model.score and r2_score giving different values for a regression model
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I am build a linear regression model and a decision tree model using sklearn. I want to compare the performance of these two models, I have calculated the r2_score for both the models. I have calculated the model.score for both the values. I am confused which is a better metric to compare the performance of these models. Also what does model.score gives?
from sklearn.metrics import r2_score
score_DT = r2_score(y_pred_DT,y_test)
dt_score = regressorDT.score(X_test,y_test)
machine-learning scikit-learn regression decision-trees linear-regression
asked 23 hours ago
ChinniChinni
103
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add a comment |
$begingroup$
I am build a linear regression model and a decision tree model using sklearn. I want to compare the performance of these two models, I have calculated the r2_score for both the models. I have calculated the model.score for both the values. I am confused which is a better metric to compare the performance of these models. Also what does model.score gives?
from sklearn.metrics import r2_score
score_DT = r2_score(y_pred_DT,y_test)
dt_score = regressorDT.score(X_test,y_test)
machine-learning scikit-learn regression decision-trees linear-regression
asked 23 hours ago
ChinniChinni
103
$endgroup$
add a comment |
$begingroup$
I am build a linear regression model and a decision tree model using sklearn. I want to compare the performance of these two models, I have calculated the r2_score for both the models. I have calculated the model.score for both the values. I am confused which is a better metric to compare the performance of these models. Also what does model.score gives?
from sklearn.metrics import r2_score
score_DT = r2_score(y_pred_DT,y_test)
dt_score = regressorDT.score(X_test,y_test)
machine-learning scikit-learn regression decision-trees linear-regression
asked 23 hours ago
ChinniChinni
103
$endgroup$
I am build a linear regression model and a decision tree model using sklearn. I want to compare the performance of these two models, I have calculated the r2_score for both the models. I have calculated the model.score for both the values. I am confused which is a better metric to compare the performance of these models. Also what does model.score gives?
from sklearn.metrics import r2_score
score_DT = r2_score(y_pred_DT,y_test)
dt_score = regressorDT.score(X_test,y_test)
machine-learning scikit-learn regression decision-trees linear-regression
machine-learning scikit-learn regression decision-trees linear-regression
asked 23 hours ago
ChinniChinni
103
asked 23 hours ago
ChinniChinni
103
asked 23 hours ago
ChinniChinni
103
asked 23 hours ago
ChinniChinni
103
asked 23 hours ago
ChinniChinni
103
103
add a comment |
add a comment |
1 Answer
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Both functions are the same r2 metric and should produce the same results.
Your usage of the r2_score function is wrong. The first argument should be the ground truth values and not the predicted values, so in your case it should be:
score_DT = r2_score(y_test, y_pred_DT)
answered 23 hours ago
Mark.FMark.F
891318
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
Both functions are the same r2 metric and should produce the same results.
Your usage of the r2_score function is wrong. The first argument should be the ground truth values and not the predicted values, so in your case it should be:
score_DT = r2_score(y_test, y_pred_DT)
answered 23 hours ago
Mark.FMark.F
891318
$endgroup$
add a comment |
$begingroup$
Both functions are the same r2 metric and should produce the same results.
Your usage of the r2_score function is wrong. The first argument should be the ground truth values and not the predicted values, so in your case it should be:
score_DT = r2_score(y_test, y_pred_DT)
answered 23 hours ago
Mark.FMark.F
891318
$endgroup$
add a comment |
$begingroup$
Both functions are the same r2 metric and should produce the same results.
Your usage of the r2_score function is wrong. The first argument should be the ground truth values and not the predicted values, so in your case it should be:
score_DT = r2_score(y_test, y_pred_DT)
answered 23 hours ago
Mark.FMark.F
891318
$endgroup$
Both functions are the same r2 metric and should produce the same results.
Your usage of the r2_score function is wrong. The first argument should be the ground truth values and not the predicted values, so in your case it should be:
score_DT = r2_score(y_test, y_pred_DT)
answered 23 hours ago
Mark.FMark.F
891318
answered 23 hours ago
Mark.FMark.F
891318
answered 23 hours ago
Mark.FMark.F
891318
answered 23 hours ago
Mark.FMark.F
891318
891318
add a comment |
add a comment |
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