The Spy EM algorithm solves exactly this problem.
S-EM is a text learning or classification system that learns from a set of positive and unlabeled examples (no negative examples). It is based on a "spy" technique, naive Bayes and EM algorithm.
The basic idea is to combine your positive set with a whole bunch of randomly crawled documents. You initially treat all the crawled documents as the negative class, and learn a naive bayes classifier on that set. Now some of those crawled documents will actually be positive, and you can conservatively relabel any documents that are scored higher than the lowest scoring true positive document. Then you iterate this process until it stablizes.
You're correct that the supervised learner can often be redundant with TF-IDF weighting. Here's the basic outline of why: In one typical form of TF-IDF weighting, the rescaling is logarithmic, so the weighting for a word $w$ in a document $d$ is
$$
\text{TF-IDF}(w,d) = (\text{no. occurrences of $w$ in $d$}) \cdot f(w)
$$ for $N$ the number of documents in the corpus and $f(w)=\log\left(\frac{N}{\text{no. documents containing $w$}}\right)$. When $f(w)>0$, TF-IDF just amounts to a rescaling of the term frequency. So if we write the matrix counting the number of occurrences of a word in each document as $X$, then a linear model has the form $X\beta$. If we use TF-IDF instead of just term frequency alone, the linear model can be written as $X(k I)\tilde{\beta}$, where $k$ is a vector storing all of our weights $k_i=f(w_i)$. The effect of $kI$ is to rescale each column of $X$. In this setting, the choice to use TF-IDF or TF alone is inconsequential, because you'll get the same predictions. Using the substitution $(kI)\tilde{\beta}=\beta$, we can see the effect is to rescale $\beta$.
But there are at least two scenarios where the choice to use TF-IDF is consequential for supervised learning.
The first case is when $f(w)=0$. This happens whenever a term occurs in every document, such as very common words like "and" or "the." In this case, TF-IDF will zero out the column in $X(kI)$, resulting in a matrix which is not full-rank. A rank-deficient matrix is often not preferred for supervised learning, so instead these words are simply dropped from $X$ because they add no information. In this way, TF-IDF provides automatic screening for the most common words.
The second case is when the matrix $X(kI)$ has its document vectors rescaled to the same norm. Since a longer document is very likely to have a much larger vocabulary than a shorter document, it can be hard to compare documents of different lengths. Rescaling each document vector will also suppress importance rare words in the document independently of how rare or common the word is in the corpus. Moreover, rescaling each document's vector to have the same norm after computing TF-IDF gives a design matrix which is not a linear transformation of $X$, so original matrix cannot be recovered using a linear scaling.
Rescaling the document vectors has a close connection to cosine similarity, since both methods involve comparing unit-length vectors.
The popularity of TF-IDF in some settings does not necessarily impose a limitation on the methods you use. Recently, it has become very common to use word and token vectors that are either pre-trained on a large corpus or trained by the researcher for their particular task. Depending on what you're doing and scale of the data, and the goal of your analysis, it might be more expedient to use TD-IDF, word2vec, or another method to represent natural language information.
A number of resources can be found here, which I reproduce for convenience.
K. Sparck Jones. "A statistical interpretation of term specificity
and its application in retrieval". Journal of Documentation, 28 (1).
1972.
G. Salton and Edward Fox and Wu Harry Wu. "Extended Boolean
information retrieval". Communications of the ACM, 26 (11). 1983.
G. Salton and M. J. McGill. "Introduction to modern information
retrieval". 1983
G. Salton and C. Buckley. "Term-weighting approaches in automatic
text retrieval". Information Processing & Management, 24 (5). 1988.
H. Wu and R. Luk and K. Wong and K. Kwok. "Interpreting TF-IDF term
weights as making relevance decisions". ACM Transactions on
Information Systems, 26 (3). 2008.
Best Answer
Eventually, the features in the Bernoulli are indeed binary. However, you can control the threshold of when a certain Tfidf will be converted to a 0 or 1.
If you use scikit-learn, the parameter for BernoulliNB is called
The more typical case for Bernoulli models is to set
binary=Truein theCountVectorizer. If you are using Tfidfs, you will probably have more success with a Multinomial model -- at least that is what I typically observe: Training naive Bayes models is cheap so I usually always compare both Bernoulli and Multinomial.