机器学习::文本特征提取(TF-IDF) - 第二部分

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

这个职位是一个延续在哪里,我们开始学习有关文本特征提取和向量空间模型表示的理论和实践的第一部分。我真的建议你阅读第一部分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.


In the first post, we learned how to use theterm-frequency以表示在矢量空间的文本信息。然而,与术语频率方法的主要问题是,它大大加快了频繁的条款和规模下降,这比高频方面经验更丰富罕见的条款。基本的直觉是,在许多文件中经常出现的一个术语不太好鉴别,真正有意义的(至少在许多实验测试);这里最重要的问题是:你为什么会在例如分类问题,强调术语,是在你的文档的整个语料库几乎礼物?


But let’s go back to our definition of the\mathrm{tf}(t,d)which is actually the term count of the termtin the documentd。The use of this simple term frequency could lead us to problems likekeyword spamming, which is when we have a repeated term in a document with the purpose of improving its ranking on an IR (信息检索)系统,甚至对创建长文档偏见,使他们看起来比他们只是因为手册中出现的高频更重要。

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 documentd4from the first part of this tutorial had this textual representation:


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

\ {VEC V_ {D_4}} =(0,2,1,0)

规范化的向量,是一样的说话g theUnit Vector矢量,而他们使用的是“帽子”符号表示:\帽子{V}。The definition of the unit vector\帽子{V}of a vector\vec{v}is:

\ displaystyle \帽子{v} = \压裂vec {v}} {\ vec {v} {\ | \ \ |_p}

Where the\帽子{V}is the unit vector, or the normalized vector, the\vec{v}在矢量将被归一化和\|\vec{v}\|_p是矢量的范数(大小,长度)\vec{v}in theL^p空间(别担心,我将所有的解释)。

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/)
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 theL^p空间,也被称为Lebesgue spaces

Lebesgue spaces


Usually, the length of a vector\vec{u} = (u_1, u_2, u_3, \ldots, u_n)is calculated using theEuclidean norma 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/)
(Source: http://processing.org/learning/pvector/)

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

但是,这不是定义长度的唯一途径,这就是为什么你看到(有时)的数ptogether 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}


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

所以,当你阅读有关L2-norm, you’re reading about theEuclidean norm,具有规范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 thep号),你有L2-norm(Euclidean norm).

When you read about aL1-norm, you’re reading about the norm withp=1, defined as:

\的DisplayStyle \ | \ VEC【U} \ | _1 =(\左| U_1 \右| + \左| U_2 \右| + \左| U_3 \右| + \ ldots + \左| u_n \右|)

Which is nothing more than a simple sum of the components of the vector, also known asTaxicab distance,也被称为曼哈顿距离。

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 length6 \times \sqrt{2} \approx 8.48,并且是唯一的最短路径。
Source:Wikipedia :: Taxicab Geometry

请注意,您也可以使用任何规范正常化的载体,但我们将使用最常用的规范,L2范数,这也是在0.9版本的默认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

现在你知道了矢量正常化进程是什么,我们可以尝试一个具体的例子,使用L2范数的过程(我们现在使用正确的术语),以规范我们的矢量\ {VEC V_ {D_4}} =(0,2,1,0)in order to get its unit vector\ {帽子V_ {D_4}}。To do that, we’ll simple plug it into the definition of the unit vector to evaluate it:

\帽子{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\ {帽子V_ {D_4}}has now a L2-norm\ | \帽子{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

现在您已经了解如何向量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):


Your document space can be defined then asD = \{ d_1, d_2, \ldots, d_n \}哪里nis the number of documents in your corpus, and in our case asD_ {火车} = \ {D_1,D_2 \}andD_ {测试} = \ {D_3,D_4 \}。The cardinality of our document space is defined by\left|{D_{train}}\right| = 2and\左| {{D_测试}} \右|= 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|}}

哪里\left|\{d : t \in d\}\right|is thenumber of documents哪里the termt出现,术语频率函数满足当\mathrm{tf}(t,d) \neq 0, we’re only adding 1 into the formula to avoid zero-division.


\ mathrm {TF \ MBOX { - } IDF}(T)= \ mathrm {TF}(T,d)\倍\ mathrm {IDF}(t)的

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处),\mathrm{idf}(t_4):

\ mathrm {IDF}(T_1)= \日志{\压裂{\左| d \右|} {1+ \左| \ {d:T_1 \在d \} \右|}} = \日志{\压裂{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处)= \日志{\压裂{\左| d \右|} {1+ \左| \ {d:t_3处\在d \} \右|}} = \日志{\压裂{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_列车}}= (0.69314718, -0.40546511, -0.40546511, 0.0)

现在,我们有我们的词频矩阵(M_{train})和表示我们的矩阵的每个特征的IDF(矢量\ {VEC {idf_列车}}), we can calculate our tf-idf weights. What we have to do is a simple multiplication of each column of the matrixM_{train}with the respective\ {VEC {idf_列车}}vector dimension. To do that, we can create a squarediagonal matrixM_{idf}with both the vertical and horizontal dimensions equal to the vector\ {VEC {idf_列车}}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}


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

Please note that the matrix multiplication isn’t commutative, the result ofA \乘以Bwill be different than the result of the乙\一个时代, and this is why theM_{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 theM_{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} = \压裂{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.


Environment Used:Python的v.2.7.2,NumPy的1.6.1,SciPy的v.0.9.0,Sklearn (Scikits.learn) v.0.9



从sklearn.feature_extraction.text进口CountVectorizer train_set =(“天空是蓝色的。”,“阳光灿烂”。)TEST_SET =(“在天空中的太阳是光明的。”,“我们可以看到闪耀的太阳,。明亮的太阳“)count_vectorizer = CountVectorizer()count_vectorizer.fit_transform(train_set)打印 ”词汇“,count_vectorizer.vocabulary#词汇:{ '蓝':0, '太阳':1, '鲜艳':2 '天空':3} freq_term_matrix = count_vectorizer.transform(TEST_SET)打印freq_term_matrix.todense()#[[0 1 1 1]#[0 2 1 0]]

Now that we have the frequency term matrix (calledfreq_term_matrix),我们可以实例化TfidfTransformer,这将是负责来计算我们的词频矩阵TF-IDF权重:

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. ]

请注意,我所指定的标准为L2,这是可选的(实际上默认为L2范数),但我已经添加了参数,使其明确向你表示,它会使用L2范数。还要注意的是,你可以通过访问称为内部属性看IDF计算权重idf_。Now thatfit()method has calculated the idf for the matrix, let’s transform thefreq_term_matrixto 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, thetf_idf_matrix其实我们以前M_{tf\mbox{-}idf}matrix. You can accomplish the same effect by using the矢量器Scikit的类。学习是一个vectorizer that automatically combines theCountVectorizerTfidfTransformerto you. See这个例子要知道如何使用它的文本分类过程。

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," inTerra Incognita,03/10/2011,//www.cpetem.com/2011/10/machine-learning-text-feature-extraction-tf-idf-part-ii/



Wikipedia :: tf-idf

The classic Vector Space Model

Sklearn text feature extraction code


13 Mar 2015格式化,固定图像的问题。
03 Oct 2011添加了有关使用Python示例环境信息

103 thoughts to “Machine Learning :: Text feature extraction (tf-idf) – Part II”

  1. Wow!
    Perfect intro in tf-idf, thank you very much! Very interesting, I’ve wanted to study this field for a long time and you posts it is a real gift. It would be very interesting to read more about use-cases of the technique. And may be you’ll be interested, please, to shed some light on other methods of text corpus representation, if they exists?
    (对不起,糟糕的英语,我努力改善它,but there is still a lot of job to do)

  2. Excellent work Christian! I am looking forward to reading your next posts on document classification, clustering and topics extraction with Naive Bayes, Stochastic Gradient Descent, Minibatch-k-Means and Non Negative Matrix factorization


    1. 十分感谢奥利弗。我真的想帮助sklearn,我只是得到一些更多的时间来做到这一点,你们都做了伟大的工作,我真的在lib中已经实现的算法量折服,保持良好的工作!

  3. I like this tutorial better for the level of new concepts i am learning here.
    That said, which version of scikits-learn are you using?.
    最新通过的easy_install安装似乎有不同的模块层次结构(即没有找到sklearn feature_extraction)。如果你能提到你使用的版本,我只是尝试用这些例子。

    1. Hello Anand, I’m glad you liked it. I’ve added the information about the environment used just before the section “Python practice”, I’m using the scikits.learn 0.9 (released a few weeks ago).

  4. 哪里是第3部分?我必须提交在4天内向量空间模型的分配。把它在周末的希望吗?

  5. 由于基督徒!与s亚洲金博宝klearn向量空间很不错的工作。我只有一个问题,假设我已经计算了“tf_idf_matrix”,我想计算成对余弦相似性(每行之间)。我是有问题的稀疏矩阵格式,你可以请给出这样的例子?也是我的基质是相当大的,由60K说25K。非常感谢!

  6. Great post… I understand what tf-idf and how to implement it with a concrete example. But I caught 2 things that I’m not sure about:
    1- You called the 2 dimensional matrix M_train, but it has the tf values of the D3 and D4 documents, so you should’ve called that matrix M_test instead of M_train. Because D3 and D4 are our test documents.
    2 - 当你计算IDF值的T2(这是“太阳”),它应该是日志(2/4)。因为文件的数目是2 D3有词“太阳” 1次,D4有它的2倍。这使得3,但是我们也加1到值摆脱0分的问题。这使得4 ...我说得对不对还是我失去了一些东西?
    Thank you.

    1. You are correct: these are excellent blog articles, but the author REALLY has a duty/responsibility to go back and correct errors, like this (and others, e.g. Part 1; …): missing training underscores; setting the stop_words parameter; also on my computer, the vocabulary indexing is different.

      As much as we appreciate the effort (kudos to the author!), it is also a significant disservice to those who struggle past those (uncorrected) errors in the original material.

      1. 回复:我“你是正确的注释”(上),我应该补充:


        ‘… what we are using is really the number of documents in which a term occurs, regardless of the number of times the term occurs in any given document. In this case, then, the denominator in the idf value for t2 (‘sun’) is indeed 2+1 (2 documents have the term ‘sun’, +1 to avoid a potential zero division error).’ “

    2. Khalid,
      This is a response to a very old question. However, I still want to respond to communicate what I understand from the article.
      Your question 2: “When you calculate the idf value for the t2 (which is ‘sun’) it should be log(2/4)”
      My understanding: The denominator in log term should be (number of documents in which the term appears + 1) and not frequency of the term. The number of documents the term “Sun” appears is 2 (1 time in D3 and 2 times in D4 — totally it appears 3 times in two documents. 3 is frequency and 2 is number of documents). Hence the denominator is 2 + 1 = 3.

  7. 优秀的帖子!
    I have some question. From the last tf-idf weight matrix, how can we get the importance of term respectively(e.g. which is the most important term?). How can we use this matrix to classify documents

  8. Thank You So Much. You explained it in such a simple way. It was really useful. Once again thanks a lot.

  9. I have same doubt as Jack(last comment). From the last tf-idf weight matrix, how can we get the importance of term respectively(e.g. which is the most important term?). How can we use this matrix to classify documents.

  10. I have a question..
    After the tf-idf operation, we get a numpy array with values. Suppose we need to get the highest 50 values from the array. How can we do that?

    1. F(IDF)的高值,表示特定载体(或文件)具有较高的局部强度和低全球实力,在这种情况下,你可以假设,在它的条款具有很高的重要性本地和不能忽视的。针对funtion(TF),其中只有长期重复大量的时间给予更多重视的那些,其中大部分时间是不正确的建模技术比较。

  11. Hey ,
    Thanx fr d code..was very helpful indeed !

    1.适用于文档聚类,计算反相的术语频率之后,shud我使用任何关联性系数等Jaccards系数,然后应用聚类算法中像k均值或shud我计算反转术语频率后直接适用d k均值到文档向量?

    2. How do u rate inverted term frequency for calcuating document vectors for document clustering ?

    谢谢a ton fr the forth coming reply!

  12. @Khalid: what you’re pointing out in 1- got me confused too for a minute (M_train vs M_test). I think you are mistaken on your second point, though, because what we are using is really the number of documents in which a term occurs, regardless of the number of times the term occurs in any given document. In this case, then, the denominator in the idf value for t2 (“sun”) is indeed 2+1 (2 documents have the term “sun”, +1 to avoid a potential zero division error).


    Thank you for the _great_ posts!

  13. Excellent article and a great introduction to td-idf normalization.

    You have a very clear and structured way of explaining these difficult concepts.


      1. very good & infomative tutorial…. please upload more tutorials related to documents clustering process.

  14. Can you provide any reference for doing cosine similarity using tfidf so we have the matrix of tf-idf how can we use that to calculate cosine. Thanks for fantastic article.

  15. 请纠正我,如果我拨错
    从“频率后的公式calculated in the first tutorial:” should Mtest not Mtrain. also after starting ‘These idf weights can be represented by a vector as:” should be idf_test not idf_train.

    Btw great series, can you give an simple approach for how to implement classification?

  16. Very good post. Congrats!!

    Showing your results, I have a question:

    I read in the wikipedia:
    The tf-idf value increases proportionally to the number of times a word appears in the document, but is offset by the frequency of the word in the corpus, which helps to control for the fact that some words are generally more common than others.

    When I read it, I understand that if a word apperars in all documents is less important that a word that only appears in one document:

    However, in the results, the word “sun” or “bright” are most important than “sky”.

    I’m not sure of understand it completly.

  17. Terrific! I was familiar with tf-idf before but I found your scikits examples helpful as I’m trying to learn that package.

  18. 优秀的帖子!一次偶然的机会找上CountVectorizer更多信息,无意中发现了这一点,但我很高兴我通过两个您的文章(第1部分和第2部分)的读取。

    Bookmarking your blog now

  19. Does not seem to fit_transform() as you describe..
    Any idea why ?
    >>> ts
    >>> v7 = CountVectorizer()
    >>> v7.fit_transform(ts)
    <2×2 sparse matrix of type '’
    >>> print v7.vocabulary_
    {u’is’: 0, u’the’: 1}

    1. 其实,还有第一个Python样本中的两个小错误。
      1. CountVectorizer should be instantiated like so:
      count_vectorizer = CountVectorizer(stop_words='english')

      print "Vocabulary:", count_vectorizer.vocabulary_

      Excellent tutorial, just small things. hoep it helps others.

      1. 谢谢ash. although the article was rather self explanatory, your comment made the entire difference.

  20. 谢谢for the great explanation.

    In the first case, you wrote idf(t1) = log(2/1), because we don’t have such term in our collection, thus, we add 1 to the denominator. Now, in case t2, you wrote log(2/3), why the denominator is equal to 3 and not to 4 (=1+2+1)? In case t3, you write: log(2/3), thus the denominator is equal 3 (=1+1+1). I see here kind of inconsistency. Could you, please, explain, how did calculate the denominator value.


    1. You got it wrong, in the denominator you don’t put the sum of the term in each document, you just sum all the documents that have at least one aparition of the term.

  21. it is good if you can provide way to know how use ft-idf in classification of document. I see that example (python code) but if there is algorithm that is best because no all people can understand this language.


  22. Nice. An explanation helps put things into perspective. Is tf-idf a good way to do clustering (e.g. use Jaccard analysis or variance against the average set from a known corpus)?


  23. Hi Christian,



    Would like to read more from you.


  24. Thank you for the good wrap up. You mention a number of papers which compare L1 and L2 norm, I plan to study that a bit more in depth. You still know their names?

  25. how can i calculate tf idf for my own text file which is located some where in my pc?

  26. Brilliant article.

    By far the easiest and most sound explanation of tf-tdf I’ve read. I really liked how you explained the mathematics behind it.

  27. 嗨,伟大的职位!我使用的是TfidVectorizer模块scikit学习产生与规范= L2的TF-IDF矩阵。我把它叫做tfidf_matrix语料的fit_transform后,我一直在检查TfidfVectorizer的输出。我总结了行,但他们并不总和为1的代码是VECT = TfidfVectorizer(use_idf =真,sublunar_tf =真,规范=” L2)。tfidf_matrix = vect.fit_transform(数据)。当我运行tfidf_matrix.sum(轴= 1)的载体是大于1也许我看错矩阵或我误解如何正常化的作品。我希望有人能澄清这一点!谢谢

  28. Can I ask when you calculated the IDF, for example, log(2/1), did you use log to base 10 (e) or some other value? I’m getting different calculations!

  29. 伟大的教程,刚开始在ML一份新工作,这很清楚,因为它应该是解释的事情。亚洲金博宝

  30. Execellent帖子...。!非常感谢这篇文章。

    But I need more information, As you show the practical with python, Can you provide it with JAVA language..

  31. I’m a little bit confused why tf-idf gives negative numbers in this case? How do we interpret them? Correct me if I am wrong, but when the vector has a positive value, it means that the magnitude of that component determines how important that word is in that document. If the it is negative, I don’t know how to interpret it. If I were to take the dot product of a vector with all positive components and one with negative components, it would mean that some components may contribute negatively to the dot product even though on of the vectors has very high importance for a particular word.

  32. Hi,
    谢谢so much for this detailed explanation on this topic, really great. Anyway, could you give me a hint what could be the source of my error that I am keep on seeing:

    freq_term_matrix = count_vectorizer.transform(TEST_SET)
    AttributeError: ‘matrix’ object has no attribute ‘transform’

    Am I using a wrong version of sklearn?

  33. 真棒简单而有效的explaination.Please发布更多的话题与这样真棒explainations.Looking着为即将到来的文章。

  34. I understood the tf-idf calculation process. But what does that matrix mean and how can we use the tfidf matrix to calculate the similarity confuse me. can you explain that how can we use the tfidf matrix .thanks

  35. best explanation.. Very helpful. Can you please tell me how to plot vectors in text classification in svm.. I am working on tweets classification. I am confused please help me.

  36. Hi, I’m sorry if i have mistaken but i could not understand how is ||Vd4||2 = 1.
    D4 =的值(0.0,0.89,0.44,0.0),因此归一化将是= SQRT(正方形(0.89)+平方(0.44))= SQRT(0.193)= 0.44

  37. Hi, it is a great blog!
    If I need to do bi-gram cases, how can I use sklearn to finish it?

  38. 这是非常大的亚洲金博宝。我喜欢你教。亚洲金博宝非常非常好

  39. I am not getting same result, when i am executing the same script.
    print (“IDF:”, tfidf.idf_) : IDF: [ 2.09861229 1. 1.40546511 1. ]


    what does i need to change? what might be the possible error?


    1. It can be many things, since you’re using a different Python interpreter version and also a different Scikit-Learn version, you should expect differences in the results since they may have changed default parameters, algorithms, rounding, etc.

  40. 完美的介绍!
    No hocus pocus. Clear and simple, as technology should be.
    Thank you very much.

  41. Why is |D| = 2, in the idf equation. Shouldn’t it be 4 since |D| denotes the number of documents considered, and we have 2 from test, 2 from train.

  42. hey , hii Christian
    your post is really helpful to me to understand tfd-idf from the basics. I’m working on a project of classification where I’m using vector space model which results in determining the categories where my test document should be present. its a part of machine learning . it would be great if you suggest me something related to that. I’m stuck at this point.

  43. 看到这个例子就知道如何使用它的文本分类过程。“这个”链接不起作用了。能否请您提供相关链接,例如。


  44. 当然有很大的了解这个问题。我真的很喜欢所有的点,你做。

  45. 1vbXlh You have brought up a very wonderful details , appreciate it for the post.

  46. I know this site provides quality based articles or
    information in quality?

  47. In the first example. idf(t1), the log (2/1) = 0.3010 by the calculator. Why they obtained 0.69.. Please What is wrong?

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