Mixing
Pass prediction method as function argument
I'd like to specify which prediction method to use via function argument. Something like:
from sklearn.linear_model import LinearRegression
def Process(data_y_train, data_x_train, data_x_test,
model=LinearRegression, predict_method=predict):
model_fit = model().fit(data_x_train, data_y_train)
predicted_values = model_fit.predict_method(data_x_test)
return predicted_values
Passing the model function via arugment model (e.g., LinearRegression, LogisticRegression) works well, but I'm having trouble passing the predict method (e.g., predict, predict_proba) via argument predict_method.
When I specify predict_method=predict, I get an error of 'name 'predict' is not defined'; if I specify predict_method=LinearRegression.predict, I get an error saying ''LinearRegression' object has no attribute 'predict_function''.
Per this discussion, I also tried
import sklearn.linear_model.LinearRegression
def Process(data_y_train, data_x_train, data_x_test,
model_module='sklearn.linear_model.LinearRegression',
model=LinearRegression, predict_method='predict'):
model_fit = model().fit(data_x_train, data_y_train)
predict_call = getattr(__import__(model_module), predict_method)
predicted_values = model_fit.predict_call(data_x_test)
return predicted_values
But here I get an error: No module named LinearRegression.
Thank you for your help!
python scikit-learn arguments predict
asked Jan 2 at 23:34
nancynancy
1
add a comment |
I'd like to specify which prediction method to use via function argument. Something like:
from sklearn.linear_model import LinearRegression
def Process(data_y_train, data_x_train, data_x_test,
model=LinearRegression, predict_method=predict):
model_fit = model().fit(data_x_train, data_y_train)
predicted_values = model_fit.predict_method(data_x_test)
return predicted_values
Passing the model function via arugment model (e.g., LinearRegression, LogisticRegression) works well, but I'm having trouble passing the predict method (e.g., predict, predict_proba) via argument predict_method.
When I specify predict_method=predict, I get an error of 'name 'predict' is not defined'; if I specify predict_method=LinearRegression.predict, I get an error saying ''LinearRegression' object has no attribute 'predict_function''.
Per this discussion, I also tried
import sklearn.linear_model.LinearRegression
def Process(data_y_train, data_x_train, data_x_test,
model_module='sklearn.linear_model.LinearRegression',
model=LinearRegression, predict_method='predict'):
model_fit = model().fit(data_x_train, data_y_train)
predict_call = getattr(__import__(model_module), predict_method)
predicted_values = model_fit.predict_call(data_x_test)
return predicted_values
But here I get an error: No module named LinearRegression.
Thank you for your help!
python scikit-learn arguments predict
asked Jan 2 at 23:34
nancynancy
1
Can you edit in the expected output and try to make this a bit more minimal? (see how to make a Minimal, Complete, and Verifiable example)
– Ethan K888
Jan 2 at 23:39
add a comment |
I'd like to specify which prediction method to use via function argument. Something like:
from sklearn.linear_model import LinearRegression
def Process(data_y_train, data_x_train, data_x_test,
model=LinearRegression, predict_method=predict):
model_fit = model().fit(data_x_train, data_y_train)
predicted_values = model_fit.predict_method(data_x_test)
return predicted_values
Passing the model function via arugment model (e.g., LinearRegression, LogisticRegression) works well, but I'm having trouble passing the predict method (e.g., predict, predict_proba) via argument predict_method.
When I specify predict_method=predict, I get an error of 'name 'predict' is not defined'; if I specify predict_method=LinearRegression.predict, I get an error saying ''LinearRegression' object has no attribute 'predict_function''.
Per this discussion, I also tried
import sklearn.linear_model.LinearRegression
def Process(data_y_train, data_x_train, data_x_test,
model_module='sklearn.linear_model.LinearRegression',
model=LinearRegression, predict_method='predict'):
model_fit = model().fit(data_x_train, data_y_train)
predict_call = getattr(__import__(model_module), predict_method)
predicted_values = model_fit.predict_call(data_x_test)
return predicted_values
But here I get an error: No module named LinearRegression.
Thank you for your help!
python scikit-learn arguments predict
asked Jan 2 at 23:34
nancynancy
1
I'd like to specify which prediction method to use via function argument. Something like:
from sklearn.linear_model import LinearRegression
def Process(data_y_train, data_x_train, data_x_test,
model=LinearRegression, predict_method=predict):
model_fit = model().fit(data_x_train, data_y_train)
predicted_values = model_fit.predict_method(data_x_test)
return predicted_values
Passing the model function via arugment model (e.g., LinearRegression, LogisticRegression) works well, but I'm having trouble passing the predict method (e.g., predict, predict_proba) via argument predict_method.
When I specify predict_method=predict, I get an error of 'name 'predict' is not defined'; if I specify predict_method=LinearRegression.predict, I get an error saying ''LinearRegression' object has no attribute 'predict_function''.
Per this discussion, I also tried
import sklearn.linear_model.LinearRegression
def Process(data_y_train, data_x_train, data_x_test,
model_module='sklearn.linear_model.LinearRegression',
model=LinearRegression, predict_method='predict'):
model_fit = model().fit(data_x_train, data_y_train)
predict_call = getattr(__import__(model_module), predict_method)
predicted_values = model_fit.predict_call(data_x_test)
return predicted_values
But here I get an error: No module named LinearRegression.
Thank you for your help!
python scikit-learn arguments predict
python scikit-learn arguments predict
asked Jan 2 at 23:34
nancynancy
1
asked Jan 2 at 23:34
nancynancy
1
asked Jan 2 at 23:34
nancynancy
1
asked Jan 2 at 23:34
nancynancy
1
asked Jan 2 at 23:34
nancynancy
1
1
Can you edit in the expected output and try to make this a bit more minimal? (see how to make a Minimal, Complete, and Verifiable example)
– Ethan K888
Jan 2 at 23:39
add a comment |
Can you edit in the expected output and try to make this a bit more minimal? (see how to make a Minimal, Complete, and Verifiable example)
– Ethan K888
Jan 2 at 23:39
Can you edit in the expected output and try to make this a bit more minimal? (see how to make a Minimal, Complete, and Verifiable example)
– Ethan K888
Jan 2 at 23:39
Can you edit in the expected output and try to make this a bit more minimal? (see how to make a Minimal, Complete, and Verifiable example)
– Ethan K888
Jan 2 at 23:39
add a comment |
1 Answer
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I notice that in your code, you're not using the predict_method parameter that you passed in anywhere in your code, so I don't think what you have written is what you were trying to do.
Currently, in your code, you are storing the output of the function model().fit(data_x_train, data_y_train) in the variable model_fit and then calling the predict_method attribute of that variable. If the above still doesn't work, that must be where the error is coming from, then.
I suspect what you want to do is the following:
def Process(data_y_train, data_x_train, data_x_test,
model=LinearRegression, predict_method=LinearRegression.predict):
model_instance = model() # create an instance of the class stored in the variable 'model'
model_instance.fit(data_x_train, data_y_train) # run the function 'fit' belonging to that instance
predicted_values = predict_method(model_instance,data_x_test) # run the method stored in the variable 'predict_method' - you have to pass the instance the method belongs to in the first parameter
return predicted_values
Some more information:
LinearRegressionis a class. It defines a bunch of methods, etc.- To create an instance of that class, you must do something like
inst = LinearRegression(). The variableinstis now an instance of the classLinearRegression
LinearRegression.predictis an example of an instance method. This means it needs an instance to run (or can be thought of as to 'operate on' in this case)- I can therefore call
inst.predict(x,y,z)but notLinearRegression.predict(x,y,z)directly. - If you want to call
LinearRegression.predict, you have to pass in the instance in the first argument:LinearRegression.predict(inst,x,y,z)
Regarding what you tried afterwards: calling a function from a string holding the function's name is not necessary in this situation and only increases the overhead, so it's probably not the correct way to go here :)
Hope this helps.
I did useLinearRegression.predictand just double checked withpredict_method=LogisticRegression.predict_proba, I'm getting a similar error ''LogisticRegression' object has no attribute 'predict_method''. I'm usingsklearn0.20.0. Thanks Abhinav!
– nancy
Jan 3 at 0:00
@nancy I edited my answer, see if that helps
– user10859576
Jan 3 at 0:22
add a comment |
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1 Answer
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1 Answer
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I notice that in your code, you're not using the predict_method parameter that you passed in anywhere in your code, so I don't think what you have written is what you were trying to do.
Currently, in your code, you are storing the output of the function model().fit(data_x_train, data_y_train) in the variable model_fit and then calling the predict_method attribute of that variable. If the above still doesn't work, that must be where the error is coming from, then.
I suspect what you want to do is the following:
def Process(data_y_train, data_x_train, data_x_test,
model=LinearRegression, predict_method=LinearRegression.predict):
model_instance = model() # create an instance of the class stored in the variable 'model'
model_instance.fit(data_x_train, data_y_train) # run the function 'fit' belonging to that instance
predicted_values = predict_method(model_instance,data_x_test) # run the method stored in the variable 'predict_method' - you have to pass the instance the method belongs to in the first parameter
return predicted_values
Some more information:
LinearRegressionis a class. It defines a bunch of methods, etc.- To create an instance of that class, you must do something like
inst = LinearRegression(). The variableinstis now an instance of the classLinearRegression
LinearRegression.predictis an example of an instance method. This means it needs an instance to run (or can be thought of as to 'operate on' in this case)- I can therefore call
inst.predict(x,y,z)but notLinearRegression.predict(x,y,z)directly. - If you want to call
LinearRegression.predict, you have to pass in the instance in the first argument:LinearRegression.predict(inst,x,y,z)
Regarding what you tried afterwards: calling a function from a string holding the function's name is not necessary in this situation and only increases the overhead, so it's probably not the correct way to go here :)
Hope this helps.
I did useLinearRegression.predictand just double checked withpredict_method=LogisticRegression.predict_proba, I'm getting a similar error ''LogisticRegression' object has no attribute 'predict_method''. I'm usingsklearn0.20.0. Thanks Abhinav!
– nancy
Jan 3 at 0:00
@nancy I edited my answer, see if that helps
– user10859576
Jan 3 at 0:22
add a comment |
I notice that in your code, you're not using the predict_method parameter that you passed in anywhere in your code, so I don't think what you have written is what you were trying to do.
Currently, in your code, you are storing the output of the function model().fit(data_x_train, data_y_train) in the variable model_fit and then calling the predict_method attribute of that variable. If the above still doesn't work, that must be where the error is coming from, then.
I suspect what you want to do is the following:
def Process(data_y_train, data_x_train, data_x_test,
model=LinearRegression, predict_method=LinearRegression.predict):
model_instance = model() # create an instance of the class stored in the variable 'model'
model_instance.fit(data_x_train, data_y_train) # run the function 'fit' belonging to that instance
predicted_values = predict_method(model_instance,data_x_test) # run the method stored in the variable 'predict_method' - you have to pass the instance the method belongs to in the first parameter
return predicted_values
Some more information:
LinearRegressionis a class. It defines a bunch of methods, etc.- To create an instance of that class, you must do something like
inst = LinearRegression(). The variableinstis now an instance of the classLinearRegression
LinearRegression.predictis an example of an instance method. This means it needs an instance to run (or can be thought of as to 'operate on' in this case)- I can therefore call
inst.predict(x,y,z)but notLinearRegression.predict(x,y,z)directly. - If you want to call
LinearRegression.predict, you have to pass in the instance in the first argument:LinearRegression.predict(inst,x,y,z)
Regarding what you tried afterwards: calling a function from a string holding the function's name is not necessary in this situation and only increases the overhead, so it's probably not the correct way to go here :)
Hope this helps.
I did useLinearRegression.predictand just double checked withpredict_method=LogisticRegression.predict_proba, I'm getting a similar error ''LogisticRegression' object has no attribute 'predict_method''. I'm usingsklearn0.20.0. Thanks Abhinav!
– nancy
Jan 3 at 0:00
@nancy I edited my answer, see if that helps
– user10859576
Jan 3 at 0:22
add a comment |
I notice that in your code, you're not using the predict_method parameter that you passed in anywhere in your code, so I don't think what you have written is what you were trying to do.
Currently, in your code, you are storing the output of the function model().fit(data_x_train, data_y_train) in the variable model_fit and then calling the predict_method attribute of that variable. If the above still doesn't work, that must be where the error is coming from, then.
I suspect what you want to do is the following:
def Process(data_y_train, data_x_train, data_x_test,
model=LinearRegression, predict_method=LinearRegression.predict):
model_instance = model() # create an instance of the class stored in the variable 'model'
model_instance.fit(data_x_train, data_y_train) # run the function 'fit' belonging to that instance
predicted_values = predict_method(model_instance,data_x_test) # run the method stored in the variable 'predict_method' - you have to pass the instance the method belongs to in the first parameter
return predicted_values
Some more information:
LinearRegressionis a class. It defines a bunch of methods, etc.- To create an instance of that class, you must do something like
inst = LinearRegression(). The variableinstis now an instance of the classLinearRegression
LinearRegression.predictis an example of an instance method. This means it needs an instance to run (or can be thought of as to 'operate on' in this case)- I can therefore call
inst.predict(x,y,z)but notLinearRegression.predict(x,y,z)directly. - If you want to call
LinearRegression.predict, you have to pass in the instance in the first argument:LinearRegression.predict(inst,x,y,z)
Regarding what you tried afterwards: calling a function from a string holding the function's name is not necessary in this situation and only increases the overhead, so it's probably not the correct way to go here :)
Hope this helps.
I notice that in your code, you're not using the predict_method parameter that you passed in anywhere in your code, so I don't think what you have written is what you were trying to do.
Currently, in your code, you are storing the output of the function model().fit(data_x_train, data_y_train) in the variable model_fit and then calling the predict_method attribute of that variable. If the above still doesn't work, that must be where the error is coming from, then.
I suspect what you want to do is the following:
def Process(data_y_train, data_x_train, data_x_test,
model=LinearRegression, predict_method=LinearRegression.predict):
model_instance = model() # create an instance of the class stored in the variable 'model'
model_instance.fit(data_x_train, data_y_train) # run the function 'fit' belonging to that instance
predicted_values = predict_method(model_instance,data_x_test) # run the method stored in the variable 'predict_method' - you have to pass the instance the method belongs to in the first parameter
return predicted_values
Some more information:
LinearRegressionis a class. It defines a bunch of methods, etc.- To create an instance of that class, you must do something like
inst = LinearRegression(). The variableinstis now an instance of the classLinearRegression
LinearRegression.predictis an example of an instance method. This means it needs an instance to run (or can be thought of as to 'operate on' in this case)- I can therefore call
inst.predict(x,y,z)but notLinearRegression.predict(x,y,z)directly. - If you want to call
LinearRegression.predict, you have to pass in the instance in the first argument:LinearRegression.predict(inst,x,y,z)
Regarding what you tried afterwards: calling a function from a string holding the function's name is not necessary in this situation and only increases the overhead, so it's probably not the correct way to go here :)
Hope this helps.
edited Jan 3 at 19:51
answered Jan 2 at 23:45
user10859576
I did useLinearRegression.predictand just double checked withpredict_method=LogisticRegression.predict_proba, I'm getting a similar error ''LogisticRegression' object has no attribute 'predict_method''. I'm usingsklearn0.20.0. Thanks Abhinav!
– nancy
Jan 3 at 0:00
@nancy I edited my answer, see if that helps
– user10859576
Jan 3 at 0:22
add a comment |
I did useLinearRegression.predictand just double checked withpredict_method=LogisticRegression.predict_proba, I'm getting a similar error ''LogisticRegression' object has no attribute 'predict_method''. I'm usingsklearn0.20.0. Thanks Abhinav!
– nancy
Jan 3 at 0:00
@nancy I edited my answer, see if that helps
– user10859576
Jan 3 at 0:22
I did use
LinearRegression.predict and just double checked with predict_method=LogisticRegression.predict_proba, I'm getting a similar error ''LogisticRegression' object has no attribute 'predict_method''. I'm using sklearn 0.20.0. Thanks Abhinav!– nancy
Jan 3 at 0:00
I did use
LinearRegression.predict and just double checked with predict_method=LogisticRegression.predict_proba, I'm getting a similar error ''LogisticRegression' object has no attribute 'predict_method''. I'm using sklearn 0.20.0. Thanks Abhinav!– nancy
Jan 3 at 0:00
@nancy I edited my answer, see if that helps
– user10859576
Jan 3 at 0:22
@nancy I edited my answer, see if that helps
– user10859576
Jan 3 at 0:22
add a comment |
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Can you edit in the expected output and try to make this a bit more minimal? (see how to make a Minimal, Complete, and Verifiable example)
– Ethan K888
Jan 2 at 23:39