Python sklearn OneVsRestClassifier : Score function gives ValueError

Question

i am working on a multilabel classification problem as

import pandas as pd
import pickle
from sklearn.feature_extraction.text import TfidfVectorizer
from sklearn.linear_model import SGDClassifier 
from sklearn.multiclass import OneVsRestClassifier
from sklearn.preprocessing import MultiLabelBinarizer
from sklearn.pipeline import Pipeline
from sklearn.linear_model import LogisticRegression
from sklearn.cross_validation import train_test_split

tdf = pd.read_csv("data.csv", index_col="DocID",error_bad_lines=False)[:8]

print tdf

gives me

DocID   Content             Tags           
1       some text here...   [70]
2       some text here...   [59]
3       some text here...  [183]
4       some text here...  [173]
5       some text here...   [71]
6       some text here...   [98]
7       some text here...  [211]
8       some text here...  [188]

then i identify and transform the columns as needed

X=tdf["Content"]
y=tdf["Tags"]

t=TfidfVectorizer()
print t.fit_transform(X).toarray()
print MultiLabelBinarizer().fit_transform(y)

gives me

[[ 0.          0.01058315  0.         ...,  0.00529157  0.          0.        ]
 [ 0.          0.00947091  0.         ...,  0.00473545  0.          0.        ]
 [ 0.01190602  0.00950931  0.         ...,  0.00475465  0.          0.        ]
 ..., 
 [ 0.          0.01314373  0.         ...,  0.00657187  0.          0.        ]
 [ 0.          0.01200425  0.37574455 ...,  0.00600212  0.01502978  0.        ]
 [ 0.          0.02206688  0.         ...,  0.01103344  0.          0.        ]]

 [[1 0 0 0 0 1 0 0 1 1]
 [0 0 0 0 1 0 0 1 1 1]
 [0 1 0 1 0 0 1 0 1 1]
 [0 1 0 1 0 1 0 0 1 1]
 [0 1 0 0 0 1 0 0 1 1]
 [0 0 0 0 0 0 1 1 1 1]
 [0 1 1 0 0 0 0 0 1 1]
 [0 1 0 0 0 0 1 0 1 1]]

Looking at my data, shouldn't there be only 8 columns here for y? why are there 10 columns?

then i split,transform,fit and score

Xtrain, Xvalidate, ytrain, yvalidate = train_test_split(X, y, test_size=.5)

Xtrain=t.fit_transform(Xtrain).toarray()
Xvalidate=t.fit_transform(Xvalidate).toarray()

ytrain=MultiLabelBinarizer().fit_transform(ytrain)
yvalidate=MultiLabelBinarizer().fit_transform(yvalidate)

clf = OneVsRestClassifier(LogisticRegression(penalty='l2', C=0.01)).fit(Xtrain, ytrain)

print "One vs rest accuracy: %.3f"  % clf.score(Xvalidate,yvalidate)

but i get the error

print "One vs rest accuracy: %.3f"  % clf.score(Xvalidate,yvalidate)
  File "X:\Anaconda2\lib\site-packages\sklearn\base.py", line 310, in score
    return accuracy_score(y, self.predict(X), sample_weight=sample_weight)
  File "X:\Anaconda2\lib\site-packages\sklearn\multiclass.py", line 325, in predict
    indices.extend(np.where(_predict_binary(e, X) > thresh)[0])
  File "X:\Anaconda2\lib\site-packages\sklearn\multiclass.py", line 83, in _predict_binary
    score = np.ravel(estimator.decision_function(X))
  File "X:\Anaconda2\lib\site-packages\sklearn\linear_model\base.py", line 249, in decision_function
    % (X.shape[1], n_features))
ValueError: X has 1546 features per sample; expecting 1354

what does this error mean? could it be the data? i have worked with the exact same algorithm with similar (same column format and data format) data and did not have a problem. Also, why does the fit function work?

What am i doing wrong here?

EDIT

so in my Tags column, the data is being read as string. hence the two extra columns in y. i tried

X=tdf["Content"]
y=tdf["Tags"]
y = [map(int, list(_y.replace(',','').replace('[','').replace(']',''))) for _y in y]

to accommodate for multiple values, but i still the same error. at least i get the correct number of columns for y.

Answer 1

When you call fit_transform() you are first adjusting the feature extractor to the data (fit part) and then transforming the data (transform part). By calling fit_transform() multiple times on the same feature extractor (with different data) you are performing different fits, e.g. your TFIDF Vectorizer might learn one vocabulary for your training set and a completely different one for the validation set, which results in a different number of columns (different number of unique words). You have to call fit_transform() on X and y first and split to training and validation set afterwards (one fit, one transform). Alternatively you can call fit_transform() to generate the training set and then just transform() to generate the validation set (one fit, multiple transforms).