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Question: Invalid parameter for sklearn estimator pipeline updated 2021 -Best answer from Stackoverflow.com

Question Description: I am implementing an example from the O’Reilly book “Introduction to Machine Learning with Python“, using Python 2.7 and sklearn 0.16.

The code I am using:

pipe = make_pipeline(TfidfVectorizer(), LogisticRegression())
param_grid = {"logisticregression_C": [0.001, 0.01, 0.1, 1, 10, 100], "tfidfvectorizer_ngram_range": [(1,1), (1,2), (1,3)]}
grid = GridSearchCV(pipe, param_grid, cv=5)
grid.fit(X_train, y_train)
print("Best cross-validation score: {:.2f}".format(grid.best_score_))

The error being returned boils down to:

ValueError: Invalid parameter logisticregression_C for estimator Pipeline

Is this an error related to using Make_pipeline from v.0.16? What is causing this error?

Expert Answer

Answer-1 (Queryfor.com recommend):

There should be two underscores between estimator name and its parameters in a Pipeline logisticregression__C. Do the same for tfidfvectorizer

Answer-2:

For a more general answer to using Pipeline in a GridSearchCV, the parameter grid for the model should start with whatever name you gave when defining the pipeline. For example:

# Pay attention to the name of the second step, i. e. 'model'
pipeline = Pipeline(steps=[
     ('preprocess', preprocess),
     ('model', Lasso())
])

# Define the parameter grid to be used in GridSearch
param_grid = {'model__alpha': np.arange(0, 1, 0.05)}

search = GridSearchCV(pipeline, param_grid)
search.fit(X_train, y_train)

In the pipeline, we used the name model for the estimator step. So, in the grid search, any hyperparameter for Lasso regression should be given with the prefix model__. The parameters in the grid depending on what name you gave in the pipeline. In plain-old GridSearchCV without a pipeline, the grid would be given like this:

param_grid = {'alpha': np.arange(0, 1, 0.05)}
search = GridSearchCV(Lasso(), param_grid)

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