CODEWITHSUNDEEP
pythonmachine-learningscikit-learnsvm
finitenessofinfinity
finitenessofinfinity

Reputation: 1013

Why does classifier.predict() method expects the number of features in the test data to be the same as in training data?

I am trying to build a simple SVM document classifier using scikit-learn and I am using the following code :

import os

import numpy as np

import scipy.sparse as sp

from sklearn.metrics import accuracy_score

from sklearn import svm

from sklearn.metrics import classification_report

from sklearn.feature_extraction.text import CountVectorizer

from sklearn.feature_extraction.text import TfidfTransformer

from sklearn.feature_extraction.text import TfidfVectorizer

from sklearn import cross_validation
from sklearn.datasets import load_svmlight_file

clf=svm.SVC()

path="C:\\Python27"


f1=[]

f2=[]
data2=['omg this is not a ship lol']

f=open(path+'\\mydata\\ACQ\\acqtot','r')

f=f.read()

f1=f.split(';',1085)

for i in range(0,1086):

    f2.append('acq')



f1.append('shipping ship')

f2.append('crude')    

from sklearn.feature_extraction.text import TfidfVectorizer

vectorizer = TfidfVectorizer(min_df=1)
counter = CountVectorizer(min_df=1)


x_train=vectorizer.fit_transform(f1)
x_test=vectorizer.fit_transform(data2)

num_sample,num_features=x_train.shape

test_sample,test_features=x_test.shape

print("#samples: %d, #features: %d" % (num_sample, num_features)) #samples: 5, #features: 25
print("#samples: %d, #features: %d" % (test_sample, test_features))#samples: 2, #features: 37

y=['acq','crude']

#print x_test.n_features

clf.fit(x_train,f2)


#den= clf.score(x_test,y)
clf.predict(x_test)

It gives the following error :

(n_features, self.shape_fit_[1]))
ValueError: X.shape[1] = 6 should be equal to 9451, the number of features at training time

But what I am not understanding is why does it expect the no. of features to be the same? If I am entering an absolutely new text data to the machine which it needs to predict, it's obviously not possible that every document will have the same number of features as the data which was used to train it. Do we have to explicitly set the no of features of the test data to be equal to 9451 in this case?

Upvotes: 4

Views: 12362

Answers (3)

emiguevara
emiguevara

Reputation: 1369

To ensure that you have the same feature representation, you should not fit_transform your test data, but only transform it.

x_train=vectorizer.fit_transform(f1)
x_test=vectorizer.transform(data2)

A similar transformation into homogeneous features should be applied to your labels.

Upvotes: 15

J. Abrahamson
J. Abrahamson

Reputation: 74334

SVM works by assuming all of your training data lives in an n-dimensional space and then performing a kind of geometric optimization on that set. To make that concrete, if n=2 then SVM is picking a line which optimally separates the (+) examples from the (-) examples.

What this means is that the result of training an SVM is tied to the dimensionality it was trained in. This dimensionality is exactly the size of your feature set (modulo kernels and other transformations, but in any case all of that information together uniquely sets the problem space). You thus cannot just apply this trained model to new data which exists in a space of a different dimensionality.

(You might suggest that we project or embed the training space into the test space—and that might work in some circumstances even—but it's invalid generally.)

This situation gets even trickier when you really analyze it, though. Not only does the test data dimensionality need to correspond with the training data dimensionality but the meaning of each dimension needs to be constant. For instance, back in our n=2 example, assume that we're classifying people's moods (happy/sad) and the x dimension is "enjoyment of life" and the y dimension is "time spent listening to sad music". We'd expect that greater x and lesser y values improve the likelihood of being happy, so a good discrimination boundary that SVM could find would be the y=x line as people closer to the x axis tend to be happy and closer to the y axis tend to be sad.

But then lets say someone bumbles and mixes up the x and y dimensions when they drop the test data in. Boom, suddenly you've got an incredibly inaccurate predictor.

So in particular, the observation space of the test data must match the observation space o the training data. Dimensionality is an important step in this regard, but the match must actually be perfect.

Which is a long way of saying that you need to either do some feature engineering or find an algorithm without this kind of dependency (which will also involve some feature engineering).

Upvotes: 4

Pedrom
Pedrom

Reputation: 3823

Do we have to explicitly set the no of features of the test data to be equal to 9451 in this case?

Yes you do. SVM needs to manage the same dimension as the training set. What people tend to do when working with documents is using a bag of words approach or select the first x less common words.

Upvotes: 0

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