Low score in Linear Regression with discrete attributes
I'm trying to do a linear regression in my dataframe. The dataframe is about apple applications, and I want to predict the notes of applications. The notes are in following format:
1.0
1.5
2.0
2.5
...
5.0
My code is:
atributos = ['size_bytes','price','rating_count_tot','cont_rating','sup_devices_num','num_screenshots','num_lang','vpp_lic']
atrib_prev = ['nota']
X = np.array(data_regress.drop(['nota'],1))
y = np.array(data_regress['nota'])
X = preprocessing.scale(X)
X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, y, test_size=0.2)
clf = LinearRegression()
clf.fit(X_train, y_train)
accuracy = clf.score(X_test, y_test)
print(accuracy)
But my accuracy is 0.046295306696438665. I think this occurs because the linear model is predicting real values, while my 'note' is real, but at intervals. I don't know how to round this values before the clf.score.
pandas jupyter-notebook sklearn-pandas
edited Nov 14 '18 at 12:56
Aqueous Carlos
293213
asked Nov 13 '18 at 0:10
Giovanni BrogiatoGiovanni Brogiato
1
add a comment |
I'm trying to do a linear regression in my dataframe. The dataframe is about apple applications, and I want to predict the notes of applications. The notes are in following format:
1.0
1.5
2.0
2.5
...
5.0
My code is:
atributos = ['size_bytes','price','rating_count_tot','cont_rating','sup_devices_num','num_screenshots','num_lang','vpp_lic']
atrib_prev = ['nota']
X = np.array(data_regress.drop(['nota'],1))
y = np.array(data_regress['nota'])
X = preprocessing.scale(X)
X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, y, test_size=0.2)
clf = LinearRegression()
clf.fit(X_train, y_train)
accuracy = clf.score(X_test, y_test)
print(accuracy)
But my accuracy is 0.046295306696438665. I think this occurs because the linear model is predicting real values, while my 'note' is real, but at intervals. I don't know how to round this values before the clf.score.
pandas jupyter-notebook sklearn-pandas
edited Nov 14 '18 at 12:56
Aqueous Carlos
293213
asked Nov 13 '18 at 0:10
Giovanni BrogiatoGiovanni Brogiato
1
add a comment |
I'm trying to do a linear regression in my dataframe. The dataframe is about apple applications, and I want to predict the notes of applications. The notes are in following format:
1.0
1.5
2.0
2.5
...
5.0
My code is:
atributos = ['size_bytes','price','rating_count_tot','cont_rating','sup_devices_num','num_screenshots','num_lang','vpp_lic']
atrib_prev = ['nota']
X = np.array(data_regress.drop(['nota'],1))
y = np.array(data_regress['nota'])
X = preprocessing.scale(X)
X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, y, test_size=0.2)
clf = LinearRegression()
clf.fit(X_train, y_train)
accuracy = clf.score(X_test, y_test)
print(accuracy)
But my accuracy is 0.046295306696438665. I think this occurs because the linear model is predicting real values, while my 'note' is real, but at intervals. I don't know how to round this values before the clf.score.
pandas jupyter-notebook sklearn-pandas
edited Nov 14 '18 at 12:56
Aqueous Carlos
293213
asked Nov 13 '18 at 0:10
Giovanni BrogiatoGiovanni Brogiato
1
I'm trying to do a linear regression in my dataframe. The dataframe is about apple applications, and I want to predict the notes of applications. The notes are in following format:
1.0
1.5
2.0
2.5
...
5.0
My code is:
atributos = ['size_bytes','price','rating_count_tot','cont_rating','sup_devices_num','num_screenshots','num_lang','vpp_lic']
atrib_prev = ['nota']
X = np.array(data_regress.drop(['nota'],1))
y = np.array(data_regress['nota'])
X = preprocessing.scale(X)
X_train, X_test, y_train, y_test = cross_validation.train_test_split(X, y, test_size=0.2)
clf = LinearRegression()
clf.fit(X_train, y_train)
accuracy = clf.score(X_test, y_test)
print(accuracy)
But my accuracy is 0.046295306696438665. I think this occurs because the linear model is predicting real values, while my 'note' is real, but at intervals. I don't know how to round this values before the clf.score.
pandas jupyter-notebook sklearn-pandas
pandas jupyter-notebook sklearn-pandas
edited Nov 14 '18 at 12:56
Aqueous Carlos
293213
asked Nov 13 '18 at 0:10
Giovanni BrogiatoGiovanni Brogiato
1
edited Nov 14 '18 at 12:56
Aqueous Carlos
293213
asked Nov 13 '18 at 0:10
Giovanni BrogiatoGiovanni Brogiato
1
edited Nov 14 '18 at 12:56
Aqueous Carlos
293213
edited Nov 14 '18 at 12:56
Aqueous Carlos
293213
edited Nov 14 '18 at 12:56
Aqueous Carlos
293213
293213
asked Nov 13 '18 at 0:10
Giovanni BrogiatoGiovanni Brogiato
1
asked Nov 13 '18 at 0:10
Giovanni BrogiatoGiovanni Brogiato
1
asked Nov 13 '18 at 0:10
Giovanni BrogiatoGiovanni Brogiato
1
1
add a comment |
add a comment |
1 Answer
1
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oldest
votes
First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)
Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:
# Get actual predictions
y_pred = clf.predict(X_test)
# You will need to implement the round function yourself
y_pred_rounded = round(y_pred)
# Call the appropriate scorer
score = r2_score(y_test, y_pred_rounded)
You can look at the sklearn documentation here for available metrics in sklearn.
answered Nov 14 '18 at 14:25
Vivek KumarVivek Kumar
15.5k41953
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)
Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:
# Get actual predictions
y_pred = clf.predict(X_test)
# You will need to implement the round function yourself
y_pred_rounded = round(y_pred)
# Call the appropriate scorer
score = r2_score(y_test, y_pred_rounded)
You can look at the sklearn documentation here for available metrics in sklearn.
answered Nov 14 '18 at 14:25
Vivek KumarVivek Kumar
15.5k41953
add a comment |
First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)
Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:
# Get actual predictions
y_pred = clf.predict(X_test)
# You will need to implement the round function yourself
y_pred_rounded = round(y_pred)
# Call the appropriate scorer
score = r2_score(y_test, y_pred_rounded)
You can look at the sklearn documentation here for available metrics in sklearn.
answered Nov 14 '18 at 14:25
Vivek KumarVivek Kumar
15.5k41953
add a comment |
First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)
Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:
# Get actual predictions
y_pred = clf.predict(X_test)
# You will need to implement the round function yourself
y_pred_rounded = round(y_pred)
# Call the appropriate scorer
score = r2_score(y_test, y_pred_rounded)
You can look at the sklearn documentation here for available metrics in sklearn.
answered Nov 14 '18 at 14:25
Vivek KumarVivek Kumar
15.5k41953
First, for regression models, clf.score() calculates R-squared value, not accuracy. So you would need to decide if you want to treat this problem as a classification problem (For some fixed number of target labels) or a regression problem (for a real-valued target)
Secondly, if you insist on using regression models and not classification, you can call clf.predict() to first get the predicted values and then round off as you want to, and then call r2_score() on actual and predicted labels. Something like:
# Get actual predictions
y_pred = clf.predict(X_test)
# You will need to implement the round function yourself
y_pred_rounded = round(y_pred)
# Call the appropriate scorer
score = r2_score(y_test, y_pred_rounded)
You can look at the sklearn documentation here for available metrics in sklearn.
answered Nov 14 '18 at 14:25
Vivek KumarVivek Kumar
15.5k41953
answered Nov 14 '18 at 14:25
Vivek KumarVivek Kumar
15.5k41953
answered Nov 14 '18 at 14:25
Vivek KumarVivek Kumar
15.5k41953
answered Nov 14 '18 at 14:25
Vivek KumarVivek Kumar
15.5k41953
15.5k41953
add a comment |
add a comment |
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