Code evaluation: Logistic regression with K fold validation. Is correct?

1

The code below is an attempt to do a logistic regression with k-fold cross-validation. The idea is to take the confusion matrices generated in each fold and then generate a mean confusion matrix with 95% confidence (confidence interval for the mean of 95%).

Is the code making sense? Any suggestions for improvement / correction?

import numpy as np
from sklearn import model_selection
from sklearn import datasets
from sklearn import svm
import pandas as pd
from sklearn.linear_model import LogisticRegression
from scipy.stats import sem, t
from scipy import mean

lista_matrizes = []


UNSW = pd.read_csv('/home/sec/Desktop/CEFET/UNSW_NB15_testing-set.csv')

previsores = UNSW.iloc[:,UNSW.columns.isin(('sload','dload',
                                                   'spkts','dpkts','swin','dwin','smean','dmean',
'sjit','djit','sinpkt','dinpkt','tcprtt','synack','ackdat','ct_srv_src','ct_srv_dst','ct_dst_ltm',
 'ct_src_ltm','ct_src_dport_ltm','ct_dst_sport_ltm','ct_dst_src_ltm')) ].values


classe= UNSW.iloc[:, -1].values

#iris = datasets.load_iris()
#print(iris.data.shape, iris.target.shape)

X_train, X_test, y_train, y_test = model_selection.train_test_split(
previsores, classe, test_size=0.4, random_state=0)

print(X_train.shape, y_train.shape)
#((90, 4), (90,))
print(X_test.shape, y_test.shape)
#((60, 4), (60,))

logmodel = LogisticRegression()
logmodel.fit(X_train,y_train)
print(previsores.shape)

#clf = svm.SVC(kernel='linear', C=1).fit(X_train, y_train)
print(logmodel.score(X_test, y_test) ) 


#Computing cross-validated metrics

logmodel = LogisticRegression()
scores = model_selection.cross_val_score(
    logmodel, previsores, classe, cv=30)

print(scores)                                             
#array([ 0.96...,  1.  ...,  0.96...,  0.96...,  1.        ])
#print("Accuracy: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2))

########K FOLD
print('########K FOLD########K FOLD########K FOLD########K FOLD')
from sklearn.model_selection import KFold
from sklearn.metrics import confusion_matrix

kf = KFold(n_splits=3, random_state=None, shuffle=False)
kf.get_n_splits(previsores)
for train_index, test_index in kf.split(previsores):

    X_train, X_test = previsores[train_index], previsores[test_index]
    y_train, y_test = classe[train_index], classe[test_index]

    logmodel.fit(X_train, y_train)
    print (confusion_matrix(y_test, logmodel.predict(X_test)))

    lista_matrizes.append(confusion_matrix(y_test, logmodel.predict(X_test)))


#print(lista_matrizes)

final = np.mean(lista_matrizes, axis=0)
print(f" Mean confidence Matrix  \n{final}")


# o intervalo de confiança
def mean_confidence_interval(data, confidence=0.95):

    #data = [1, 2, 3, 4, 5]

    n = len(data)
    m = mean(data)
    std_err = sem(data)
    h = std_err * t.ppf((1 + confidence) / 2, n - 1)

    start = m - h
    #print (start)


    end = m + h
    #print (end)

    return start, end




print()
print(f"Intervalo de confiança: \n{mean_confidence_interval(final)}")
    
asked by anonymous 29.06.2018 / 01:36

1 answer

2

Your code is correct. But a little messy. so you explained in the question statement you do not need the first part above k-fold. Another advice is to import the libs all at the beginning of the code as this does not cause confusion.

I usually translate these procedures into functions to create a Pipeline to generate reports for future models. Another interesting thing to evaluate in your code is:

sklearn.metrics.classification_report

This function returns a report containing, precission, recall, and f1-score only very important metrics for classification problems. Variable names and functional comments: It is important to define well, as your code grows, you may end up failing to enjoy what you have done.

    
07.08.2018 / 15:49