Distance and dissimilarities

knitr::opts_chunk$set(echo = TRUE)

Definition of a distance

• A distance function or a metric on \(\mathbb{R}^n,\:n\geq 1\), is a function \(d:\mathbb{R}^n\times\mathbb{R}^n\rightarrow \mathbb{R}\).

• A distance function must satisfy some required properties or axioms.

• There are three main axioms.

• A1. \(d(\mathbf{x},\mathbf{y})= 0\iff \mathbf{x}=\mathbf{y}\) (identity of indiscernibles);

• A2. \(d(\mathbf{x},\mathbf{y})= d(\mathbf{y},\mathbf{x})\) (symmetry);

• A3. \(d(\mathbf{x},\mathbf{z})\leq d(\mathbf{x},\mathbf{y})+d(\mathbf{y},\mathbf{z})\) (triangle inequality), where \(\mathbf{x}=(x_1,\cdots,x_n)\), \(\mathbf{y}=(y_1,\cdots,y_n)\) and \(\mathbf{z}=(z_1,\cdots,z_n)\) are all vectors of \(\mathbb{R}^n\).

• We should use the term dissimilarity rather than distance when not all the three axioms A1-A3 are valid.

• Most of the time, we shall use, with some abuse of vocabulary, the term distance.

Exercice 1

• Prove that the three axioms A1-A3 imply the non-negativity condition: \[d(\mathbf{x},\mathbf{y})\geq 0.\]

Euclidean distance

• It is defined by:

\[d(\mathbf{x},\mathbf{y})=\sqrt{\sum_{i=1}^n (x_i-y_i)^2}.\] * A1A2 ae onbvious. * The proof of A3 is provided below.

Manhattan distance

• The Manhattan distance also called taxi-cab metric or city-block metric is defined by:

\[d(\mathbf{x},\mathbf{y}) =\sum_{i=1}^n |x_i-y_i|. \]

• A1-A2 hold.
• A3 also holds using the fact that \(|a+b|\leq |a|+|b|\) for any reals \(a,b\).
• There exists also a weighted version of the Manhattan distance called the Canberra distance.

Manhattan distance vs Euclidean distance Graph

x = c(0, 0)
y = c(6,6)
dist(rbind(x, y), method = "euclidian")

         x
y 8.485281

6*sqrt(2)

[1] 8.485281

dist(rbind(x, y), method = "manhattan")

   x
y 12

Canberra distance

• It is defined by:

\[d(\mathbf{x},\mathbf{y}) =\sum_{i=1}^n \frac{|x_i-y_i|}{|x_i|+|y_i|}.\]

• Note that the term \(|x_i − y_i|/(|x_i|+|y_i|)\) is not properly defined when \(x_i=y_i=0\).
• By convention we set the ratio to be zero in that case.
• The Canberra distance is specially sensitive to small changes near zero.

x = c(0, 0)
y = c(6,6)
dist(rbind(x, y), method = "canberra")

  x
y 2

6/6+6/6

[1] 2

Exercice 2

• Prove that the Canberra distance is a true distance.

Minkowski distance

• Both the Euclidian and the Manattan distances are special cases of the Minkowski distance which is defined, for \(p\geq 1\), by: \[ d(\mathbf{x},\mathbf{y})= \left[\sum_{i=1} |x_i-y_i|^{p}\right]^{1/p}. \]
• For \(p=1\), we get the Manhattan distance.
• For \(p=2\), we get the Euclidian distance.
• Let us also define: \[\|\mathbf{x}\|_p\equiv\left[\sum_{i=1}^n |x_i|^{p}\right]^{1/p},\] where \(\|\mathbf{\cdot}\|_p\) is known as the \(p\)-norm or Minkowski norm.
• Note that the Minkowski distance and norm are related by: \[ d(\mathbf{x},\mathbf{y})=\|\mathbf{x}-\mathbf{y}\|_p. \]
• Conversely, we have: \[ \|\mathbf{x}\|_p=d(\mathbf{x},\mathbf{0}), \] where \(\mathbf{0}\) is the null-vetor of \(\mathbb{R}^n\).

library("ggplot2")
x = c(0, 0)
y = c(6,6)
MinkowDist=c()
for (p in seq(1,30,.01))
{
MinkowDist=c(MinkowDist,dist(rbind(x, y), method = "minkowski", p = p))     
}
ggplot(data =data.frame(x = seq(1,30,.01), y=MinkowDist ) , mapping = aes(x = x, y = y))+geom_point(size=.1,color="red")+xlim(1,11)+xlab("p")+ylab("Minkowski Distance")+ggtitle("Minkowski distance wrt p")

Chebyshev distance

• At the limit, we get the Chebyshev distance which is defined by: \[ d(\mathbf{x},\mathbf{y})=\max_{i=1,\cdots,n}(|x_i-y_i|)=\lim_{p\rightarrow\infty} \left[\sum_{i=1} |x_i-y_i|^{p}\right]^{1/p}. \]
• The corresponding norm is: \[ \|\mathbf{x}|_\infty=\max_{i=1,\cdots,n}(|x_i|). \]

Minkowski inequality

• The proof of the triangular inequality A3 is based on the Minkowski inequality:

• For any nonnegative real numbers \(a_1,\cdots,a_n\); \(b_1,\cdots,b_n\), and for any \(p\geq 1\), we have: \[ \left[\sum_{i=1}^n (a_i+b_i)^{p}\right]^{1/p}\leq \left[\sum_{i=1}^n a_i^{p}\right]^{1/p} +\left[\sum_{i=1}^n b_i^{p}\right]^{1/p}. \]

• To prove that the Minkowski distance satisfies A3, notice that \[ \sum_{i=1}^n|x_i-z_i|^{p}= \sum_{i=1}^n|(x_i-y_i)+(y_i-z_i)|^{p}. \]

• Since for any reals \(x,y\), we have: \(|x+y|\leq |x|+|y|\), and using the fact that \(x^p\) is increasing in \(x\geq 0\), we obtain: \[ \sum_{i=1}^n|x_i-z_i|^{p}\leq \sum_{i=1}^n(|x_i-y_i|+|y_i-z_i|)^{p}. \]

• Applying the Minkowski inequality with \(a_i=|x_i-y_i|\) and \(b_i=|y_i-z_i|\), \(i=1,\cdots,n\), we get: \[ \sum_{i=1}^n|x_i-z_i|^{p}\leq \left(\sum_{i=1}^n |x_i-y_i|^{p}\right)^{1/p}+\left(\sum_{i=1}^n |y_i-z_i|^{p}\right)^{1/p}. \]

Hölder inequality

• The proof of the Minkowski inequality itself requires the Hölder inequality:
• For any nonnegative real numbers \(a_1,\cdots,a_n\); \(b_1,\cdots,b_n\), and any \(p,q>1\) with \(1/p+1/q=1\), we have: \[ \sum_{i=1}^n a_ib_i\leq \left[\sum_{i=1}^n a_i^{p}\right]^{1/p} \left[\sum_{i=1}^n b_i^{q}\right]^{1/q} \]
• The proof of the Hölder inequality relies on the Young inequality:
• For any \(a,b>0\), we have \[ ab\leq \frac{a^p}{p}+\frac{b^q}{q}, \] with equality occuring iff: \(a^p=b^q\).
• To prove the Young inequality, one can use the (strict) convexity of the exponential function.
• For any reals \(x,y\), we have: \[ e^{\frac{x}{p}+\frac{y}{q} }\leq \frac{e^{x}}{p}+\frac{e^{y}}{q}. \]
• We then set: \(x=p\ln a\) and \(y=q\ln b\) to get the Young inequality.
• A good reference on inequalities is: Z. Cvetkovski, Inequalities: theorems, techniques and selected problems, 2012, Springer Science & Business Media. # Cauchy-Schwartz inequality
• Note that the triangular inequality for the Minkowski distance implies: \[ \sum_{i=1}^n |x_i|\leq \left[\sum_{i=1}^n |x_i|^{p}\right]^{1/p}. \]
• Note that for \(p=2\), we have \(q=2\). The Hölder inequality implies for that special case \[ \sum_{i=1}^n|x_iy_i|\leq\sqrt{\sum_{i=1}^n x_i^2}\sqrt{\sum_{i=1}^n y_i^2}. \]
• Since the LHS od thes above inequality is greater then \(|\sum_{i=1}^nx_iy_i|\), we get the Cauchy-Schwartz inequality

\[ |\sum_{i=1}^nx_iy_i|\leq\sqrt{\sum_{i=1}^n x_i^2}\sqrt{\sum_{i=1}^n y_i^2}. \] * Using the dot product notation called also scalar product noation: \(\mathbf{x\cdot y}=\sum_{i=1}^nx_iy_i\), and the norm notation \(\|\mathbf{\cdot}\|_2 \|\), the Cauchy-Schwart inequality is: \[ |\mathbf{x\cdot y} | \leq \|\mathbf{x}\|_2 \| \mathbf{y}\|_2. \]

Pearson correlation distance

• The Pearson correlation coefficient is a similarity measure on \(\mathbb{R}^n\) defined by: \[ \rho(\mathbf{x},\mathbf{y})= \frac{\sum_{i=1}^n (x_i-\bar{\mathbf{x}})(y_i-\bar{\mathbf{y}})}{{\sqrt{\sum_{i=1}^n (x_i-\bar{\mathbf{x}})^2\sum_{i=1}^n (y_i-\bar{\mathbf{y}})^2}}}, \] where \(\bar{\mathbf{x}}\) is the mean of the vector \(\mathbf{x}\) defined by: \[\bar{\mathbf{x}}=\frac{1}{n}\sum_{i=1}^n x_i,\]

• Note that the Pearson correlation coefficient satisfies P2 and is invariant to any positive linear transformation, i.e.: \[\rho(\alpha\mathbf{x},\mathbf{y})=\rho(\mathbf{x},\mathbf{y}),\] for any \(\alpha>0\).

• The Pearson distance (or correlation distance) is defined by: \[ d(\mathbf{x},\mathbf{y})=1-\rho(\mathbf{x},\mathbf{y}).\]

• Note that the Pearson distance does not satisfy A1 since \(d(\mathbf{x},\mathbf{x})=0\) for any non-zero vector \(\mathbf{x}\). It neither satisfies the triangle inequality. However, the symmetry property is fullfilled.

Cosine correlation distance

• The cosine of the angle \(\theta\) between two vectors \(\mathbf{x}\) and \(\mathbf{y}\) is a measure of similarity given by: \[ \cos(\theta)=\frac{\mathbf{x}\cdot \mathbf{y}}{\|\mathbf{x}\|_2\|\mathbf{y}\|_2}=\frac{\sum_{i=1}^n x_i y_i}{{\sqrt{\sum_{i=1}^n x_i^2\sum_{i=1}^n y_i^2}}}. \]
• Note that the cosine of the angle between the two centred vectors \((x_1-\bar{\mathbf{x}},\cdots,x_n-\bar{\mathbf{x}})\) and \((y_1-\bar{\mathbf{y}},\cdots,y_n-\bar{\mathbf{y}})\) coincides with the Pearson correlation coefficient of \(\mathbf{x}\) and \(\mathbf{y}\).
  
• The cosine correlation distance is defined by: \[ d(\mathbf{x},\mathbf{y})=1-\cos(\theta). \]
• It shares similar properties than the Pearson correlation distance. Likewise, Axioms A1 and A3 are not satisfied.

Spearman correlation distance

• To calculate the Spearman’s rank-order correlation, we need to map seperately each of the vectors to ranked data values: \[\mathbf{x}\rightarrow \text{rank}(\mathbf{x})=(x_1^r,\cdots,x_n^r).\]
• Here, \(x_i^r\) is the rank of \(x_i\) among the set of values of \(\mathbf{x}\).
• We illustrate this transformation with a simple example:
• If \(\mathbf{x}=(3, 1, 4, 15, 92)\), then the rank-order vector is \(\text{rank}(\mathbf{x})=(2,1,3,4,5)\).

x=c(3, 1, 4, 15, 92)
rank(x)

[1] 2 1 3 4 5

• The Spearman’s rank correlation of two numerical variables \(\mathbf{x}\) and \(\mathbf{y}\) is simply the Pearson correlation of the two correspnding rank-order variables \(\text{rank}(\mathbf{x})\) and \(\text{rank}(\mathbf{y})\), i.e. \(\rho(\text{rank}(\mathbf{x}),\text{rank}(\mathbf{y}))\). This measure is is useful because it is more robust against outliers than the Pearson correlation.
• If all the \(n\) ranks are distinct, it can be computed using the following formula: \[ \rho(\text{rank}(\mathbf{x}),\text{rank}(\mathbf{y}))=1-\frac{6\sum_{i=1}^n d_i^2}{n(n^2-1)}, \] where \(d_i=x_i^r-y_i^r,\:i=1,\cdots,n\).
• The spearman distance is then defined by: \[ d(\mathbf{x},\mathbf{y})=1-\rho(\text{rank}(\mathbf{x}),\text{rank}(\mathbf{y})). \]
• It can be shown that easaly that it is not a proper distance.
• If all the \(n\) ranks are distinct, we get: \[ d(\mathbf{x},\mathbf{y})=\frac{6\sum_{i=1}^n d_i^2}{n(n^2-1)}. \]

x=c(3, 1, 4, 15, 92)
rank(x)

[1] 2 1 3 4 5

y=c(30,2 , 9, 20, 48)
rank(y)

[1] 4 1 2 3 5

d=rank(x)-rank(y)
d

[1] -2  0  1  1  0

cor(rank(x),rank(y))

[1] 0.7

1-6*sum(d^2)/(5*(5^2-1))

[1] 0.7

Kendall tau distance

• The Kendall rank correlation coefficient is calculated from the number of correspondances between the rankings of \(\mathbf{x}\) and the rankings of \(\mathbf{y}\).
• The number of pairs of observations among \(n\) observations or values is: \[{n \choose 2} =\frac{n(n-1)}{2}.\]
• The pairs of observations \((x_{i},x_{j})\) and \((y_{i},y_{j})\) are said to be concordant if: \[\text{sign}(x_j-x_i)=\text{sign}(y_j-y_i),\] and to be discordant if: \[\text{sign}(x_j-x_i)=-\text{sign}(y_j-y_i),\] where \(\text{sign}(\cdot)\) returns \(1\) for positive numbers and \(-1\) negative numbers and \(0\) otherwise.
• If \(x_i=x_j\) or \(y_i=y_j\) (or both), there is a tie.
• The Kendall \(\tau\) coefficient is defined by (neglecting ties): \[\tau =\frac {1}{n(n-1)}\sum_{i=1}^{n}\sum_{j=1}^n\text{sign}(x_j-x_i)\text{sign}(y_j-y_i).\]
• Let \(n_c\) (resp. \(n_d\)) be the number of concordant (resp. discordant) pairs, we have \[\tau =\frac {2(n_c-n_d)}{n(n-1)}.\]
• The Kendall tau distance is then: \[d(\mathbf{x},\mathbf{y})=1-\tau. \]
• Remark: the triangular inequality may fail in cases where there are ties.

x=c(3, 1, 4, 15, 92)
y=c(30,2 , 9, 20, 48)
tau=0
for (i in 1:5)
{  
tau=tau+sign(x -x[i])%*%sign(y -y[i])
}
tau=tau/(5*4)
tau

     [,1]
[1,]  0.6

cor(x,y, method="kendall")

[1] 0.6

Variables standardization

• Variables are often standardized before measuring dissimilarities.
• Standardization converts the original variables into uniteless variables.
• A well known method is the z-score transformation: \[ \mathbf{x}\rightarrow (\frac{x_1-\bar{\mathbf{x}}}{s_\mathbf{x}},\cdots,\frac{x_n-\bar{\mathbf{x}}}{s_\mathbf{x}}), \] where \(s_\mathbf{x}\) is the sample standard deviation given by: \[ s_\mathbf{x}=\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{\mathbf{x}})^2. \]
• The transformed variable will have a mean of \(0\) and a variance of \(1\).
• The result obtained with Pearson correlation measures and standardized Euclidean distances are comparable.
• For other methods, see: Milligan, G. W., & Cooper, M. C. (1988). A study of standardization of variables in cluster analysis. Journal of classification, 5(2), 181-204.

x=c(3, 1, 4, 15, 92)
y=c(30,2 , 9, 20, 48)
(x-mean(x))/sd(x)

[1] -0.5134116 -0.5647527 -0.4877410 -0.2053646  1.7712699

scale(x)

           [,1]
[1,] -0.5134116
[2,] -0.5647527
[3,] -0.4877410
[4,] -0.2053646
[5,]  1.7712699
attr(,"scaled:center")
[1] 23
attr(,"scaled:scale")
[1] 38.9551

(y-mean(y))/sd(y)

[1]  0.45263128 -1.09293895 -0.70654639 -0.09935809
[5]  1.44621214

scale(y)

            [,1]
[1,]  0.45263128
[2,] -1.09293895
[3,] -0.70654639
[4,] -0.09935809
[5,]  1.44621214
attr(,"scaled:center")
[1] 21.8
attr(,"scaled:scale")
[1] 18.11629

Distance matrix computation

• We’ll use a subset of the data USArrests
• We’ll use only a by taking 15 random rows among the 50 rows in the data set.
• Next, we standardize the data using the function scale():

install.packages("FactoMineR")

Error in install.packages : Updating loaded packages

library("FactoMineR")
data("USArrests") # Loading
head(USArrests, 3) # Print the first 3 rows

set.seed(123)
ss <- sample(1:50, 15) # Take 15 random rows
df <- USArrests[ss, ] # Subset the 15 rows
df.scaled <- scale(df) # Standardize the variables