crf

看到不错的介绍crf的资料

转自：http://blog.echen.me/2012/01/03/introduction-to-conditional-random-fields/

原来crf是通过下降法调节参数的

又了解到无约束的最优化问题比较容易求解的，只要是可以求导的，不管是二次、三次.....可以用最速下降法，牛顿法.....

Introduction

Imagine you have a sequence of snapshots from a day in Justin Bieber’s life, and you want to label each image with the activity it represents (eating, sleeping, driving, etc.). How can you do this?

One way is to ignore the sequential nature of the snapshots, and build a per-image classifier. For example, given a month’s worth of labeled snapshots, you might learn that dark images taken at 6am tend to be about sleeping, images with lots of bright colors tend to be about dancing, images of cars are about driving, and so on.

By ignoring this sequential aspect, however, you lose a lot of information. For example, what happens if you see a close-up picture of a mouth – is it about singing or eating? If you know that the previous image is a picture of Justin Bieber eating or cooking, then it’s more likely this picture is about eating; if, however, the previous image contains Justin Bieber singing or dancing, then this one probably shows him singing as well.

Thus, to increase the accuracy of our labeler, we should incorporate the labels of nearby photos, and this is precisely what a conditional random field does.

Part-of-Speech Tagging

Let’s go into some more detail, using the more common example of part-of-speech tagging.

In POS tagging, the goal is to label a sentence (a sequence of words or tokens) with tags like ADJECTIVE, NOUN, PREPOSITION, VERB, ADVERB, ARTICLE.

For example, given the sentence “Bob drank coffee at Starbucks”, the labeling might be “Bob (NOUN) drank (VERB) coffee (NOUN) at (PREPOSITION) Starbucks (NOUN)”.

So let’s build a conditional random field to label sentences with their parts of speech. Just like any classifier, we’ll first need to decide on a set of feature functions  fi .

Feature Functions in a CRF

In a CRF, each feature function is a function that takes in as input:

• a sentence s
• the position i of a word in the sentence
• the label  li  of the current word
• the label  li−1  of the previous word

and outputs a real-valued number (though the numbers are often just either 0 or 1).

(Note: by restricting our features to depend on only the current and previous labels, rather than arbitrary labels throughout the sentence, I’m actually building the special case of a linear-chain CRF. For simplicity, I’m going to ignore general CRFs in this post.)

For example, one possible feature function could measure how much we suspect that the current word should be labeled as an adjective given that the previous word is “very”.

Features to Probabilities

Next, assign each feature function  fj  a weight  λj  (I’ll talk below about how to learn these weights from the data). Given a sentence s, we can now score a labeling l of s by adding up the weighted features over all words in the sentence:

score(l|s)=∑mj=1∑ni=1λjfj(s,i,li,li−1)

(The first sum runs over each feature function  j , and the inner sum runs over each position  i  of the sentence.)

Finally, we can transform these scores into probabilities  p(l|s)  between 0 and 1 by exponentiating and normalizing:

p(l|s)=exp[score(l|s)]∑l′exp[score(l′|s)]=exp[∑mj=1∑ni=1λjfj(s,i,li,li−1)]∑l′exp[∑mj=1∑ni=1λjfj(s,i,l′i,l′i−1)]

Example Feature Functions

So what do these feature functions look like? Examples of POS tagging features could include:

• f1(s,i,li,li−1)=1  if  li=  ADVERB and the ith word ends in “-ly”; 0 otherwise.
  • If the weight  λ1  associated with this feature is large and positive, then this feature is essentially saying that we prefer labelings where words ending in -ly get labeled as ADVERB.
• f2(s,i,li,li−1)=1  if  i=1 ,  li=  VERB, and the sentence ends in a question mark; 0 otherwise.
  • Again, if the weight  λ2  associated with this feature is large and positive, then labelings that assign VERB to the first word in a question (e.g., “Is this a sentence beginning with a verb?”) are preferred.
• f3(s,i,li,li−1)=1  if  li−1=  ADJECTIVE and  li=  NOUN; 0 otherwise.
  • Again, a positive weight for this feature means that adjectives tend to be followed by nouns.
• f4(s,i,li,li−1)=1  if  li−1=  PREPOSITION and  li=  PREPOSITION.
  • A negative weight  λ4  for this function would mean that prepositions don’t tend to follow prepositions, so we should avoid labelings where this happens.

And that’s it! To sum up: to build a conditional random field, you just define a bunch of feature functions (which can depend on the entire sentence, a current position, and nearby labels), assign them weights, and add them all together, transforming at the end to a probability if necessary.

Now let’s step back and compare CRFs to some other common machine learning techniques.

Smells like Logistic Regression…

The form of the CRF probabilities  p(l|s)=exp[∑mj=1∑ni=1fj(s,i,li,li−1)]∑l′exp[∑mj=1∑ni=1fj(s,i,l′i,l′i−1)]  might look familiar.

That’s because CRFs are indeed basically the sequential version of logistic regression: whereas logistic regression is a log-linear model for classification, CRFs are a log-linear model for sequential labels.

Looks like HMMs…

Recall that Hidden Markov Models are another model for part-of-speech tagging (and sequential labeling in general). Whereas CRFs throw any bunch of functions together to get a label score, HMMs take a generative approach to labeling, defining

p(l,s)=p(l1)∏ip(li|li−1)p(wi|li)

where

• p(li|li−1)  are transition probabilities (e.g., the probability that a preposition is followed by a noun);
• p(wi|li−1)  are emission probabilities (e.g., the probability that a noun emits the word “dad”).

So how do HMMs compare to CRFs? CRFs are more powerful – they can model everything HMMs can and more. One way of seeing this is as follows.

Note that the log of the HMM probability is  logp(l,s)=logp(l0)+∑ilogp(li|li−1)+∑ilogp(wi|li) . This has exactly the log-linear form of a CRF if we consider these log-probabilities to be the weights associated to binary transition and emission indicator features.

That is, we can build a CRF equivalent to any HMM by…

• For each HMM transition probability  p(li=y|li−1=x) , define a set of CRF transition features of the form  fx,y(s,i,li,li−1)=1  if  li=y  and  li−1=x . Give each feature a weight of wx,y=logp(li=y|li−1=x) .
• Similarly, for each HMM emission probability  p(wi=z|li=x) , define a set of CRF emission features of the form  gx,y(s,i,li,li−1)=1  if  wi=z  and  li=x . Give each feature a weight of  wx,z=logp(wi=z|li=x) .

Thus, the score  p(l|s)  computed by a CRF using these feature functions is precisely proportional to the score computed by the associated HMM, and so every HMM is equivalent to some CRF.

However, CRFs can model a much richer set of label distributions as well, for two main reasons:

• CRFs can define a much larger set of features. Whereas HMMs are necessarily local in nature (because they’re constrained to binary transition and emission feature functions, which force each word to depend only on the current label and each label to depend only on the previous label), CRFs can use more global features. For example, one of the features in our POS tagger above increased the probability of labelings that tagged the first word of a sentence as a VERB if the end of the sentence contained a question mark.
• CRFs can have arbitrary weights. Whereas the probabilities of an HMM must satisfy certain constraints (e.g.,  0<=p(wi|li)<=1,∑wp(wi=w|l1)=1) , the weights of a CRF are unrestricted (e.g.,  logp(wi|li)  can be anything it wants).

Learning Weights

Let’s go back to the question of how to learn the feature weights in a CRF. One way is (surprise) to use gradient ascent.

Assume we have a bunch of training examples (sentences and associated part-of-speech labels). Randomly initialize the weights of our CRF model. To shift these randomly initialized weights to the correct ones, for each training example…

• Go through each feature function  fi , and calculate the gradient of the log probability of the training example with respect to  λi :  ∂∂wjlogp(l|s)=∑mj=1fi(s,j,lj,lj−1)−∑l′p(l′|s)∑mj=1fi(s,j,l′j,l′j−1)
• Note that the first term in the gradient is the contribution of feature