监督学习与情感分析

Supervised ML(training)

V维特征

出现为1，否则为0，得出V维向量

计数器

包含四个推文的Corpus（语料库）

I am happy because I am learning NLP I am happy I am sad,I am not learning NLP

I am sad

得到vocabulary

I

am happy because learning NLP sad not

已经有的分类

Positive tweets	negative tweets
I am happy because I am learning NLP	I am sad,I am not learning NLP
I am happy	I am sad

计数

freq: dictionary mapping from (word,class) to frequency

vocabulary	PosFreq(1)	NegFreq(0)
I	3	3
am	3	3
happy	2	0
because	1	0
learning	1	1
NLP	1	1
sad	0	2
not	0	1

特征提取得向量

例如I am sad,I am not learning NLP

vocabulary	PosFreq(1)	NegFreq(0)
I	3	3
am	3	3
learning	1	1
NLP	1	1
sad	0	2
not	0	1

计算

\[\sum_{w}freqs(w,1)=3+3+1+1+0+0=8 \]

\[\sum_w{freqs(w,0)=3+3+1+1+1+2+1=11} \]

\[X_m=[1,8,11] \]

预处理

停用词和标点符号

Stop words	Punctuation
and is are at has for a	, . ; ! " '

将@YMourri and @AndrewYNg are tuninga GREAT AI modelat https://deeplearning. ai!!!

去掉停用词@YMourri @AndrewYNg tuning GREAT AI model https://deeplearning. ai!!!

去掉标点符号``@YMourri @AndrewYNg tuning GREAT AI model https://deeplearning. ai`

Handles and urls

去掉handles和urls 后tuning GREAT AI model

stemming and lowercasing

stemming词干提取：去除单词的前后缀得到词根的过程

Preprocessed tweet

[tun,great,ai,model]

代码

#建立频率词典
freqs=build_freqs(tweets,labels)#build freqs dicitonary
#初始化X矩阵
X=np.zeros((m,3))
for i in range(m):#For every tweet
    p_tweet=process_tweet(tweets[i])
    X[i,:]=extract_features(p_tweet,freqs)#提取特征

逻辑回归

公式

左下角预测为negative,右上角为positive

@YMourri and @AndrewYNg are tuning a GREAT AI model

去掉标点符号和停用词后，转化为词干

[tun,ai,great,model]

LR

梯度下降

测试

\[ X_{val} Y_{val} \theta \]

\[pred=h(X_{val},\theta)>=0.5 \]

得到如上预测向量，用验证集来计算

\[\sum_{i=1}^{m}\frac{pred^{(i)}==y^{(i)}_{val}}{m} \]

预测结果和验证集比较，如果相等就为1，如

\[Y_{val}=\left[\begin{matrix}0\\1\\1\\0\\1\end{matrix}\right] pred=\left[\begin{matrix}0\\1\\0\\0\\1\end{matrix}\right] (Y_{val}==pred)=\left[\begin{matrix}1\\1\\0\\1\\1\end{matrix}\right] \]

计算

\[accuracy=\frac{4}{5}=0.8 \]

cost function损失函数

\[J(\theta)=-\frac{1}{m}\sum_{i=1}^{m}[y^{(i)}logh(x^{(i)},\theta)+(1-y^{(i)})log(1-h(x^{(i)},\theta))] \]

m:样本数，负号使结果为正数

当标签为1时，与下面相关

\[y^{(i)}logh(x^{(i)},\theta) \]

y^i	h(x^i,theta)
0	any	0
1	0.99	~0 约等于0
1	~0	-inf 负无穷

可以看出，当标签为1，预测1，损失很小，预测为0损失很大

当标签为0，与下面相关

\[(1-y^{(i)})log(1-h(x^{(i)},\theta)) \]

y^i	h(x^i,theta)
1	any	0
0	0.01	~0
0	~1	-inf

情感分析与朴素贝叶斯

朴素贝叶斯

介绍

某类别推特总数除以语料库中的推文总数

\[A\rightarrow Positive tweet\\ P(A)=P(Positive)=N_{pos}/N \]

如

$$ P(A)=N_{pos}/N=13/20=0.65\\ P(Negative)=1-P(Positive)=0.35 $$

Probabilities

包含happy的推特

$$ B\rightarrow tweet contains "happy"\\ P(B)=P(happy)=N_{happy}/N\\ P(B)=4/20=0.2 $$ $$ P(A\cap B)=P(A,B)=3/20=0.15 $$

Conditional Probabilities条件概率

P(AB)=P(A|B)*P(B)

P（AB）是AB同时发生，P(A|B)是B发生条件下A发生的概率，乘以P（B)即AB同时发生.或在A集合中一个元素同时也属于B的概率

\[P(A|B)=P(Positive|"happy")\\ P(A|B)=3/4=0.75 \]

$$ P(B|A)=P("happy"|Positive)\\ P(B|A)=3/313=0.231 $$ $$ P(Positive|"happy")=\frac{P(Positive\cap"happy")}{P("happy")} $$

Bayes' Rule

\[P(Positive|"happy")=\frac{P(Positive\cap"happy")}{P("happy")}\\ P("happy"|Positive)=\frac{P("happy"\cap Positive)}{P(Positive)} \]

而

\[P("happy"\cap Positive)和P(Positive\cap"happy")相等\\在等式中可以删除 \]

得

\[P(Positive|"happy")=P("happy"|Positive)*\frac{P(Positive)}{P("happy")} \]

即

\[P(X|Y)=P(Y|X)*\frac{P(X)}{P(Y)} \]

naive Bayes for sentiment analysis

naive:因为假设X和Y是独立的，但是很多情况并不是

step 1 频率表

Positive tweets:

I am happy because I am learning NLP

I am happy, not sad

Negative:

I am sad, I am not learning NLP

I am sad, not happy

进行计数

word	PosFreq(1)	NegFreq(0)
I	3	3
am	3	3
happy	2	1
because	1	0
learning	1	1
NLP	1	1
sad	1	2
not	1	2
N_class	13	12

step 2 概率表

word	Pos	Neg
I	0.24	0.25
am	0.24	0.25
happy	0.15	0.08
because	0.08	0
learning	0.08	0.08
NLP	0.08	0.08
sad	0.08	0.17
not	0.08	0.17
sum	1	1

像I am lerning之类差值很小的值为中性词，而happy是power word，becuase的Neg为0，造成计算问题，为避免这种情况，我们使概率函数平滑

word	Pos	Neg
I	0.20	0.20
am	0.20	0.20
happy	0.14	0.10
because	0.10	0.05
learning	0.10	0.10
NLP	0.10	0.10
sad	0.10	0.15
not	0.10	0.15

naive Bayes inference condition rule for binary classification

Tweet:

I am happy today; I am learning.

\[\prod_{i=1}^m\frac{P(w_i|pos)}{P(w_i|neg)} \\将tweet中的单词依次累乘,today没有就不算 \\\frac{0.20}{0.20}*\frac{0.20}{0.20}*\frac{0.14}{0.10}*\frac{0.20}{0.20}*\frac{0.20}{0.20}*\frac{0.10}{0.10}\\ 将\frac{0.20}{0.20}这类中性词去掉 \\得 \frac{0.14}{0.10}=1.4>1 \\所以我们得出推文是positive \]

Laplacian Smoothing 拉普拉斯平滑

避免概率为0

\[P(w_i|class)=\frac{freq(w_i,class)}{N_{class}}\\ class \in \{Positive,Negative\}\\ P(w_i|class)=\frac{freq(w_i,class)+1}{N_{class}+V_{class}}\\ N_{class}=frequency\ of\ all\ words\ in\ class\\ V_{class}=number\ of\ unique\ words\ in\ class \]

+1:防止概率为0，为了+1后的归一化，分母加V，词汇表中去重后单词的数量

四舍五入后得Pos和Neg,接下来利用

\[\begin{align}ratio(w_i)&=\frac{P(w_i|Pos)}{P(w_i|Neg)} \\&\approx\frac{frq(w_i,1)+1}{freq(w_i,0)+1} \end{align} \]

word	Pos	Neg	ratio
I	0.19	0.20	1
am	0.19	0.20	1
happy	0.14	0.10	1.4
because	0.10	0.05	1
learning	0.10	0.10	1
NLP	0.10	0.10	1
sad	0.10	0.15	0.6
not	0.10	0.15	0.6
sum	1	1

积极的词>1，越大说明越积极，消极的词小于1，越接近0说明越消极，

Navie Bayes' inference 推论

\[class\in \{pos,neg\} \\w\rightarrow set\ of\ m\ words\ in\ a\ tweet\\ \prod_{i=1}^m\frac{P(w_i|pos)}{P(w_i|neg)}>1\ likelihood \\>1说推文是积极的，<1说是消极的，叫似然估计 \\前面加上pos和neg的比率 \\\frac{P(pos)}{P(neg)}\prod_{i=1}^m\frac{P(w_i|pos)}{P(w_i|neg)}>1\\ \frac{P(pos)}{P(neg)}\ prior\ probability\ 先验概率 \]

先验概率对不均衡的数据集很重要

Log likelihood

连续相乘面临下溢出风险，太小而无法存储。

使用数学技巧先log

\[log(a*b)=log(a)+log(b) \\log(\frac{P(pos)}{P(neg)}\prod_{i=1}^m\frac{P(w_i|pos)}{P(w_i|neg)}) \\\Longrightarrow log\frac{P(pos)}{P(neg)}+\sum_{i=1}^nlog\frac{P(w_i|pos)}{P(w_i|neg)} \]

log prior + log likelihood

Calculating Lambda

lambda为比率的对数

\[\lambda(w)=log\frac{P(w|pos)}{P(w|neg)} \]

$$ \lambda(I)=log\frac{0.05}{0.05}=log(1)=0 $$ 得

doc:I am happy because I am learning.

log likelihood=0+0+2.2+0+0+0+1.1=3.3

\[\prod_{i=1}^m\frac{P(w_i|pos)}{P(w_i|neg)}>1 \]

如右图

\[\sum_{i=1}^nlog\frac{P(w_i|pos)}{P(w_i|neg)} \]

如右图

3.3>0得出推文为正

summary

\[log\prod_{i=1}^mratio(w_i)=\sum_{i=1}^m\lambda(w_i)>0 \\log likelihood 对数似然 \]

naive Bayes model

step0: collect and annotate corpus

step1: preprocess

• lowercase

• remove punctuation, urls, names

• remove stops words

• stemming

• tokenize sentences

  

step2: word count

step3: P(w|class)

\[V_{class}=6 \\\frac{freq(w,class)+1}{N_{class}+V_{class}} \]

step4: get lambda

step5: get the log prior

\[D_{pos}=number\ of\ positive tweets\\ D_{neg}=number\ of\ negative\ tweets\\ logprior=log\frac{D_{pos}}{D_{neg}}\\ if\ dataset\ is\ balanced,\ D_{pos}=D_{neg}\ and\ logprior=0 \]

summary

• get or annotate a dataset with positive and negative tweets

• preprocess the tweets: process_tweet(tweet)->[w1,w2,w3,...]

• compute freq(w,class)

• get P(w|pos),P(w|neg)

• get lambda(w)

• compute logprior=log(P(pos)/P(neg))

test navie baye's

• predict using naive bayes model

• using your validation set to compute model accuray

• log-likehood dictionary

  \[\lambda(w)=log\frac{P(w|pos)}{P(w|neg)} \]
  

• \[logprior=log\frac{D_{pos}}{D_{neg}}=0 \]

• tweet: [I,pass,the,NLP,interview]

  依次累加分数，表格没有的单词为中性词不需要操作,添加logprior平衡数据集

  score=-0..01+0.5-0.01+0+logprior=0.48

  pred=score>0积极

• \[X_{val}\ Y_{val}\ \lambda_{logprior}\\ score=predict(X_{val},\lambda,logprior)\\ pred=score>0\\ \left[\begin{matrix}0.5\\-1\\1.3\\...\\score_m\end{matrix}\right]>0 =\left[\begin{matrix}0.5>0\\-1>0\\1.3>0\\...\\socre_m>0\end{matrix}\right] =\left[\begin{matrix}1\\0\\1\\...\\pred_m\end{matrix}\right] \]

首先，计算Xval中每列的分数，计算每个分数是否大于0,得到pred矩阵，1为积极，0为消极

\[\frac{1}{m}\sum_{i=1}^{m}(pred_i==Y{val_i})\\ 计算accuray \]

summary

• \[X_{val}\ Y_{val}\longrightarrow Performance\ on\ unseen\ data \]

• \[Predict\ using\ \lambda and logprior for each new tweet \]

• \[Accuracy\ \longrightarrow \frac{1}{m}\sum_{i=1}^m(pred_i==Y_{val_i}) \]

• \[what\ about\ words\ that\ do\ not\ appear\ in\ \lambda (w)? \]

Application of naive bayes

\[P(pos|tweet)\approx P(pos)P(tweet|pos)\\ P(neg|tweet)\approx P(neg)P(tweet|neg)\\ \frac{P(pos|tweet)}{P(neg|tweet)}=\frac{P(pos)}{P(neg)} \prod_{i=1}^m\frac{P(w_i|pos)}{P(w_i|neg)} \]

applicatons:

• 作者识别

  \[\frac{P(莎士比亚|book)}{P(海明威|book)} \]

• 垃圾邮件过滤

  \[\frac{P(spam|email)}{P(nonspam|email)} \]

• Information retrieval

  \[P(document_k|query)\varpropto \prod_{i=0}^{|query|}P(query_i|document_k)\\ Retrieve\ document\ if\ P(document_k|query)>threshold \]

  最早应用于查找数据库中相关和不相关的文档

• word disambiguation消除单词歧义

Bank:河岸或银行

\[ \frac{P(river|text)}{P(money|text)} \]

navie bayes assumptions假设

Independence

预测变量或特征之间的独立性

It is sunnuy and hot in the Sahara desert

假设文本中的单词是独立的，但通常情况并非如此，sunny 和 hot 经常同时出现，可能会导致低估或者高估单个单词的条件概率

It's always cold and snowy in _

spring?summer?fall?winter?

贝叶斯认为他们相等，但是上下文得是winter

Relative frequency in corpus

依赖与数据集的分布。实际上推文中发送正面的推文频率高于负面推文的频率

错误分析

• Removing punctuation and stop words 预处理过程失去语义

• word order 单词顺序影响句子的含义

• adversarial attaks 人类有些自然语言的怪癖

Processing as a Source of errors: Punctuation

• 去掉标点符号

  Tweet: My beloved grandmother :(

  去掉:(

  processed_tweet: [belov,grandmoth]

• 去停顿词

  Tweet: This is not good, because your attitude is not even close to being nice.

  prcessed_tweet:[good,attitude,close,nice]

• 单词顺序

  tweet:I am happy because I do not go.

  tweet:I am not happy because I did go.

  not被贝叶斯分类器忽略

• Adversarial attacks

  对抗攻击，Sarcasm, Irony and Euphemisms 面对讽刺和委婉语

  tweet:This is a ridiculously powerful movie. The plot was gripping and I cried through until the ending!

  processed_tweet:[ridicul,power,movi,ploy,grip,cry,end]

  积极的推文处理获得大量否定的词汇