Dimensions problem linear regression Python scikit learn
0
I'm implementing a function in which I have to perform a linear regression using scikit learn.
What I have when running it with an example:
X_train.shape=(34,3)
X_test.shape=(12,3)
Y_train.shape=(34,1)
Y_test.shape=(12,1)
Then
lm.fit(X_train,Y_train)
Y_pred = lm.predict(X_test)
However Python tells me there is a mistake at this line
dico['R2 value']=lm.score(Y_test, Y_pred)
What Python tells me:
ValueError: shapes (12,1) and (3,1) not aligned: 1 (dim 1) != 3 (dim 0)
Thanks in advance for the help anyone could bring me :)
Alex
python-2.7 scikit-learn linear-regression
edited Nov 16 '18 at 11:42
Vivek Kumar
16.8k42156
asked Nov 16 '18 at 11:36
Alex Alex
175
add a comment |
0
I'm implementing a function in which I have to perform a linear regression using scikit learn.
What I have when running it with an example:
X_train.shape=(34,3)
X_test.shape=(12,3)
Y_train.shape=(34,1)
Y_test.shape=(12,1)
Then
lm.fit(X_train,Y_train)
Y_pred = lm.predict(X_test)
However Python tells me there is a mistake at this line
dico['R2 value']=lm.score(Y_test, Y_pred)
What Python tells me:
ValueError: shapes (12,1) and (3,1) not aligned: 1 (dim 1) != 3 (dim 0)
Thanks in advance for the help anyone could bring me :)
Alex
python-2.7 scikit-learn linear-regression
edited Nov 16 '18 at 11:42
Vivek Kumar
16.8k42156
asked Nov 16 '18 at 11:36
Alex Alex
175
add a comment |
0
0
0
I'm implementing a function in which I have to perform a linear regression using scikit learn.
What I have when running it with an example:
X_train.shape=(34,3)
X_test.shape=(12,3)
Y_train.shape=(34,1)
Y_test.shape=(12,1)
Then
lm.fit(X_train,Y_train)
Y_pred = lm.predict(X_test)
However Python tells me there is a mistake at this line
dico['R2 value']=lm.score(Y_test, Y_pred)
What Python tells me:
ValueError: shapes (12,1) and (3,1) not aligned: 1 (dim 1) != 3 (dim 0)
Thanks in advance for the help anyone could bring me :)
Alex
python-2.7 scikit-learn linear-regression
edited Nov 16 '18 at 11:42
Vivek Kumar
16.8k42156
asked Nov 16 '18 at 11:36
Alex Alex
175
I'm implementing a function in which I have to perform a linear regression using scikit learn.
What I have when running it with an example:
X_train.shape=(34,3)
X_test.shape=(12,3)
Y_train.shape=(34,1)
Y_test.shape=(12,1)
Then
lm.fit(X_train,Y_train)
Y_pred = lm.predict(X_test)
However Python tells me there is a mistake at this line
dico['R2 value']=lm.score(Y_test, Y_pred)
What Python tells me:
ValueError: shapes (12,1) and (3,1) not aligned: 1 (dim 1) != 3 (dim 0)
Thanks in advance for the help anyone could bring me :)
Alex
python-2.7 scikit-learn linear-regression
python-2.7 scikit-learn linear-regression
edited Nov 16 '18 at 11:42
Vivek Kumar
16.8k42156
asked Nov 16 '18 at 11:36
Alex Alex
175
edited Nov 16 '18 at 11:42
Vivek Kumar
16.8k42156
asked Nov 16 '18 at 11:36
Alex Alex
175
edited Nov 16 '18 at 11:42
Vivek Kumar
16.8k42156
edited Nov 16 '18 at 11:42
Vivek Kumar
16.8k42156
edited Nov 16 '18 at 11:42
Vivek Kumar
16.8k42156
16.8k42156
asked Nov 16 '18 at 11:36
Alex Alex
175
asked Nov 16 '18 at 11:36
Alex Alex
175
asked Nov 16 '18 at 11:36
Alex Alex
175
175
add a comment |
add a comment |
1 Answer
1
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oldest
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For using lm.score() you need to pass X_test, y_test.
dico['R2 value']=lm.score(X_test, Y_test)
See the documentation here:
score(X, y, sample_weight=None)
X : array-like, shape = (n_samples, n_features) Test samples.
For some estimators this may be a precomputed kernel matrix instead,
shape = (n_samples, n_samples_fitted], where n_samples_fitted is the
number of samples used in the fitting for the estimator.
y : array-like, shape = (n_samples) or (n_samples, n_outputs) True values for X.
sample_weight : array-like, shape = [n_samples], optional Sample weights.
You are trying to use the score method as a metric method, which is wrong. A score() method on any estimator will itself calculate the predictions and then send them to appropriate metric scorer.
If you want to use Y_test and Y_pred yourself, then you can do this:
from sklearn.metrics import r2_score
dico['R2 value'] = r2_score(Y_test, Y_pred)
answered Nov 16 '18 at 11:42
Vivek KumarVivek Kumar
16.8k42156
Thanks a lot for your help ! Seems I was a bit confused :) However now I don't get it why the r2 score is really low (0.11) whereas the dataset I used is the iris one...
– Alex
Nov 16 '18 at 11:58
@Alex Iris is a classification dataset and you are using regression model (LinearRegression with R-squared) and hence not working. Use models which haveClassifierin their names
– Vivek Kumar
Nov 16 '18 at 12:04
Hmm I don't see why because I only kept the setosa type of iris so that the regression would have a sense. My features were SepalLengthCm, SepalWidthCm, PetalLengthCm and I wanted to predict PetalWidthCm. So why wouldn't the linear regression be legit?
– Alex
Nov 16 '18 at 12:59
@Alex Well, in that case the regression makes sense. But then you need to consider if it actually makes sense to predict petalwidth from other features. Regression will only perform good, if the dependent variable (petalwidth in this case) is actually a dependent of other variables. Which I dont think it is.
– Vivek Kumar
Nov 19 '18 at 10:02
One last question: Can I still use that LinearRegression from sklearn if there are also some nominal/ordinal features? (Obviously I would encode them before performing the regression)
– Alex
Nov 22 '18 at 9:21
add a comment |
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1 Answer
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1 Answer
1
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1
For using lm.score() you need to pass X_test, y_test.
dico['R2 value']=lm.score(X_test, Y_test)
See the documentation here:
score(X, y, sample_weight=None)
X : array-like, shape = (n_samples, n_features) Test samples.
For some estimators this may be a precomputed kernel matrix instead,
shape = (n_samples, n_samples_fitted], where n_samples_fitted is the
number of samples used in the fitting for the estimator.
y : array-like, shape = (n_samples) or (n_samples, n_outputs) True values for X.
sample_weight : array-like, shape = [n_samples], optional Sample weights.
You are trying to use the score method as a metric method, which is wrong. A score() method on any estimator will itself calculate the predictions and then send them to appropriate metric scorer.
If you want to use Y_test and Y_pred yourself, then you can do this:
from sklearn.metrics import r2_score
dico['R2 value'] = r2_score(Y_test, Y_pred)
answered Nov 16 '18 at 11:42
Vivek KumarVivek Kumar
16.8k42156
Thanks a lot for your help ! Seems I was a bit confused :) However now I don't get it why the r2 score is really low (0.11) whereas the dataset I used is the iris one...
– Alex
Nov 16 '18 at 11:58
@Alex Iris is a classification dataset and you are using regression model (LinearRegression with R-squared) and hence not working. Use models which haveClassifierin their names
– Vivek Kumar
Nov 16 '18 at 12:04
Hmm I don't see why because I only kept the setosa type of iris so that the regression would have a sense. My features were SepalLengthCm, SepalWidthCm, PetalLengthCm and I wanted to predict PetalWidthCm. So why wouldn't the linear regression be legit?
– Alex
Nov 16 '18 at 12:59
@Alex Well, in that case the regression makes sense. But then you need to consider if it actually makes sense to predict petalwidth from other features. Regression will only perform good, if the dependent variable (petalwidth in this case) is actually a dependent of other variables. Which I dont think it is.
– Vivek Kumar
Nov 19 '18 at 10:02
One last question: Can I still use that LinearRegression from sklearn if there are also some nominal/ordinal features? (Obviously I would encode them before performing the regression)
– Alex
Nov 22 '18 at 9:21
add a comment |
1
For using lm.score() you need to pass X_test, y_test.
dico['R2 value']=lm.score(X_test, Y_test)
See the documentation here:
score(X, y, sample_weight=None)
X : array-like, shape = (n_samples, n_features) Test samples.
For some estimators this may be a precomputed kernel matrix instead,
shape = (n_samples, n_samples_fitted], where n_samples_fitted is the
number of samples used in the fitting for the estimator.
y : array-like, shape = (n_samples) or (n_samples, n_outputs) True values for X.
sample_weight : array-like, shape = [n_samples], optional Sample weights.
You are trying to use the score method as a metric method, which is wrong. A score() method on any estimator will itself calculate the predictions and then send them to appropriate metric scorer.
If you want to use Y_test and Y_pred yourself, then you can do this:
from sklearn.metrics import r2_score
dico['R2 value'] = r2_score(Y_test, Y_pred)
answered Nov 16 '18 at 11:42
Vivek KumarVivek Kumar
16.8k42156
Thanks a lot for your help ! Seems I was a bit confused :) However now I don't get it why the r2 score is really low (0.11) whereas the dataset I used is the iris one...
– Alex
Nov 16 '18 at 11:58
@Alex Iris is a classification dataset and you are using regression model (LinearRegression with R-squared) and hence not working. Use models which haveClassifierin their names
– Vivek Kumar
Nov 16 '18 at 12:04
Hmm I don't see why because I only kept the setosa type of iris so that the regression would have a sense. My features were SepalLengthCm, SepalWidthCm, PetalLengthCm and I wanted to predict PetalWidthCm. So why wouldn't the linear regression be legit?
– Alex
Nov 16 '18 at 12:59
@Alex Well, in that case the regression makes sense. But then you need to consider if it actually makes sense to predict petalwidth from other features. Regression will only perform good, if the dependent variable (petalwidth in this case) is actually a dependent of other variables. Which I dont think it is.
– Vivek Kumar
Nov 19 '18 at 10:02
One last question: Can I still use that LinearRegression from sklearn if there are also some nominal/ordinal features? (Obviously I would encode them before performing the regression)
– Alex
Nov 22 '18 at 9:21
add a comment |
1
1
1
For using lm.score() you need to pass X_test, y_test.
dico['R2 value']=lm.score(X_test, Y_test)
See the documentation here:
score(X, y, sample_weight=None)
X : array-like, shape = (n_samples, n_features) Test samples.
For some estimators this may be a precomputed kernel matrix instead,
shape = (n_samples, n_samples_fitted], where n_samples_fitted is the
number of samples used in the fitting for the estimator.
y : array-like, shape = (n_samples) or (n_samples, n_outputs) True values for X.
sample_weight : array-like, shape = [n_samples], optional Sample weights.
You are trying to use the score method as a metric method, which is wrong. A score() method on any estimator will itself calculate the predictions and then send them to appropriate metric scorer.
If you want to use Y_test and Y_pred yourself, then you can do this:
from sklearn.metrics import r2_score
dico['R2 value'] = r2_score(Y_test, Y_pred)
answered Nov 16 '18 at 11:42
Vivek KumarVivek Kumar
16.8k42156
For using lm.score() you need to pass X_test, y_test.
dico['R2 value']=lm.score(X_test, Y_test)
See the documentation here:
score(X, y, sample_weight=None)
X : array-like, shape = (n_samples, n_features) Test samples.
For some estimators this may be a precomputed kernel matrix instead,
shape = (n_samples, n_samples_fitted], where n_samples_fitted is the
number of samples used in the fitting for the estimator.
y : array-like, shape = (n_samples) or (n_samples, n_outputs) True values for X.
sample_weight : array-like, shape = [n_samples], optional Sample weights.
You are trying to use the score method as a metric method, which is wrong. A score() method on any estimator will itself calculate the predictions and then send them to appropriate metric scorer.
If you want to use Y_test and Y_pred yourself, then you can do this:
from sklearn.metrics import r2_score
dico['R2 value'] = r2_score(Y_test, Y_pred)
answered Nov 16 '18 at 11:42
Vivek KumarVivek Kumar
16.8k42156
edited Nov 16 '18 at 11:48
answered Nov 16 '18 at 11:42
Vivek KumarVivek Kumar
16.8k42156
answered Nov 16 '18 at 11:42
Vivek KumarVivek Kumar
16.8k42156
answered Nov 16 '18 at 11:42
Vivek KumarVivek Kumar
16.8k42156
16.8k42156
Thanks a lot for your help ! Seems I was a bit confused :) However now I don't get it why the r2 score is really low (0.11) whereas the dataset I used is the iris one...
– Alex
Nov 16 '18 at 11:58
@Alex Iris is a classification dataset and you are using regression model (LinearRegression with R-squared) and hence not working. Use models which haveClassifierin their names
– Vivek Kumar
Nov 16 '18 at 12:04
Hmm I don't see why because I only kept the setosa type of iris so that the regression would have a sense. My features were SepalLengthCm, SepalWidthCm, PetalLengthCm and I wanted to predict PetalWidthCm. So why wouldn't the linear regression be legit?
– Alex
Nov 16 '18 at 12:59
@Alex Well, in that case the regression makes sense. But then you need to consider if it actually makes sense to predict petalwidth from other features. Regression will only perform good, if the dependent variable (petalwidth in this case) is actually a dependent of other variables. Which I dont think it is.
– Vivek Kumar
Nov 19 '18 at 10:02
One last question: Can I still use that LinearRegression from sklearn if there are also some nominal/ordinal features? (Obviously I would encode them before performing the regression)
– Alex
Nov 22 '18 at 9:21
add a comment |
Thanks a lot for your help ! Seems I was a bit confused :) However now I don't get it why the r2 score is really low (0.11) whereas the dataset I used is the iris one...
– Alex
Nov 16 '18 at 11:58
@Alex Iris is a classification dataset and you are using regression model (LinearRegression with R-squared) and hence not working. Use models which haveClassifierin their names
– Vivek Kumar
Nov 16 '18 at 12:04
Hmm I don't see why because I only kept the setosa type of iris so that the regression would have a sense. My features were SepalLengthCm, SepalWidthCm, PetalLengthCm and I wanted to predict PetalWidthCm. So why wouldn't the linear regression be legit?
– Alex
Nov 16 '18 at 12:59
@Alex Well, in that case the regression makes sense. But then you need to consider if it actually makes sense to predict petalwidth from other features. Regression will only perform good, if the dependent variable (petalwidth in this case) is actually a dependent of other variables. Which I dont think it is.
– Vivek Kumar
Nov 19 '18 at 10:02
One last question: Can I still use that LinearRegression from sklearn if there are also some nominal/ordinal features? (Obviously I would encode them before performing the regression)
– Alex
Nov 22 '18 at 9:21
Thanks a lot for your help ! Seems I was a bit confused :) However now I don't get it why the r2 score is really low (0.11) whereas the dataset I used is the iris one...
– Alex
Nov 16 '18 at 11:58
Thanks a lot for your help ! Seems I was a bit confused :) However now I don't get it why the r2 score is really low (0.11) whereas the dataset I used is the iris one...
– Alex
Nov 16 '18 at 11:58
@Alex Iris is a classification dataset and you are using regression model (LinearRegression with R-squared) and hence not working. Use models which have
Classifier in their names– Vivek Kumar
Nov 16 '18 at 12:04
@Alex Iris is a classification dataset and you are using regression model (LinearRegression with R-squared) and hence not working. Use models which have
Classifier in their names– Vivek Kumar
Nov 16 '18 at 12:04
Hmm I don't see why because I only kept the setosa type of iris so that the regression would have a sense. My features were SepalLengthCm, SepalWidthCm, PetalLengthCm and I wanted to predict PetalWidthCm. So why wouldn't the linear regression be legit?
– Alex
Nov 16 '18 at 12:59
Hmm I don't see why because I only kept the setosa type of iris so that the regression would have a sense. My features were SepalLengthCm, SepalWidthCm, PetalLengthCm and I wanted to predict PetalWidthCm. So why wouldn't the linear regression be legit?
– Alex
Nov 16 '18 at 12:59
@Alex Well, in that case the regression makes sense. But then you need to consider if it actually makes sense to predict petalwidth from other features. Regression will only perform good, if the dependent variable (petalwidth in this case) is actually a dependent of other variables. Which I dont think it is.
– Vivek Kumar
Nov 19 '18 at 10:02
@Alex Well, in that case the regression makes sense. But then you need to consider if it actually makes sense to predict petalwidth from other features. Regression will only perform good, if the dependent variable (petalwidth in this case) is actually a dependent of other variables. Which I dont think it is.
– Vivek Kumar
Nov 19 '18 at 10:02
One last question: Can I still use that LinearRegression from sklearn if there are also some nominal/ordinal features? (Obviously I would encode them before performing the regression)
– Alex
Nov 22 '18 at 9:21
One last question: Can I still use that LinearRegression from sklearn if there are also some nominal/ordinal features? (Obviously I would encode them before performing the regression)
– Alex
Nov 22 '18 at 9:21
add a comment |
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