My r-squared score is coming negative but my accur

2020-06-06 02:36发布

For the code below, my r-squared score is coming out to be negative but my accuracies score using k-fold cross validation is coming out to be 92%. How's this possible? Im using random forest regression algorithm to predict some data. The link to the dataset is given in the link below: https://www.kaggle.com/ludobenistant/hr-analytics

import numpy as np
import pandas as pd
from sklearn.preprocessing import LabelEncoder,OneHotEncoder

dataset = pd.read_csv("HR_comma_sep.csv")
x = dataset.iloc[:,:-1].values   ##Independent variable
y = dataset.iloc[:,9].values     ##Dependent variable

##Encoding the categorical variables

le_x1 = LabelEncoder()
x[:,7] = le_x1.fit_transform(x[:,7])
le_x2 = LabelEncoder()
x[:,8] = le_x1.fit_transform(x[:,8])
ohe = OneHotEncoder(categorical_features = [7,8])
x = ohe.fit_transform(x).toarray()


##splitting the dataset in training and testing data

from sklearn.cross_validation import train_test_split
y = pd.factorize(dataset['left'].values)[0].reshape(-1, 1)
x_train, x_test, y_train, y_test = train_test_split(x, y, test_size = 0.2, random_state = 0)

from sklearn.preprocessing import StandardScaler
sc_x = StandardScaler()
x_train = sc_x.fit_transform(x_train)
x_test = sc_x.transform(x_test)
sc_y = StandardScaler()
y_train = sc_y.fit_transform(y_train)

from sklearn.ensemble import RandomForestRegressor
regressor = RandomForestRegressor(n_estimators = 10, random_state = 0)
regressor.fit(x_train, y_train)

y_pred = regressor.predict(x_test)
print(y_pred)
from sklearn.metrics import r2_score
r2_score(y_test , y_pred)

from sklearn.model_selection import cross_val_score
accuracies = cross_val_score(estimator = regressor, X = x_train, y = y_train, cv = 10)
accuracies.mean()
accuracies.std()

1条回答
家丑人穷心不美
2楼-- · 2020-06-06 03:36

There are several issues with your question...

For starters, you are doing a very basic mistake: you think you are using accuracy as a metric, while you are in a regression setting and the actual metric used underneath is the mean squared error (MSE).

Accuracy is a metric used in classification, and it has to do with the percentage of the correctly classified examples - check the Wikipedia entry for more details.

The metric used internally in your chosen regressor (Random Forest) is included in the verbose output of your regressor.fit(x_train, y_train) command - notice the criterion='mse' argument:

RandomForestRegressor(bootstrap=True, criterion='mse', max_depth=None,
           max_features='auto', max_leaf_nodes=None,
           min_impurity_split=1e-07, min_samples_leaf=1,
           min_samples_split=2, min_weight_fraction_leaf=0.0,
           n_estimators=10, n_jobs=1, oob_score=False, random_state=0,
           verbose=0, warm_start=False)

MSE is a positive continuous quantity, and it is not upper-bounded by 1, i.e. if you got a value of 0.92, this means... well, 0.92, and not 92%.

Knowing that, it is good practice to include explicitly the MSE as the scoring function of your cross-validation:

cv_mse = cross_val_score(estimator = regressor, X = x_train, y = y_train, cv = 10, scoring='neg_mean_squared_error')
cv_mse.mean()
# -2.433430574463703e-28

For all practical purposes, this is zero - you fit the training set almost perfectly; for confirmation, here is the (perfect again) R-squared score on your training set:

train_pred = regressor.predict(x_train)
r2_score(y_train , train_pred)
# 1.0

But, as always, the moment of truth comes when you apply your model on the test set; your second mistake here is that, since you train your regressor with scaled y_train, you should also scale y_test before evaluating:

y_test = sc_y.fit_transform(y_test)
r2_score(y_test , y_pred)
# 0.9998476914664215

and you get a very nice R-squared in the test set (close to 1).

What about the MSE?

from sklearn.metrics import mean_squared_error
mse_test = mean_squared_error(y_test, y_pred)
mse_test
# 0.00015230853357849051
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