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Lecture 8: Feature Selection and Analysis
Introduction to Machine Learning Semester 1, 2022
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Features in Machine Learning

Machine Learning Workflow
Evaluation Task
??? ? Model
Depends on features
Evaluation Task

Data Preparation vs Feature Selection
GIGO: Garbage In, Garbage Out
Data Preparation and Cleaning (discussed before)
• Data Cleaning
• Data Aggregation
• Dealing with missing values
• Transformation (e.g., log transform) • Binarization
• Scaling or Normalization
Feature Selection (this lecture)
• Wrapper methods (aka recursive elimination) • Filtering (aka univariate filtering)
• Glance into some other common approaches

Data Preparation vs Feature Selection
Our job as Machine Learning experts:
• Inspect / clean the data
• Choose a model suitable for classifying the data according to the attributes
• Choose attributes suitable for classifying the data according to the model
• Inspection • Intuition

Data Preparation vs Feature Selection
Our job as Machine Learning experts:
• Inspect / clean the data
• Choose a model suitable for classifying the data according to the attributes
• Choose attributes suitable for classifying the data according to the model
• Inspection
• Intuition
• Neither possible in practice

Feature Selection

What makes features good?
Lead to better models
• Better performance according to some evaluation metric
Side-goal 1
• Seeing important features can suggest other important features • Tell us interesting things about the problem
Side-goal 2
• Fewer features → smaller models → faster answer • More accurate answer >> faster answer

Iterative feature selection: Wrappers

Choosing a good feature set
“Wrapper” methods
• Choose subset of attributes that give best performance on the development data
• For example: for the Weather data set:
• Train model on {Outlook}
• Train model on {Temperature}
• Train model on {Outlook, Temperature}
• Train model on {Outlook, Temperature, Humidity}
• Train model on {Outlook, Temperature, Humidity, Windy}

Choosing a good feature set
“Wrapper” methods
• Choose subset of attributes that give best performance on the development data
• For example: for the Weather data set:
• Evaluate model on {Outlook}
• Evaluate model on {Temperature}
• Evaluate model on {Outlook, Temperature}
• Evaluate model on {Outlook, Temperature, Humidity}
• Evaluate model on {Outlook, Temperature, Humidity, Windy}
• Best performance on data set → best feature set

Choosing a good feature set
“Wrapper” methods
• Choose subset of attributes that give best performance on the development data
• Advantages:
• Feature set with optimal performance on development data
• Disadvantages:
• Takes a long time

Aside: how long does the full wrapper method take?
Assume we have a fast method (e.g. Naive Bayes) over a data set of non-trivial size (∼10K instances):
• Assume: train–evaluate cycle takes 10 sec to complete How many cycles? For m features:
• 2m subsets = 2m minutes 6
• m=10→3hours
• m = 60 → heat death of universe
Only practical for very small data sets.

More practical wrapper methods: Greedy search
Greedy approach
• Train and evaluate model on each single attribute
• Choose best attribute
• Until convergence:
• Train and evaluate model on best attribute(s), plus each remaining
single attribute
• Choose best attribute out of the remaining set
• Iterate until performance (e.g. accuracy) stops increasing

More practical wrapper methods: Greedy search
Greedy approach
• Bad news:
• Takes 21 m2 cycles, for m attributes • In theory, 386 attributes → days
• Good news:
• In practice, converges much more quickly than this
• Bad news again:
• Convergences to a sub-optimal (and often very bad) solution

More practical wrapper methods: Ablation
“Ablation” approach
• Start with all attributes
• Remove one attribute, train and evaluate model
• Until divergence:
• From remaining attributes, remove each attribute, train and evaluate model
• Remove attribute that causes least performance degradation
• Termination condition usually: performance (e.g. accuracy) starts to degrade by more than ε

More practical wrapper methods: Ablation
“Ablation” approach
• Good news:
• Mostly removes irrelevant attributes (at the start)
• Bad news:
• Assumes independence of attributes
(Actually, both approaches do this)
• Takes O(m2) time; cycles are slower with more attributes • Not feasible on non-trivial data sets.

Feature Filtering

Feature filtering
Intuition: Evaluate the “goodness” of each feature, separate from other features
• Consider each feature separately: linear time in number of attributes • Possible (but difficult) to control for inter-dependence of features
• Typically most popular strategy

Feature “goodness”
What makes a feature set single feature good?

Toy example
a1 a2 c YYY YNY NYN NNN
Which of a1, a2 is good?

Toy example
a1 a2 c YYY YNY NYN NNN

Toy example
a1 a2 c YYY YNY NYN NNN

Pointwise Mutual Information
Discrepancy between the observed joint probability of two random variables A and C and the expected joint probability if A and C were independent.
Recall independence: P(C|A) = P(C)

Pointwise Mutual Information
Discrepancy between the observed joint probability of two random variables A and C and the expected joint probability if A and C were independent.
Recall independence: P(C|A) = P(C)
PMI is defined as
We want to find attributes that are not independent of the class.
• If PMI >> 0, attribute and class occur together much more often than randomly.
• If RHS ∼ 0, attribute and class occur together as often as we would expect from random chance
• If RHS << 0, attribute and class are negatively correlated. (More on that later!) Attributes with greatest PMI: best attributes PMI(A, C) = log P(A, C) 2 P(A)P(C) Toy example, revisited a1 a2 c YYY YNY NYN NNN Calculate PMI of a1, a2 with respect to c Toy example, revisited a1 a2 c YYY YNY NYN NNN P(a1) = P(c) = P(a1,c) = PMI(a1,c) = Toy example, revisited a1 a2 c YYY YNY NYN NNN P(a1) = P(c) = P(a1,c) = PMI(a1,c) = Toy example, revisited a1 a2 c YYY YNY NYN NNN P ( a 2 ) = 42 P ( c ) = 42 P ( a 2 , c ) = 14 Toy example, revisited a1 a2 c YYY YNY NYN NNN P ( a 2 ) = 42 P ( c ) = 42 P ( a 2 , c ) = 14 PMI(a2,c) = = log2(1) = 0 Feature “goodness”, revisited What makes a single feature good? • Well correlated with class • Knowing a lets us predict c with more confidence • Reverse correlated with class • Knowing a ̄ lets us predict c with more confidence • Well correlated (or reverse correlated) with not class • Knowing a lets us predict c ̄ with more confidence • Usually not quite as good, but still useful Mutual Information • Expected value of PMI over all possible events • For our example: Combine PMI of all possible combinations: a, a ̄, c, c ̄ Aside: Contingency tables Contingency tables: compact representation of these frequency counts a ̄ Total σ(c) σ(c ̄) σ(a, c) σ(a ̄, c) σ(a, c ̄) σ(a ̄, c ̄) P(a, c) = σ(a,c) , etc. N Aside: Contingency tables Contingency tables for toy example: a1 a=Y a=N Total c=Y 2 c=N 2 Total 4 Total c=Y 1 1 2 c=N 1 1 2 Total 2 2 4 22 a=Y a=N Mutual Information Combine PMI of all possible combinations: a, a ̄, c, c ̄ MI(A, C) =P(a, c)PMI(a, c) + P(a ̄, c)PMI(a ̄, c)+ P(a, c ̄)PMI(a, c ̄) + P(a ̄, c ̄)PMI(a ̄, c ̄) MI(A, C) =P(a, c) log P(a, c) 2 P(a)P(c) P(a, c ̄) log P(a, c ̄) 2 P(a)P(c ̄) + P(a ̄, c) log + P(a ̄, c ̄) log P(a ̄, c) + 2 P(a ̄)P(c) P(a ̄, c ̄) 2 P(a ̄)P(c ̄) Mutual Information Combine PMI of all possible combinations: a, a ̄, c, c ̄ MI(A, C) =P(a, c)PMI(a, c) + P(a ̄, c)PMI(a ̄, c)+ P(a, c ̄)PMI(a, c ̄) + P(a ̄, c ̄)PMI(a ̄, c ̄) MI(A, C) =P(a, c) log P(a, c) 2 P(a)P(c) P(a, c ̄) log P(a, c ̄) 2 P(a)P(c ̄) Often written more compactly as: MI(A,C) = 􏰂 􏰂 i∈{a,a ̄} j∈{c,c ̄} We define that 0log0 ≡ 0. + P(a ̄, c) log + P(a ̄, c ̄) log P(a ̄, c) + 2 P(a ̄)P(c) P(a ̄, c ̄) 2 P(a ̄)P(c ̄) P(i,j) 2 P(i)P(j) Mutual Information Example Contingency Table for attribute a1 a1 a=Y a=N Total 2 Total 2 2 4 P(a,c)= 24; P(a)= 42; P(c)= 42; P(a,c ̄)=0 P(a ̄,c ̄) = 42; P(a ̄) = 42; P(c ̄) = 42; P(a ̄,c) = 0 Mutual Information Example Contingency Table for attribute a1 a1 a=Y a=N Total 2 Total 2 2 4 P(a,c)= 24; P(a)= 42; P(c)= 42; P(a,c ̄)=0 P(a ̄,c ̄) = 42; P(a ̄) = 42; P(c ̄) = 42; P(a ̄,c) = 0 MI(A1, C) = P(a1, c) log P(a1, c) 2 P(a1)P(c) P(a1, c ̄) log P(a1, c ̄) 2 P(a1)P(c ̄) + P(a ̄1, c) log + P(a ̄1, c ̄) log P(a ̄1, c) + 2 P(a ̄1)P(c) P(a ̄1, c ̄) 2 P(a ̄1)P(c ̄) Mutual Information Example Contingency Table for attribute a1 a1 a=Y a=N Total 2 Total 2 2 4 P(a,c)= 24; P(a)= 42; P(c)= 42; P(a,c ̄)=0 P(a ̄,c ̄) = 42; P(a ̄) = 42; P(c ̄) = 42; P(a ̄,c) = 0 MI(a1, C) = P(a1, c) log P(a1, c) 2 P(a1)P(c) P(a1, c ̄) log P(a1, c ̄) 2 P(a1)P(c ̄) + P(a ̄1, c) log + P(a ̄1, c ̄) log P(a ̄1, c) + 2 P(a ̄1)P(c) P(a ̄1, c ̄) 2 P(a ̄1)P(c ̄) 112 00112 = 2log2 11 +0log2 11 +0log2 11 +2log2 11 22222222 = 12(1)+0+0+12(1)=1 Mutual Information Example continued Contingency Table for attribute a2 a2 a=Y a=N Total 2 Total 2 2 4 Mutual Information Example continued Contingency Table for attribute a2 a2 a=Y a=N Total 2 Total 2 2 4 P(a) = 42; P(c) = 42; P(a ̄,c) = 41 P(a ̄) = 42; P(c ̄) = 42; P(a,c ̄) = 41 P(a,c) = 14; P(a ̄,c ̄) = 41; Mutual Information Example continued Contingency Table for attribute a2 a2 a=Y a=N Total 2 Total 2 2 4 P(a) = 42; P(c) = 42; P(a ̄,c) = 41 P(a ̄) = 42; P(c ̄) = 42; P(a,c ̄) = 41 P(a,c) = 14; P(a ̄,c ̄) = 41; = P(a2, c) log P(a2, c) 2 P(a2)P(c) P(a2, c ̄) log P(a2, c ̄) 2 P(a2)P(c ̄) + P(a ̄2, c) log + P(a ̄2, c ̄) log P(a ̄2, c) + 2 P(a ̄2)P(c) P(a ̄2, c ̄) 2 P(a ̄2)P(c ̄) 1 14 1 14 1 41 1 14 = 4log2 11 +4log2 11 +4log2 11 +4log2 11 22222222 = 41(0)+ 14(0)+ 14(0)+ 41(0) = 0 Mutual Information Example continued Contingency Table for attribute a2 a2 a=Y a=N Total 2 Total 2 2 4 P(a) = 42; P(c) = 42; P(a ̄,c) = 41 P(a ̄) = 42; P(c ̄) = 42; P(a,c ̄) = 41 P(a,c) = 14; P(a ̄,c ̄) = 41; χ2 (“Chi-square”) Similar idea, different solution: Contingency table (shorthand): σ(c) σ(c ̄) N σ(a, c) σ(a ̄, c) σ(a, c ̄) σ(a ̄, c ̄) c W+X c ̄ Y+Z Total W+Y X+Z N=W+X+Y+Z If a, c were independent (uncorrelated), what value would you expect in W ? Denote the expected value as E(W). χ2 (“Chi-square”) If a, c were independent, then P(a, c) = P(a)P(c) P(a, c) = P(a)P(c) σ(a, c) = σ(a) σ(c) NNN σ(a,c) = E(W) = σ(a)σ(c) N (W +Y)(W +X) W+X+Y+Z χ2 (“Chi-square”) Compare the value we actually observed O(W) with the expected value E(W): • If the observed value is much greater than the expected value, a occurs more often with c than we would expect at random — predictive • If the observed value is much smaller than the expected value, a occurs less often with c than we would expect at random — predictive • If the observed value is close to the expected value, a occurs as often with c as we would expect randomly — not predictive Similarly with X, Y, Z χ2 (“Chi-square”) Actual calculation (to fit to a chi-square distribution) 2 (O(W) − E(W))2 (O(X) − E(X))2 χ = E(W) + E(X) + (O(Y) − E(Y))2 (O(Z) − E(Z))2 E(Y) + E(Z) r c (Oi,j −Ei,j)2 i=1 j=1 Ei,j • i sums over rows and j sums over columns. • Because the values are squared, χ2 becomes much greater when |O−E |islarge,evenifE isalsolarge. Chi-square Example Contingency table for toy example (observed values): a1 a=Y a=N Total c =Y 2 Total 2 2 4 Contingency table for toy example (expected values): a1 a=Y a=N Total 2 Total 2 2 4 Chi-square Example (Oa,c −Ea,c)2 Ea,c (Oa,c ̄ − Ea,c ̄)2 Ea,c ̄ (Oa ̄,c −Ea ̄,c)2 Ea ̄,c (Oa ̄,c ̄ − Ea ̄,c ̄)2 Ea ̄,c ̄ (2−1)2 (0−1)2 (0−1)2 (2−1)2 =1+1+1+1 = 1+1+1+1=4 χ2(A2, C) is obviously 0, because all observed values are equal to expected values. Common Issues Types of Attribute So far, we’ve only looked at binary (Y/N) attributes: • Nominal attributes • Continuous attributes • Ordinal attributes Types of Attributes: Nominal Two common strategies 1. Treat as multiple binary attributes: • e.g. sunny=Y, overcast=N, rainy=N, etc. • Can just use the formulae as given • Results sometimes difficult to interpret • Forexample,Outlook=sunnyisuseful,butOutlook=overcast and Outlook=rainy are not useful... Should we use Outlook? 2. Modify contingency tables (and formulae) Osor c=Y U V W c=N X Y Z Types of Attributes: Nominal Modified MI: MI(O,C) = 􏰂 􏰂 P(i,j)log P(i,j) i∈{s,o,r} j∈{c,c ̄} = P(s,c)log P(s,c) 2 P(i)P(j) + P(s,c ̄)log + P(o, c ̄) log 2 P(s)P(c) P(o, c) log P(o, c) P(r, c) log P(s,c ̄ + 2 P(s)P(c ̄) P(o, c ̄) + 2 P(o)P(c ̄) P(r, c ̄) 2 P(r)P(c ̄) 2 P(o)P(c) P(r, c) + P(r, c ̄) log 2 P(r)P(c) • Biased towards attributes with many values. Types of Attributes: Nominal Chi-square can be used as normal, with 6 observed/expected values. • To control for score inflation, we need to consider “number of degrees of freedom”, and then use the significance test explicitly (beyond the scope of this subject) Types of Attributes: Continuous Continuous attributes • Usually dealt with by estimating probability based on a Gaussian (normal) distribution • With a large number of values, most random variables are normally distributed due to the Central Limit Theorem • For small data sets or pathological features, we may need to use binomial/multinomial distributions All of this is beyond the scope of this subject Types of Attributes: Ordinal Three possibilities, roughly in order of popularity: 1. Treat as binary • Particularly appropriate for frequency counts where events are low-frequency (e.g. words in tweets) 2. Treat as continuous • The fact that we haven’t seen any intermediate values is usually not important • Does have all of the technical downsides of continuous attributes, however 3. Treat as nominal (i.e. throw away ordering) Multi-class problems So far, we’ve only looked at binary (Y/N) classification tasks. Multiclass (e.g. LA, NY, C, At, SF) classification tasks are usually much more difficult. What makes a single feature good? • Highly correlated with class • Highly reverse correlated with class • Highly correlated (or reverse correlated) with not class ... What if there are many classes? What makes a feature bad? • Irrelevant • Correlated with other features • Good at only predicting one class (but is this truly bad?) Multi-class problems So far, we’ve only looked at binary (Y/N) classification tasks. Multiclass (e.g. LA, NY, C, At, SF) classification tasks are usually much more difficult. • PMI, MI, χ2 are all calculated per-class • (Some other feature selection metrics, e.g. Information Gain, work for all classes at once) • Need to make a point of selecting (hopefully uncorrelated) features for each class to give our classifier the best chance of predicting everything correctly. Multi-class problems So far, we’ve only looked at binary (Y/N) classification tasks. Multiclass (e.g. LA, NY, C, At, SF) classification tasks are usually much more difficult. Actual example (MI): LA NY C At SF chicago atlanta sf hollywood atlanta atlanta yankees httpbitlyczmk lol san cubs u u la georgia lol chi chicago save bears atl il ga httpdealnaycom Multi-class problems So far, we’ve only looked at binary (Y/N) classification tasks. Multiclass (e.g. LA, NY, C, At, SF) classification tasks are usually much more difficult. Intuitive features: LA NY C At SF la angeles los chicago hollywood atlanta lakers nyc chicago atlanta sf york bears atl ny il ga httpdealnaycom francisco chicago httpbitlyczmk lol san atlanta cubs u u yankees la sf chi georgia lol chicago save Multi-class problems So far, we’ve only looked at binary (Y/N) classification tasks. Multiclass (e.g. LA, NY, C, At, SF) classification tasks are usually much more difficult. Features for predicting not class: LA NY C At SF la angeles los chicago hollywood atlanta lakers nyc chicago atlanta sf york bears atl ny il ga httpdealnaycom chicago httpbitlyczmk lol san atlanta cubs u u yankees la sf chi georgia lol chicago save Multi-class problems So far, we’ve only looked at binary (Y/N) classification tasks. Multiclass (e.g. LA, NY, C, At, SF) classification tasks are usually much more difficult. Unintuitive features: LA NY C At SF chicago atlanta sf hollywood atlanta atlanta yankees cubs u u la georgia lol chi chicago save bears atl il ga httpdealnaycom francisco httpbitlyczmk lol san What’s going on with MI? Mutual Information is biased toward rare, uninformative features • All probabilities: no notion of the raw frequency of events • If a feature is seen rarely, but always with a given class, it will be seen as “good” • Best features in the Twitter dataset only had MI of about 0.01 bits; 100th best for a given class had MI of about 0.002 bits Glance into a few other common approaches to feature selection A common (unsupervised) alternative Term Frequency Inverse Document Frequency (TFIDF) • Detect important words / Natural Language Processing • Find words that are relevant to a document in a given document collection • To be relevant, a word should be • Frequent enough in the corpus (TF). A word that occurs only 5 times in a corpus of 5,000,000 words is probably not too interesting • Special enough (IDF). A very common word (“the”, “you”, ...) that occurs in (almost) every document is probably not too interesting A common (unsupervised) alternative Term Frequency Inverse Document Frequency (TFIDF) • Detect important words / Natural Language Processing • Find words that are relevant to a document in a given document collection • To be relevant, a word should be • Frequent enough in the corpus (TF). A word that occurs only 5 times in a corpus of 5,000,000 words is probably not too interesting • Special enough (IDF). A very common word (“the”, “you”, ...) that occurs in (almost) every document is probably not too interesting tfidf(d,t,D)=tf +idf tf =log(1+freq(t,d)) idf = log􏰃 |D| 􏰄 count(d ∈ D : t ∈ d) d=document, t=term, D=document collection; |D|=number of documents in D Embedded Methods Some ML models include feature selection inherently 1. Decision trees: Generalization of 1-R 2. Regression models with regularization house price = β0 +β1 ×size+β2 ×location+β3 ×age Regularization (or ‘penalty’) nudges the weight β of unimportant features towards zero https://towardsdatascience.com/a- beginners- guide- to- decision- tree- classification- 6d3209353ea?gi=e0 程序代写 CS代考 加微信: cscodehelp QQ: 2235208643 Email: kyit630461@163.com

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