What does a negative coefficient of determination mean for evaluating ridge regression?
$begingroup$
Judging by the negative result being displayed from my ridge.score() I am guessing that I am doing something wrong. Maybe someone could point me in the right direction?
# Create a practice data set for exploring Ridge Regression
data_2 = np.array([[1, 2, 0], [3, 4, 1], [5, 6, 0], [1, 3, 1],
[3, 5, 1], [1, 7, 0], [1, 8, 1]], dtype=np.float64)
# Separate X and Y
x_2 = data_2[:, [0, 1]]
y_2 = data_2[:, 2]
# Train Test Split
x_2_train, x_2_test, y_2_train, y_2_test = train_test_split(x_2, y_2, random_state=0)
# Scale the training data
scaler_2 = StandardScaler()
scaler_2.fit(x_2_train)
x_2_transformed = scaler_2.transform(x_2_train)
# Ridge Regression
ridge_2 = Ridge().fit(x_2_transformed, y_2_train)
x_2_test_scaled = scaler_2.transform(x_2_test)
ridge_2.score(x_2_test_scaled, y_2_test)
Output is: -4.47
EDIT: From reading the scikit learn docs this value is the R^2 value. I guess the question is though, how do we interpret this?
machine-learning scikit-learn ridge-regression
edited 41 secs ago
Wes
16810
asked 1 hour ago
EthanEthan
15015
$endgroup$
add a comment |
$begingroup$
Judging by the negative result being displayed from my ridge.score() I am guessing that I am doing something wrong. Maybe someone could point me in the right direction?
# Create a practice data set for exploring Ridge Regression
data_2 = np.array([[1, 2, 0], [3, 4, 1], [5, 6, 0], [1, 3, 1],
[3, 5, 1], [1, 7, 0], [1, 8, 1]], dtype=np.float64)
# Separate X and Y
x_2 = data_2[:, [0, 1]]
y_2 = data_2[:, 2]
# Train Test Split
x_2_train, x_2_test, y_2_train, y_2_test = train_test_split(x_2, y_2, random_state=0)
# Scale the training data
scaler_2 = StandardScaler()
scaler_2.fit(x_2_train)
x_2_transformed = scaler_2.transform(x_2_train)
# Ridge Regression
ridge_2 = Ridge().fit(x_2_transformed, y_2_train)
x_2_test_scaled = scaler_2.transform(x_2_test)
ridge_2.score(x_2_test_scaled, y_2_test)
Output is: -4.47
EDIT: From reading the scikit learn docs this value is the R^2 value. I guess the question is though, how do we interpret this?
machine-learning scikit-learn ridge-regression
edited 41 secs ago
Wes
16810
asked 1 hour ago
EthanEthan
15015
$endgroup$
$begingroup$
That's a very small data set. How many points are in your test set? Also, how did you generate those points? There are some good facilities with sklearn to generate synthetic data sets that might make more sense.
$endgroup$
– Wes
4 mins ago
add a comment |
$begingroup$
Judging by the negative result being displayed from my ridge.score() I am guessing that I am doing something wrong. Maybe someone could point me in the right direction?
# Create a practice data set for exploring Ridge Regression
data_2 = np.array([[1, 2, 0], [3, 4, 1], [5, 6, 0], [1, 3, 1],
[3, 5, 1], [1, 7, 0], [1, 8, 1]], dtype=np.float64)
# Separate X and Y
x_2 = data_2[:, [0, 1]]
y_2 = data_2[:, 2]
# Train Test Split
x_2_train, x_2_test, y_2_train, y_2_test = train_test_split(x_2, y_2, random_state=0)
# Scale the training data
scaler_2 = StandardScaler()
scaler_2.fit(x_2_train)
x_2_transformed = scaler_2.transform(x_2_train)
# Ridge Regression
ridge_2 = Ridge().fit(x_2_transformed, y_2_train)
x_2_test_scaled = scaler_2.transform(x_2_test)
ridge_2.score(x_2_test_scaled, y_2_test)
Output is: -4.47
EDIT: From reading the scikit learn docs this value is the R^2 value. I guess the question is though, how do we interpret this?
machine-learning scikit-learn ridge-regression
edited 41 secs ago
Wes
16810
asked 1 hour ago
EthanEthan
15015
$endgroup$
Judging by the negative result being displayed from my ridge.score() I am guessing that I am doing something wrong. Maybe someone could point me in the right direction?
# Create a practice data set for exploring Ridge Regression
data_2 = np.array([[1, 2, 0], [3, 4, 1], [5, 6, 0], [1, 3, 1],
[3, 5, 1], [1, 7, 0], [1, 8, 1]], dtype=np.float64)
# Separate X and Y
x_2 = data_2[:, [0, 1]]
y_2 = data_2[:, 2]
# Train Test Split
x_2_train, x_2_test, y_2_train, y_2_test = train_test_split(x_2, y_2, random_state=0)
# Scale the training data
scaler_2 = StandardScaler()
scaler_2.fit(x_2_train)
x_2_transformed = scaler_2.transform(x_2_train)
# Ridge Regression
ridge_2 = Ridge().fit(x_2_transformed, y_2_train)
x_2_test_scaled = scaler_2.transform(x_2_test)
ridge_2.score(x_2_test_scaled, y_2_test)
Output is: -4.47
EDIT: From reading the scikit learn docs this value is the R^2 value. I guess the question is though, how do we interpret this?
machine-learning scikit-learn ridge-regression
machine-learning scikit-learn ridge-regression
edited 41 secs ago
Wes
16810
asked 1 hour ago
EthanEthan
15015
edited 41 secs ago
Wes
16810
asked 1 hour ago
EthanEthan
15015
edited 41 secs ago
Wes
16810
edited 41 secs ago
Wes
16810
edited 41 secs ago
Wes
16810
16810
asked 1 hour ago
EthanEthan
15015
asked 1 hour ago
EthanEthan
15015
asked 1 hour ago
EthanEthan
15015
15015
$begingroup$
That's a very small data set. How many points are in your test set? Also, how did you generate those points? There are some good facilities with sklearn to generate synthetic data sets that might make more sense.
$endgroup$
– Wes
4 mins ago
add a comment |
$begingroup$
That's a very small data set. How many points are in your test set? Also, how did you generate those points? There are some good facilities with sklearn to generate synthetic data sets that might make more sense.
$endgroup$
– Wes
4 mins ago
$begingroup$
That's a very small data set. How many points are in your test set? Also, how did you generate those points? There are some good facilities with sklearn to generate synthetic data sets that might make more sense.
$endgroup$
– Wes
4 mins ago
$begingroup$
That's a very small data set. How many points are in your test set? Also, how did you generate those points? There are some good facilities with sklearn to generate synthetic data sets that might make more sense.
$endgroup$
– Wes
4 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A negative value means you're getting a terrible fit - which makes sense if you create a test set that doesn't have the same distribution as the training set.
From the sklearn documentation:
The coefficient $R^2$ is defined as (1 - u/v), where u is the residual sum of squares ((y_true - y_pred) ** 2).sum() and v is the total sum of squares ((y_true - y_true.mean()) ** 2).sum(). The best possible score is 1.0 and it can be negative (because the model can be arbitrarily worse). A constant model that always predicts the expected value of y, disregarding the input features, would get a $R^2$ score of 0.0.
answered 2 mins ago
WesWes
16810
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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votes
$begingroup$
A negative value means you're getting a terrible fit - which makes sense if you create a test set that doesn't have the same distribution as the training set.
From the sklearn documentation:
The coefficient $R^2$ is defined as (1 - u/v), where u is the residual sum of squares ((y_true - y_pred) ** 2).sum() and v is the total sum of squares ((y_true - y_true.mean()) ** 2).sum(). The best possible score is 1.0 and it can be negative (because the model can be arbitrarily worse). A constant model that always predicts the expected value of y, disregarding the input features, would get a $R^2$ score of 0.0.
answered 2 mins ago
WesWes
16810
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
A negative value means you're getting a terrible fit - which makes sense if you create a test set that doesn't have the same distribution as the training set.
From the sklearn documentation:
The coefficient $R^2$ is defined as (1 - u/v), where u is the residual sum of squares ((y_true - y_pred) ** 2).sum() and v is the total sum of squares ((y_true - y_true.mean()) ** 2).sum(). The best possible score is 1.0 and it can be negative (because the model can be arbitrarily worse). A constant model that always predicts the expected value of y, disregarding the input features, would get a $R^2$ score of 0.0.
answered 2 mins ago
WesWes
16810
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
A negative value means you're getting a terrible fit - which makes sense if you create a test set that doesn't have the same distribution as the training set.
From the sklearn documentation:
The coefficient $R^2$ is defined as (1 - u/v), where u is the residual sum of squares ((y_true - y_pred) ** 2).sum() and v is the total sum of squares ((y_true - y_true.mean()) ** 2).sum(). The best possible score is 1.0 and it can be negative (because the model can be arbitrarily worse). A constant model that always predicts the expected value of y, disregarding the input features, would get a $R^2$ score of 0.0.
answered 2 mins ago
WesWes
16810
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A negative value means you're getting a terrible fit - which makes sense if you create a test set that doesn't have the same distribution as the training set.
From the sklearn documentation:
The coefficient $R^2$ is defined as (1 - u/v), where u is the residual sum of squares ((y_true - y_pred) ** 2).sum() and v is the total sum of squares ((y_true - y_true.mean()) ** 2).sum(). The best possible score is 1.0 and it can be negative (because the model can be arbitrarily worse). A constant model that always predicts the expected value of y, disregarding the input features, would get a $R^2$ score of 0.0.
answered 2 mins ago
WesWes
16810
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 mins ago
WesWes
16810
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 mins ago
WesWes
16810
answered 2 mins ago
WesWes
16810
16810
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
That's a very small data set. How many points are in your test set? Also, how did you generate those points? There are some good facilities with sklearn to generate synthetic data sets that might make more sense.
$endgroup$
– Wes
4 mins ago