Discrete&Continuous Data

Question 1

Types of Attribute

Answer 1

Continuous
Ordinal
Nominal

Question 2

Instances aren’t labelled

Answer 2

Unsupervised ML

Question 3

Not enough instances are labelled

Answer 3

Semi-supervised ML

Question 4

instances are all labelled

Answer 4

Supervised ML

Question 5

instances are ordered

Answer 5

sequence learning

Question 6

nominal learners

Answer 6

NB
1-R
DT

Question 7

continuous learners

Answer 7

KNN
NP
SVM

Question 8

Nominal Attributes, but Numeric Learner

Answer 8

(1) For k-NN and NP: Hamming distance
(2) randomly assign numbers to attribute values
• If scale is constant between attributes, this is not as bad an idea as it sounds! (But still undesirable)
• Worse with higher-arity attributes (more attribute values)
• Imposes an attribute ordering which may not exist
(3) one–hot encoding

Question 9

Hamming distances

Answer 9

the minimum number of substitutions required to change one string into the other, or the minimum number of errors that could have transformed one string into the other

Question 10

one–hot encoding

Answer 10

If nominal attribute takes m values, replace it with m
Boolean attributes

Example:
hot = [1, 0, 0]
mild = [0, 1, 0]
cool = [0, 0, 1]

Question 11

Pros & Cons of one-hot encoding

Answer 11

Pro: solve the nominal attribute in continuous learner issue

Con: massively increase the feature space

Question 12

Numeric Attributes, but Nominal Learner

Answer 12

(1) NB
(2) DT
(3) 1-R

Discretization

Question 13

Types of Naive Bayes

Answer 13

• Multivariate NB: attributes are nominal, and can take any
(fixed) number of values

• Binomal (or Bernoulli) NB: attributes are binary (special
case of MV)

• Multinomial NB: attributes are natural numbers,
corresponding to frequencies

• Gaussian NB: attributes are real numbers, use Probability Density Function

Question 14

numeric attributes for DT

Answer 14

(1) Binarisation

(2) Range

Question 15

Binarisation

Answer 15

Each node is labelled with ak , and has two branches: one
branch is ak ≤ m, one branch is ak > m.

Info Gain/Gain Ratio must be calculated for each non-trivial “split point” for each attribute

Question 16

Con of Binarisation

Answer 16

leads to arbitrarily large trees

Question 17

Discretisation

Answer 17

the translation of continuous attributes onto nominal attributes

Steps:

1. decide on the interval (out-of-scope)
2. map each continuous value onto a discrete value

Types:

1. Unsupervised (does not know/use the class label)
2. Supervised (know/use the class label)

Question 18

Unsupervised Discretisation

Answer 18

(1) Naive
(2) Equal Size
(3) Equal Frequency
(4) K-Means Clustering

Question 19

Naive Unsupervised Discretisation

Answer 19

treat each unique value as a discrete nominal value

Question 20

Pros & Cons of Naive Unsupervised Discretisation

Answer 20

Advantages:
• simple to implement

Disadvantages:
• loss of generality
• no sense of ordering
• describes the training data, but nothing more (overfitting)

Question 21

Equal Size Unsupervised Discretisation

Answer 21

Identify the upper and lower bounds and partition the overall space into n equal intervals = equal width

min = 64
max = 83
interval: 64-70, 71-75; 80-83;

Question 22

Pros & Cons of Equal Size Unsupervised Discretisation

Answer 22

Advantages:
• simple

Disadvantages:
• badly effected by outliers
• arbitrary n

Question 23

Equal Frequency Unsupervised Discretisation

Answer 23

sort the values, and identify breakpoints which
produce n (roughly) equal-sized partitions = equal
frequency

1st bin: 1st-4th instances
2nd bin: 5th-8th instances

Question 24

Pros & Cons of Equal Frequency Unsupervised Discretisation

Answer 24

Advantages:
• simple

Disadvantages:
• arbitrary n

Question 25

K-Means Clustering

Answer 25

(1) Select k points at random (or otherwise) to act as seed
clusters
(2) Assign each instance to the cluster with the nearest
centroid
(3) Compute seed points as the centroids of the clusters of
the current partition (the centroid is the centre, i.e.,
mean point, of the cluster)
(4) Go back to 2, stop when no reassignments (converge)

It may or may not converge at the end but fast converge is fairly typical

One typical improvement runs k-means multiple times (with random seeds), looking for a common clustering and simply ignore runs which don’t converge within τ
iterations

Question 26

Pros & Cons of K-Means Clustering

Answer 26

Strengths:
• relatively efficient: O(tkn), where n is # instances, k is #
clusters, and t is # iterations; normally k,t <= n

Weaknesses:
• tends to converge to local minimum; sensitive to seed
instances
• need to specify k in advance
• not able to handle non-convex clusters
• “mean” ill-defined for nominal attributes

Question 27

Supervised Discretisation

Answer 27

(1) Naive
(2) v1 improvement
(3) v2 improvement

Question 28

Naive Supervised Discretisation

Answer 28

“Group” values into class-contiguous intervals

Steps:
1. Sort the values, and identify breakpoints in class
membership
2. Reposition any breakpoints where there is no change in numeric value
3. Set the breakpoints midway between the neighbouring values

Question 29

Pros & Cons of Naive Supervised Discretisation

Answer 29

Advantages:
• simple to implement

Disadvantages:
• no sense of ordering
• usually creates too many categories (overfitting)

Question 30

Improvement on Naive Supervised Discretisation

Answer 30

v1:
delay inserting a breakpoint until each “cluster” contains at least n instances of the majority class

v2:
merge neighbouring clusters until they reach a certain size/at least n instances of the majority class

Question 31

Probability Mass Functions (PMF)

Answer 31

For a discrete random variable X that takes on a finite or countably infinite number of possible values, we determined P(X = x) for all of the possible values of X, and called it the probability mass function.

Question 32

Probability Density Function (PDF)

Answer 32

For continuous random variables, the probability that X takes on any particular value x is 0. That is, finding P(X = x) for a continuous random variable X is not going to work. Instead, we’ll need to find the probability that X falls in some interval (a, b), that is, we’ll need to find P(a < X < b). We’ll do that using a probability density function

Question 33

a popular PDF

Answer 33

Gaussian/normal distribution

Question 34

Gaussian distribution

Answer 34

• symmetric about the mean
• area under the curve = 1
• to estimate the probability, we need mean µ and standard deviation σ of a distribution X

Question 35

Why Gaussians?

Answer 35

• In practice, a normal distribution is a reasonable
approximation for many events

• This is a consequence of the Central Limit Theorem
• More careful analysis shows that the mean is almost always normally distributed, but outliers can wreak havoc on our probability estimates