Distances

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Euclidean distance

Scalar notation

The distance between two points can be described using scalar notation as follows

\[ d_{ij} = \sqrt{\sum_{k=1}^m (x_{ik}-x_{jk})^2} \]

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Let’s take a table containing 4 rows (observations) and 4 columns (variables)

df=data.frame(c(1,2,3,4),c(3,4,5,6),c(5,6,7,8),c(7,8,9,10))
df2=t(df)
row.names(df2) = c('a','b','c','d')
colnames(df2) = c('x1','x2','x3','x4')
df3 = data.frame(df2)

#DT::datatable(df2, option=list(dom='t')) 
knitr::kable(df2)

	x1	x2	x3	x4
a	1	2	3	4
b	3	4	5	6
c	5	6	7	8
d	7	8	9	10

NA

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For example, isolate the first two rows, a and b and calculate the deviations for each variable

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DT::datatable(df3[1:2,], options = list(dom='t'))

NA

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In this case the discards are all equal and are equal to 2, therefore the sum of their squares will be

\[ 2^2+2^2+2^2+2^2 = 16 \]

Then the distance will be \(d = \sqrt{16}\), or \(4\)  

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Matrix notation

Similarly we can calculate the distance between two vectors using the scalar product

\[ d_{ij} = \sqrt{{\bf (x_i - x_j)\cdot (x_i - x_j)}^T} \]

From the table above we always isolate the two rows a and b but considering them as two vectors \(\bf a\) and \(\bf b\) of a space in \(m=4\) dimensions

a = as.numeric(df2[1,])
b = as.numeric(df2[2,])

a

[1] 1 2 3 4

b

[1] 3 4 5 6

z= b-a

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Let’s call \(\bf z\) the vector resulting from differences in vector components \(\bf a\) and \(\bf b\).

z

[1] 2 2 2 2

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Now do the dot product between this vector and its transpose: \({\bf z \cdot z}^T\)

z%*%z

     [,1]
[1,]   16

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I get, as before, the value of \(16\), whose square root is precisely \(4\)

Manhattan distance

The Manhattan distance between two vectors, a and b, for \(m\) variables is calculated as:

\[ d_{ab} = \sum_{i=1}^m |a_{i}-b_{k}| \]

manhattan_dist <- function(a, b){
     dist <- abs(a-b)
     dist <- sum(dist)
     return(dist)
}

manhattan_dist(a,b)

[1] 8