Machine Learning | Terra Incognita

Machine Learning, Python

Luigi’s Codex Seraphinianus deep learning dreams

Some time ago I received the Luigi’s Codex Seraphinianus book, for those who still didn’t had a chance to take a look, I really recommend you to buy the book, this is the kind of book that will certainly leave a weirdness in your senses. Codex looks like a treatise of a obscure world, and the Codex alone is by far the most weirdest book I’ve ever seen, so I decided to use the GoogleNet model to create the inceptionisms (using the code based on Caffe, that Google kindly released) with some selected images from the Codex. The result of this process created some very interesting images that I decided to share below (click to enlarge):

I hope you liked it !

– Christian S. Perone

Cite this article as: Christian S. Perone, "Luigi’s Codex Seraphinianus deep learning dreams," in Terra Incognita, 11/08/2015, https://blog.christianperone.com/2015/08/luigis-codex-seraphinianus-deep-learning-dreams/.

11/08/201519/01/2020 by Christian S. Perone

Machine Learning, Programming, Python

Machine Learning :: Cosine Similarity for Vector Space Models (Part III)

* It has been a long time since I wrote the TF-IDF tutorial (Part I and Part II) and as I promissed, here is the continuation of the tutorial. Unfortunately I had no time to fix the previous tutorials for the newer versions of the scikit-learn (sklearn) package nor to answer all the questions, but I hope to do that in a close future.

So, on the previous tutorials we learned how a document can be modeled in the Vector Space, how the TF-IDF transformation works and how the TF-IDF is calculated, now what we are going to learn is how to use a well-known similarity measure (Cosine Similarity) to calculate the similarity between different documents.

The Dot Product

Let’s begin with the definition of the dot product for two vectors: $\vec{a} = (a_1, a_2, a_3, \ldots)$ and $\vec{b} = (b_1, b_2, b_3, \ldots)$ , where $a_n$ and $b_n$ are the components of the vector (features of the document, or TF-IDF values for each word of the document in our example) and the $\mathit{n}$ is the dimension of the vectors:

$\vec{a} \cdot \vec{b} = \sum_{i=1}^n a_ib_i = a_1b_1 + a_2b_2 + \cdots + a_nb_n$

As you can see, the definition of the dot product is a simple multiplication of each component from the both vectors added together. See an example of a dot product for two vectors with 2 dimensions each (2D):

$\vec{a} = (0, 3) \\ \vec{b} = (4, 0) \\ \vec{a} \cdot \vec{b} = 0*4 + 3*0 = 0$

The first thing you probably noticed is that the result of a dot product between two vectors isn’t another vector but a single value, a scalar.

This is all very simple and easy to understand, but what is a dot product ? What is the intuitive idea behind it ? What does it mean to have a dot product of zero ? To understand it, we need to understand what is the geometric definition of the dot product:

$\vec{a} \cdot \vec{b} = \|\vec{a}\|\|\vec{b}\|\cos{\theta}$

Rearranging the equation to understand it better using the commutative property, we have:

$\vec{a} \cdot \vec{b} = \|\vec{b}\|\|\vec{a}\|\cos{\theta}$

So, what is the term $\displaystyle \|\vec{a}\|\cos{\theta}$ ? This term is the projection of the vector $\vec{a}$ into the vector $\vec{b}$ as shown on the image below:

Dot_Product — The projection of the vector A into the vector B. By Wikipedia.

Now, what happens when the vector $\vec{a}$ is orthogonal (with an angle of 90 degrees) to the vector $\vec{b}$ like on the image below ?

vectors1 — Two orthogonal vectors (with 90 degrees angle).

There will be no adjacent side on the triangle, it will be equivalent to zero, the term $\displaystyle \|\vec{a}\|\cos{\theta}$ will be zero and the resulting multiplication with the magnitude of the vector $\vec{b}$ will also be zero. Now you know that, when the dot product between two different vectors is zero, they are orthogonal to each other (they have an angle of 90 degrees), this is a very neat way to check the orthogonality of different vectors. It is also important to note that we are using 2D examples, but the most amazing fact about it is that we can also calculate angles and similarity between vectors in higher dimensional spaces, and that is why math let us see far than the obvious even when we can’t visualize or imagine what is the angle between two vectors with twelve dimensions for instance.

The Cosine Similarity

The cosine similarity between two vectors (or two documents on the Vector Space) is a measure that calculates the cosine of the angle between them. This metric is a measurement of orientation and not magnitude, it can be seen as a comparison between documents on a normalized space because we’re not taking into the consideration only the magnitude of each word count (tf-idf) of each document, but the angle between the documents. What we have to do to build the cosine similarity equation is to solve the equation of the dot product for the $\cos{\theta}$ :

$\displaystyle \vec{a} \cdot \vec{b} = \|\vec{a}\|\|\vec{b}\|\cos{\theta} \\ \\ \cos{\theta} = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\|\|\vec{b}\|}$

And that is it, this is the cosine similarity formula. Cosine Similarity will generate a metric that says how related are two documents by looking at the angle instead of magnitude, like in the examples below:

cosinesimilarityfq1 — The Cosine Similarity values for different documents, 1 (same direction), 0 (90 deg.), -1 (opposite directions).

Note that even if we had a vector pointing to a point far from another vector, they still could have an small angle and that is the central point on the use of Cosine Similarity, the measurement tends to ignore the higher term count on documents. Suppose we have a document with the word “sky” appearing 200 times and another document with the word “sky” appearing 50, the Euclidean distance between them will be higher but the angle will still be small because they are pointing to the same direction, which is what matters when we are comparing documents.

Now that we have a Vector Space Model of documents (like on the image below) modeled as vectors (with TF-IDF counts) and also have a formula to calculate the similarity between different documents in this space, let’s see now how we do it in practice using scikit-learn (sklearn).

Practice Using Scikit-learn (sklearn)

* In this tutorial I’m using the Python 2.7.5 and Scikit-learn 0.14.1.

The first thing we need to do is to define our set of example documents:

documents = (
"The sky is blue",
"The sun is bright",
"The sun in the sky is bright",
"We can see the shining sun, the bright sun"
)

And then we instantiate the Sklearn TF-IDF Vectorizer and transform our documents into the TF-IDF matrix:

from sklearn.feature_extraction.text import TfidfVectorizer
tfidf_vectorizer = TfidfVectorizer()
tfidf_matrix = tfidf_vectorizer.fit_transform(documents)
print tfidf_matrix.shape
(4, 11)

Now we have the TF-IDF matrix (tfidf_matrix) for each document (the number of rows of the matrix) with 11 tf-idf terms (the number of columns from the matrix), we can calculate the Cosine Similarity between the first document (“The sky is blue”) with each of the other documents of the set:

from sklearn.metrics.pairwise import cosine_similarity
cosine_similarity(tfidf_matrix[0:1], tfidf_matrix)
array([[ 1.        ,  0.36651513,  0.52305744,  0.13448867]])

The tfidf_matrix[0:1] is the Scipy operation to get the first row of the sparse matrix and the resulting array is the Cosine Similarity between the first document with all documents in the set. Note that the first value of the array is 1.0 because it is the Cosine Similarity between the first document with itself. Also note that due to the presence of similar words on the third document (“The sun in the sky is bright”), it achieved a better score.

If you want, you can also solve the Cosine Similarity for the angle between vectors:

$\cos{\theta} = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\|\|\vec{b}\|}$

We only need to isolate the angle ( $\theta$ ) and move the $\cos$ to the right hand of the equation:

$\theta = \arccos{\frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\|\|\vec{b}\|}}$

The $\arccos$ is the same as the inverse of the cosine ( $\cos^-1$ ).

Lets for instance, check the angle between the first and third documents:

import math
# This was already calculated on the previous step, so we just use the value
cos_sim = 0.52305744
angle_in_radians = math.acos(cos_sim)
print math.degrees(angle_in_radians)
58.462437107432784

And that angle of ~58.5 is the angle between the first and the third document of our document set.

That is it, I hope you liked this third tutorial !

Cite this article as: Christian S. Perone, "Machine Learning :: Cosine Similarity for Vector Space Models (Part III)," in Terra Incognita, 12/09/2013, https://blog.christianperone.com/2013/09/machine-learning-cosine-similarity-for-vector-space-models-part-iii/.

Related Material

A video about Dot Product on The Khan Academy

Wikipedia: Dot Product

Wikipedia: Cosine Similarity

Scikit-learn (sklearn) – The de facto Machine Learning package for Python

12/09/201319/01/2020 by Christian S. Perone

Machine Learning, Python

Machine Learning :: Text feature extraction (tf-idf) – Part II

Read the first part of this tutorial: Text feature extraction (tf-idf) – Part I.

This post is a continuation of the first part where we started to learn the theory and practice about text feature extraction and vector space model representation. I really recommend you to read the first part of the post series in order to follow this second post.

Since a lot of people liked the first part of this tutorial, this second part is a little longer than the first.

Introduction

In the first post, we learned how to use the term-frequency to represent textual information in the vector space. However, the main problem with the term-frequency approach is that it scales up frequent terms and scales down rare terms which are empirically more informative than the high frequency terms. The basic intuition is that a term that occurs frequently in many documents is not a good discriminator, and really makes sense (at least in many experimental tests); the important question here is: why would you, in a classification problem for instance, emphasize a term which is almost present in the entire corpus of your documents ?

The tf-idf weight comes to solve this problem. What tf-idf gives is how important is a word to a document in a collection, and that’s why tf-idf incorporates local and global parameters, because it takes in consideration not only the isolated term but also the term within the document collection. What tf-idf then does to solve that problem, is to scale down the frequent terms while scaling up the rare terms; a term that occurs 10 times more than another isn’t 10 times more important than it, that’s why tf-idf uses the logarithmic scale to do that.

But let’s go back to our definition of the $\mathrm{tf}(t,d)$ which is actually the term count of the term $t$ in the document $d$ . The use of this simple term frequency could lead us to problems like keyword spamming, which is when we have a repeated term in a document with the purpose of improving its ranking on an IR (Information Retrieval) system or even create a bias towards long documents, making them look more important than they are just because of the high frequency of the term in the document.

To overcome this problem, the term frequency $\mathrm{tf}(t,d)$ of a document on a vector space is usually also normalized. Let’s see how we normalize this vector.

Vector normalization

Suppose we are going to normalize the term-frequency vector $\vec{v_{d_4}}$ that we have calculated in the first part of this tutorial. The document $d4$ from the first part of this tutorial had this textual representation:

d4: We can see the shining sun, the bright sun.

And the vector space representation using the non-normalized term-frequency of that document was:

$\vec{v_{d_4}} = (0,2,1,0)$

To normalize the vector, is the same as calculating the Unit Vector of the vector, and they are denoted using the “hat” notation: $\hat{v}$ . The definition of the unit vector $\hat{v}$ of a vector $\vec{v}$ is:

$\displaystyle \hat{v} = \frac{\vec{v}}{\|\vec{v}\|_p}$

Where the $\hat{v}$ is the unit vector, or the normalized vector, the $\vec{v}$ is the vector going to be normalized and the $\|\vec{v}\|_p$ is the norm (magnitude, length) of the vector $\vec{v}$ in the $L^p$ space (don’t worry, I’m going to explain it all).

The unit vector is actually nothing more than a normalized version of the vector, is a vector which the length is 1.

The normalization process (Source: http://processing.org/learning/pvector/)

But the important question here is how the length of the vector is calculated and to understand this, you must understand the motivation of the $L^p$ spaces, also called Lebesgue spaces.

Lebesgue spaces

How long is this vector ? (Source: Source: http://processing.org/learning/pvector/)

Usually, the length of a vector $\vec{u} = (u_1, u_2, u_3, \ldots, u_n)$ is calculated using the Euclidean norm – a norm is a function that assigns a strictly positive length or size to all vectors in a vector space -, which is defined by:

(Source: http://processing.org/learning/pvector/)

$\|\vec{u}\| = \sqrt{u^2_1 + u^2_2 + u^2_3 + \ldots + u^2_n}$

But this isn’t the only way to define length, and that’s why you see (sometimes) a number $p$ together with the norm notation, like in $\|\vec{u}\|_p$ . That’s because it could be generalized as:

$\displaystyle \|\vec{u}\|_p = ( \left|u_1\right|^p + \left|u_2\right|^p + \left|u_3\right|^p + \ldots + \left|u_n\right|^p )^\frac{1}{p}$

and simplified as:

$\displaystyle \|\vec{u}\|_p = (\sum\limits_{i=1}^{n}\left|\vec{u}_i\right|^p)^\frac{1}{p}$

So when you read about a L2-norm, you’re reading about the Euclidean norm, a norm with $p=2$ , the most common norm used to measure the length of a vector, typically called “magnitude”; actually, when you have an unqualified length measure (without the $p$ number), you have the L2-norm (Euclidean norm).

When you read about a L1-norm, you’re reading about the norm with $p=1$ , defined as:

$\displaystyle \|\vec{u}\|_1 = ( \left|u_1\right| + \left|u_2\right| + \left|u_3\right| + \ldots + \left|u_n\right|)$

Which is nothing more than a simple sum of the components of the vector, also known as Taxicab distance, also called Manhattan distance.

Taxicab geometry versus Euclidean distance: In taxicab geometry all three pictured lines have the same length (12) for the same route. In Euclidean geometry, the green line has length $6 \times \sqrt{2} \approx 8.48$ , and is the unique shortest path.
Source: Wikipedia :: Taxicab Geometry

Note that you can also use any norm to normalize the vector, but we’re going to use the most common norm, the L2-Norm, which is also the default in the 0.9 release of the scikits.learn. You can also find papers comparing the performance of the two approaches among other methods to normalize the document vector, actually you can use any other method, but you have to be concise, once you’ve used a norm, you have to use it for the whole process directly involving the norm (a unit vector that used a L1-norm isn’t going to have the length 1 if you’re going to take its L2-norm later).

Back to vector normalization

Now that you know what the vector normalization process is, we can try a concrete example, the process of using the L2-norm (we’ll use the right terms now) to normalize our vector $\vec{v_{d_4}} = (0,2,1,0)$ in order to get its unit vector $\hat{v_{d_4}}$ . To do that, we’ll simple plug it into the definition of the unit vector to evaluate it:

$\hat{v} = \frac{\vec{v}}{\|\vec{v}\|_p} \\ \\ \hat{v_{d_4}} = \frac{\vec{v_{d_4}}}{||\vec{v_{d_4}}||_2} \\ \\ \\ \hat{v_{d_4}} = \frac{(0,2,1,0)}{\sqrt{0^2 + 2^2 + 1^2 + 0^2}} \\ \\ \hat{v_{d_4}} = \frac{(0,2,1,0)}{\sqrt{5}} \\ \\ \small \hat{v_{d_4}} = (0.0, 0.89442719, 0.4472136, 0.0)$

And that is it ! Our normalized vector $\hat{v_{d_4}}$ has now a L2-norm $\|\hat{v_{d_4}}\|_2 = 1.0$ .

Note that here we have normalized our term frequency document vector, but later we’re going to do that after the calculation of the tf-idf.

The term frequency – inverse document frequency (tf-idf) weight

Now you have understood how the vector normalization works in theory and practice, let’s continue our tutorial. Suppose you have the following documents in your collection (taken from the first part of tutorial):

Train Document Set:

d1: The sky is blue.
d2: The sun is bright.

Test Document Set:

d3: The sun in the sky is bright.
d4: We can see the shining sun, the bright sun.

Your document space can be defined then as $D = \{ d_1, d_2, \ldots, d_n \}$ where $n$ is the number of documents in your corpus, and in our case as $D_{train} = \{d_1, d_2\}$ and $D_{test} = \{d_3, d_4\}$ . The cardinality of our document space is defined by $\left|{D_{train}}\right| = 2$ and $\left|{D_{test}}\right| = 2$ , since we have only 2 two documents for training and testing, but they obviously don’t need to have the same cardinality.

Let’s see now, how idf (inverse document frequency) is then defined:

$\displaystyle \mathrm{idf}(t) = \log{\frac{\left|D\right|}{1+\left|\{d : t \in d\}\right|}}$

where $\left|\{d : t \in d\}\right|$ is the number of documents where the term $t$ appears, when the term-frequency function satisfies $\mathrm{tf}(t,d) \neq 0$ , we’re only adding 1 into the formula to avoid zero-division.

The formula for the tf-idf is then:

$\mathrm{tf\mbox{-}idf}(t) = \mathrm{tf}(t, d) \times \mathrm{idf}(t)$

and this formula has an important consequence: a high weight of the tf-idf calculation is reached when you have a high term frequency (tf) in the given document (local parameter) and a low document frequency of the term in the whole collection (global parameter).

Now let’s calculate the idf for each feature present in the feature matrix with the term frequency we have calculated in the first tutorial:

$M_{train} = \begin{bmatrix} 0 & 1 & 1 & 1\\ 0 & 2 & 1 & 0 \end{bmatrix}$

Since we have 4 features, we have to calculate $\mathrm{idf}(t_1)$ , $\mathrm{idf}(t_2)$ , $\mathrm{idf}(t_3)$ , $\mathrm{idf}(t_4)$ :

$\mathrm{idf}(t_1) = \log{\frac{\left|D\right|}{1+\left|\{d : t_1 \in d\}\right|}} = \log{\frac{2}{1}} = 0.69314718$

$\mathrm{idf}(t_2) = \log{\frac{\left|D\right|}{1+\left|\{d : t_2 \in d\}\right|}} = \log{\frac{2}{3}} = -0.40546511$

$\mathrm{idf}(t_3) = \log{\frac{\left|D\right|}{1+\left|\{d : t_3 \in d\}\right|}} = \log{\frac{2}{3}} = -0.40546511$

$\mathrm{idf}(t_4) = \log{\frac{\left|D\right|}{1+\left|\{d : t_4 \in d\}\right|}} = \log{\frac{2}{2}} = 0.0$

These idf weights can be represented by a vector as:

$\vec{idf_{train}} = (0.69314718, -0.40546511, -0.40546511, 0.0)$

Now that we have our matrix with the term frequency ( $M_{train}$ ) and the vector representing the idf for each feature of our matrix ( $\vec{idf_{train}}$ ), we can calculate our tf-idf weights. What we have to do is a simple multiplication of each column of the matrix $M_{train}$ with the respective $\vec{idf_{train}}$ vector dimension. To do that, we can create a square diagonal matrix called $M_{idf}$ with both the vertical and horizontal dimensions equal to the vector $\vec{idf_{train}}$ dimension:

$M_{idf} = \begin{bmatrix} 0.69314718 & 0 & 0 & 0\\ 0 & -0.40546511 & 0 & 0\\ 0 & 0 & -0.40546511 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$

and then multiply it to the term frequency matrix, so the final result can be defined then as:

$M_{tf\mbox{-}idf} = M_{train} \times M_{idf}$

Please note that the matrix multiplication isn’t commutative, the result of $A \times B$ will be different than the result of the $B \times A$ , and this is why the $M_{idf}$ is on the right side of the multiplication, to accomplish the desired effect of multiplying each idf value to its corresponding feature:

$\begin{bmatrix} \mathrm{tf}(t_1, d_1) & \mathrm{tf}(t_2, d_1) & \mathrm{tf}(t_3, d_1) & \mathrm{tf}(t_4, d_1)\\ \mathrm{tf}(t_1, d_2) & \mathrm{tf}(t_2, d_2) & \mathrm{tf}(t_3, d_2) & \mathrm{tf}(t_4, d_2) \end{bmatrix} \times \begin{bmatrix} \mathrm{idf}(t_1) & 0 & 0 & 0\\ 0 & \mathrm{idf}(t_2) & 0 & 0\\ 0 & 0 & \mathrm{idf}(t_3) & 0\\ 0 & 0 & 0 & \mathrm{idf}(t_4) \end{bmatrix} \\ = \begin{bmatrix} \mathrm{tf}(t_1, d_1) \times \mathrm{idf}(t_1) & \mathrm{tf}(t_2, d_1) \times \mathrm{idf}(t_2) & \mathrm{tf}(t_3, d_1) \times \mathrm{idf}(t_3) & \mathrm{tf}(t_4, d_1) \times \mathrm{idf}(t_4)\\ \mathrm{tf}(t_1, d_2) \times \mathrm{idf}(t_1) & \mathrm{tf}(t_2, d_2) \times \mathrm{idf}(t_2) & \mathrm{tf}(t_3, d_2) \times \mathrm{idf}(t_3) & \mathrm{tf}(t_4, d_2) \times \mathrm{idf}(t_4) \end{bmatrix}$

Let’s see now a concrete example of this multiplication:

$M_{tf\mbox{-}idf} = M_{train} \times M_{idf} = \\ \begin{bmatrix} 0 & 1 & 1 & 1\\ 0 & 2 & 1 & 0 \end{bmatrix} \times \begin{bmatrix} 0.69314718 & 0 & 0 & 0\\ 0 & -0.40546511 & 0 & 0\\ 0 & 0 & -0.40546511 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \\ = \begin{bmatrix} 0 & -0.40546511 & -0.40546511 & 0\\ 0 & -0.81093022 & -0.40546511 & 0 \end{bmatrix}$

And finally, we can apply our L2 normalization process to the $M_{tf\mbox{-}idf}$ matrix. Please note that this normalization is “row-wise” because we’re going to handle each row of the matrix as a separated vector to be normalized, and not the matrix as a whole:

$M_{tf\mbox{-}idf} = \frac{M_{tf\mbox{-}idf}}{\|M_{tf\mbox{-}idf}\|_2}$ $= \begin{bmatrix} 0 & -0.70710678 & -0.70710678 & 0\\ 0 & -0.89442719 & -0.4472136 & 0 \end{bmatrix}$

And that is our pretty normalized tf-idf weight of our testing document set, which is actually a collection of unit vectors. If you take the L2-norm of each row of the matrix, you’ll see that they all have a L2-norm of 1.

Python practice

Environment Used: Python v.2.7.2, Numpy 1.6.1, Scipy v.0.9.0, Sklearn (Scikits.learn) v.0.9.

Now the section you were waiting for ! In this section I’ll use Python to show each step of the tf-idf calculation using the Scikit.learn feature extraction module.

The first step is to create our training and testing document set and computing the term frequency matrix:

from sklearn.feature_extraction.text import CountVectorizer

train_set = ("The sky is blue.", "The sun is bright.")
test_set = ("The sun in the sky is bright.",
"We can see the shining sun, the bright sun.")

count_vectorizer = CountVectorizer()
count_vectorizer.fit_transform(train_set)
print "Vocabulary:", count_vectorizer.vocabulary

# Vocabulary: {'blue': 0, 'sun': 1, 'bright': 2, 'sky': 3}

freq_term_matrix = count_vectorizer.transform(test_set)
print freq_term_matrix.todense()

#[[0 1 1 1]
#[0 2 1 0]]

Now that we have the frequency term matrix (called freq_term_matrix), we can instantiate the TfidfTransformer, which is going to be responsible to calculate the tf-idf weights for our term frequency matrix:

from sklearn.feature_extraction.text import TfidfTransformer

tfidf = TfidfTransformer(norm="l2")
tfidf.fit(freq_term_matrix)

print "IDF:", tfidf.idf_

# IDF: [ 0.69314718 -0.40546511 -0.40546511  0.        ]

Note that I’ve specified the norm as L2, this is optional (actually the default is L2-norm), but I’ve added the parameter to make it explicit to you that it it’s going to use the L2-norm. Also note that you can see the calculated idf weight by accessing the internal attribute called idf_. Now that fit() method has calculated the idf for the matrix, let’s transform the freq_term_matrix to the tf-idf weight matrix:

tf_idf_matrix = tfidf.transform(freq_term_matrix)
print tf_idf_matrix.todense()

# [[ 0.         -0.70710678 -0.70710678  0.        ]
# [ 0.         -0.89442719 -0.4472136   0.        ]]

And that is it, the tf_idf_matrix is actually our previous $M_{tf\mbox{-}idf}$ matrix. You can accomplish the same effect by using the Vectorizer class of the Scikit.learn which is a vectorizer that automatically combines the CountVectorizer and the TfidfTransformer to you. See this example to know how to use it for the text classification process.

I really hope you liked the post, I tried to make it simple as possible even for people without the required mathematical background of linear algebra, etc. In the next Machine Learning post I’m expecting to show how you can use the tf-idf to calculate the cosine similarity.

If you liked it, feel free to comment and make suggestions, corrections, etc.

Cite this article as: Christian S. Perone, "Machine Learning :: Text feature extraction (tf-idf) – Part II," in Terra Incognita, 03/10/2011, https://blog.christianperone.com/2011/10/machine-learning-text-feature-extraction-tf-idf-part-ii/.

References

Understanding Inverse Document Frequency: on theoretical arguments for IDF

Wikipedia :: tf-idf

The classic Vector Space Model

Sklearn text feature extraction code

Updates

13 Mar 2015 – Formating, fixed images issues.
03 Oct 2011 – Added the info about the environment used for Python examples

03/10/201119/01/2020 by Christian S. Perone

Machine Learning, Python

Machine Learning :: Text feature extraction (tf-idf) – Part I

Short introduction to Vector Space Model (VSM)

In information retrieval or text mining, the term frequency – inverse document frequency (also called tf-idf), is a well know method to evaluate how important is a word in a document. tf-idf are is a very interesting way to convert the textual representation of information into a Vector Space Model (VSM), or into sparse features, we’ll discuss more about it later, but first, let’s try to understand what is tf-idf and the VSM.

VSM has a very confusing past, see for example the paper The most influential paper Gerard Salton Never Wrote that explains the history behind the ghost cited paper which in fact never existed; in sum, VSM is an algebraic model representing textual information as a vector, the components of this vector could represent the importance of a term (tf–idf) or even the absence or presence (Bag of Words) of it in a document; it is important to note that the classical VSM proposed by Salton incorporates local and global parameters/information (in a sense that it uses both the isolated term being analyzed as well the entire collection of documents). VSM, interpreted in a lato sensu, is a space where text is represented as a vector of numbers instead of its original string textual representation; the VSM represents the features extracted from the document.

Let’s try to mathematically define the VSM and tf-idf together with concrete examples, for the concrete examples I’ll be using Python (as well the amazing scikits.learn Python module).

Going to the vector space

The first step in modeling the document into a vector space is to create a dictionary of terms present in documents. To do that, you can simple select all terms from the document and convert it to a dimension in the vector space, but we know that there are some kind of words (stop words) that are present in almost all documents, and what we’re doing is extracting important features from documents, features do identify them among other similar documents, so using terms like “the, is, at, on”, etc.. isn’t going to help us, so in the information extraction, we’ll just ignore them.

Let’s take the documents below to define our (stupid) document space:

Train Document Set:

d1: The sky is blue.
d2: The sun is bright.

Test Document Set:

d3: The sun in the sky is bright.
d4: We can see the shining sun, the bright sun.

Now, what we have to do is to create a index vocabulary (dictionary) of the words of the train document set, using the documents $d1$ and $d2$ from the document set, we’ll have the following index vocabulary denoted as $\mathrm{E}(t)$ where the $t$ is the term:

$\mathrm{E}(t) = \begin{cases} 1, & \mbox{if } t\mbox{ is ``blue''} \\ 2, & \mbox{if } t\mbox{ is ``sun''} \\ 3, & \mbox{if } t\mbox{ is ``bright''} \\ 4, & \mbox{if } t\mbox{ is ``sky''} \\ \end{cases}$

Note that the terms like “is” and “the” were ignored as cited before. Now that we have an index vocabulary, we can convert the test document set into a vector space where each term of the vector is indexed as our index vocabulary, so the first term of the vector represents the “blue” term of our vocabulary, the second represents “sun” and so on. Now, we’re going to use the term-frequency to represent each term in our vector space; the term-frequency is nothing more than a measure of how many times the terms present in our vocabulary $\mathrm{E}(t)$ are present in the documents $d3$ or $d4$ , we define the term-frequency as a couting function:

$\mathrm{tf}(t,d) = \sum\limits_{x\in d} \mathrm{fr}(x, t)$

where the $\mathrm{fr}(x, t)$ is a simple function defined as:

$\mathrm{fr}(x,t) = \begin{cases} 1, & \mbox{if } x = t \\ 0, & \mbox{otherwise} \\ \end{cases}$

So, what the $tf(t,d)$ returns is how many times is the term $t$ is present in the document $d$ . An example of this, could be $tf(``sun'', d4) = 2$ since we have only two occurrences of the term “sun” in the document $d4$ . Now you understood how the term-frequency works, we can go on into the creation of the document vector, which is represented by:

$\displaystyle \vec{v_{d_n}} =(\mathrm{tf}(t_1,d_n), \mathrm{tf}(t_2,d_n), \mathrm{tf}(t_3,d_n), \ldots, \mathrm{tf}(t_n,d_n))$

Each dimension of the document vector is represented by the term of the vocabulary, for example, the $\mathrm{tf}(t_1,d_2)$ represents the frequency-term of the term 1 or $t_1$ (which is our “blue” term of the vocabulary) in the document $d_2$ .

Let’s now show a concrete example of how the documents $d_3$ and $d_4$ are represented as vectors:

$\vec{v_{d_3}} = (\mathrm{tf}(t_1,d_3), \mathrm{tf}(t_2,d_3), \mathrm{tf}(t_3,d_3), \ldots, \mathrm{tf}(t_n,d_3)) \\ \vec{v_{d_4}} = (\mathrm{tf}(t_1,d_4), \mathrm{tf}(t_2,d_4), \mathrm{tf}(t_3,d_4), \ldots, \mathrm{tf}(t_n,d_4))$

which evaluates to:

$\vec{v_{d_3}} = (0, 1, 1, 1) \\ \vec{v_{d_4}} = (0, 2, 1, 0)$

As you can see, since the documents $d_3$ and $d_4$ are:

d3: The sun in the sky is bright.
d4: We can see the shining sun, the bright sun.

The resulting vector $\vec{v_{d_3}}$ shows that we have, in order, 0 occurrences of the term “blue”, 1 occurrence of the term “sun”, and so on. In the $\vec{v_{d_3}}$ , we have 0 occurences of the term “blue”, 2 occurrences of the term “sun”, etc.

But wait, since we have a collection of documents, now represented by vectors, we can represent them as a matrix with $|D| \times F$ shape, where $|D|$ is the cardinality of the document space, or how many documents we have and the $F$ is the number of features, in our case represented by the vocabulary size. An example of the matrix representation of the vectors described above is:

$M_{|D| \times F} = \begin{bmatrix} 0 & 1 & 1 & 1\\ 0 & 2 & 1 & 0 \end{bmatrix}$

As you may have noted, these matrices representing the term frequencies tend to be very sparse (with majority of terms zeroed), and that’s why you’ll see a common representation of these matrix as sparse matrices.

Python practice

Environment Used: Python v.2.7.2, Numpy 1.6.1, Scipy v.0.9.0, Sklearn (Scikits.learn) v.0.9.

Since we know the theory behind the term frequency and the vector space conversion, let’s show how easy is to do that using the amazing scikit.learn Python module.

Scikit.learn comes with lots of examples as well real-life interesting datasets you can use and also some helper functions to download 18k newsgroups posts for instance.

Since we already defined our small train/test dataset before, let’s use them to define the dataset in a way that scikit.learn can use:

train_set = ("The sky is blue.", "The sun is bright.")
test_set = ("The sun in the sky is bright.",
    "We can see the shining sun, the bright sun.")

In scikit.learn, what we have presented as the term-frequency, is called CountVectorizer, so we need to import it and create a news instance:

from sklearn.feature_extraction.text import CountVectorizer
vectorizer = CountVectorizer()

The CountVectorizer already uses as default “analyzer” called WordNGramAnalyzer, which is responsible to convert the text to lowercase, accents removal, token extraction, filter stop words, etc… you can see more information by printing the class information:

print vectorizer

CountVectorizer(analyzer__min_n=1,
analyzer__stop_words=set(['all', 'six', 'less', 'being', 'indeed', 'over', 'move', 'anyway', 'four', 'not', 'own', 'through', 'yourselves', (...)

Let’s create now the vocabulary index:

vectorizer.fit_transform(train_set)
print vectorizer.vocabulary
{'blue': 0, 'sun': 1, 'bright': 2, 'sky': 3}

See that the vocabulary created is the same as $E(t)$ (except because it is zero-indexed).

Let’s use the same vectorizer now to create the sparse matrix of our test_set documents:

smatrix = vectorizer.transform(test_set)

print smatrix

(0, 1)        1
(0, 2)        1
(0, 3)        1
(1, 1)        2
(1, 2)        1

Note that the sparse matrix created called smatrix is a Scipy sparse matrix with elements stored in a Coordinate format. But you can convert it into a dense format:

smatrix.todense()

matrix([[0, 1, 1, 1],
........[0, 2, 1, 0]], dtype=int64)

Note that the sparse matrix created is the same matrix $M_{|D| \times F}$ we cited earlier in this post, which represents the two document vectors $\vec{v_{d_3}}$ and $\vec{v_{d_4}}$ .

We’ll see in the next post how we define the idf (inverse document frequency) instead of the simple term-frequency, as well how logarithmic scale is used to adjust the measurement of term frequencies according to its importance, and how we can use it to classify documents using some of the well-know machine learning approaches.

I hope you liked this post, and if you really liked, leave a comment so I’ll able to know if there are enough people interested in these series of posts in Machine Learning topics.

As promised, here is the second part of this tutorial series.

Cite this article as: Christian S. Perone, "Machine Learning :: Text feature extraction (tf-idf) – Part I," in Terra Incognita, 18/09/2011, https://blog.christianperone.com/2011/09/machine-learning-text-feature-extraction-tf-idf-part-i/.

References

The classic Vector Space Model

The most influential paper Gerard Salton never wrote

Wikipedia: tf-idf

Wikipedia: Vector space model

Scikits.learn Examples

Updates

21 Sep 11 – fixed some typos and the vector notation
22 Sep 11 – fixed import of sklearn according to the new 0.9 release and added the environment section
02 Oct 11 – fixed Latex math typos
18 Oct 11 – added link to the second part of the tutorial series
04 Mar 11 – Fixed formatting issues

18/09/201119/01/2020 by Christian S. Perone

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