of the data before beginning the model building process
  • Locate the best model using cross-validation on the remaining data, and test it using the hold-out set
  • More reliable estimate of out-of-sample performance since hold-out set is truly out-of-sample

Feature engineering and selection within cross-validation iterations

  • Normally, feature engineering and selection occurs before cross-validation
  • Instead, perform all feature engineering and selection within each cross-validation iteration
  • More reliable estimate of out-of-sample performance since it better mimics the application of the model to out-of-sample data

Comments or Questions?


In [1]:
from IPython.core.display import HTML
def css_styling():
    styles = open("styles/custom.css", "r").read()
    return HTML(styles)
css_styling()


Out[1]: