I have a simple scikit-learn Pipeline
of two steps: a TfIdfVectorizer
followed by a LinearSVC
.
I have fit the pipeline using my data. All good.
Now I want to transform (not predict!) an item, using my fitted pipeline
.
I tried pipeline.transform([item])
, but it gives a different result compared to pipeline.named_steps['tfidf'].transform([item])
. Even the shape and type of the result is different: the first is a 1x3000 CSR matrix, the second a 1x15000 CSC matrix. Which one is correct? Why do they differ?
How do I transform items, i.e. get an item's vector representation before the final estimator, when using scikit-learn's Pipeline
?
You can't call a transform method on a pipeline which contains Non-transformer on last step. If you wan't to call transfrom on such pipeline last estimator must be a transformer.
Even transform
method doc says so:
Applies transforms to the data, and the transform method of the final estimator. Valid only if the final estimator implements transform.
Also, there is no method to use every estimator except last one. Thou you can make your own Pipeline, and inherit everything from scikit-learn's Pipeline, but add one method, something like:
def just_transforms(self, X):
"""Applies all transforms to the data, without applying last
estimator.
Parameters
----------
X : iterable
Data to predict on. Must fulfill input requirements of first step of
the pipeline.
"""
Xt = X
for name, transform in self.steps[:-1]:
Xt = transform.transform(Xt)
return Xt
The reason why the results are different (and why calling transform
even workds) is that LinearSVC
also has a transform (now deprecated) that does feature selection
If you want to transform using just the first step, pipeline.named_steps['tfidf'].transform([item])
is the right thing to do.
If you would like to transform using all but the last step, olologin's answer provides the code.
By default, all steps of the pipeline are executed, so also the transform on the last step, which is the feature selection performed by the LinearSVC.
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