# Calculating Jaccard or other association coefficient for binary data using matrix multiplication

I want to know if is there any possible way to calculate Jaccard coefficient using matrix multiplication.

I used this code

jaccard_sim <- function(x) {
# initialize similarity matrix
m <- matrix(NA, nrow=ncol(x),ncol=ncol(x),dimnames=list(colnames(x),colnames(x)))
jaccard <- as.data.frame(m)

for(i in 1:ncol(x)) {
for(j in i:ncol(x)) {
jaccard[i,j]= length(which(x[,i] & x[,j])) / length(which(x[,i] | x[,j]))
jaccard[j,i]=jaccard[i,j]
}
}

This is quite ok to implement in R. I have done the dice similarity one, but got stuck with Tanimoto/Jaccard. Anybody can help?

• Looks like @ttnphns has this covered, but since you're using R, I thought I'd also point out that a number of similarity indices (including Jaccard's) are already implemented in the vegan package. I think they tend to be pretty well-optimized for speed, too. Apr 28, 2013 at 7:50

We know that Jaccard (computed between any two columns of binary data $\bf{X}$) is $\frac{a}{a+b+c}$, while Rogers-Tanimoto is $\frac{a+d}{a+d+2(b+c)}$, where

• a - number of rows where both columns are 1
• b - number of rows where this and not the other column is 1
• c - number of rows where the other and not this column is 1
• d - number of rows where both columns are 0

$a+b+c+d=n$, the number of rows in $\bf{X}$

Then we have:

$\bf X'X=A$ is the square symmetric matrix of $a$ between all columns.

$\bf (not X)'(not X)=D$ is the square symmetric matrix of $d$ between all columns ("not X" is converting 1->0 and 0->1 in X).

So, $\frac{\bf A}{n-\bf D}$ is the square symmetric matrix of Jaccard between all columns.

$\frac{\bf A+D}{\bf A+D+2(n-(A+D))}=\frac{\bf A+D}{2n-\bf A-D}$ is the square symmetric matrix of Rogers-Tanimoto between all columns.

I checked numerically if these formulas give correct result. They do.

Upd. You can also obtain matrices $\bf B$ and $\bf C$:

$\bf B= [1]'X-A$, where "[1]" denotes matrix of ones, sized as $\bf X$. $\bf B$ is the square asymmetric matrix of $b$ between all columns; its element ij is the number of rows in $\bf X$ with 0 in column i and 1 in column j.

Consequently, $\bf C=B'$.

Matrix $\bf D$ can be also computed this way, of course: $n \bf -A-B-C$.

Knowing matrices $\bf A, B, C, D$, you are able to calculate a matrix of any pairwise (dis)similarity coefficient invented for binary data.

• Fractions make no sense for matrices unless they commute: multiplying on the right by an inverse will otherwise give a different result than multiplying on the left. Moreover, it usually is not the case that a product of two symmetric matrices is symmetric. Do you perhaps mean component-by-component division? Could you fix up your notation to reflect what you intend is the correct formula?
– whuber
Feb 7, 2013 at 7:19
• @whuber I don't use inversion nor multiplication of square symmetric matrices. X is the binary data matrix and X'X is its SSCP matrix. not X is X where 1->0, 0->1. And any division here is elementwise division. Please correct my notation if you see it is not appropriate. Feb 7, 2013 at 7:29
• How to calculate inner product (notX)′(notX) in R? Apr 28, 2013 at 1:34
• @user4959, I don't know R. Here !X is recommended; however the result is boolean TRUE/FALSE, not numeric 1/0. Note that I updated my answer where I say that there is also another way to arrive at D matrix. Apr 28, 2013 at 7:21

The above solution is not very good if X is sparse. Because taking !X will make a dense matrix, taking huge amount of memory and computation.

A better solution is to use formula Jaccard[i,j] = #common / (#i + #j - #common). With sparse matrixes you can do it as follows (note the code also works for non-sparse matrices):

library(Matrix)
jaccard <- function(m) {
## common values:
A = tcrossprod(m)
## indexes for non-zero common values
im = which(A > 0, arr.ind=TRUE)
## counts for each row
b = rowSums(m)

## only non-zero values of common
Aim = A[im]

## Jacard formula: #common / (#i + #j - #common)
J = sparseMatrix(
i = im[,1],
j = im[,2],
x = Aim / (b[im[,1]] + b[im[,2]] - Aim),
dims = dim(A)
)

return( J )
}

This may or may not be useful to you, depending on what your needs are. Assuming that you're interested in similarity between clustering assignments:

The Jaccard Similarity Coefficient or Jaccard Index can be used to calculate the similarity of two clustering assignments.

Given the labelings L1 and L2, Ben-Hur, Elisseeff, and Guyon (2002) have shown that the Jaccard index can be calculated using dot-products of an intermediate matrix. The code below leverages this to quickly calculate the Jaccard Index without having to store the intermediate matrices in memory.

The code is written in C++, but can be loaded into R using the sourceCpp command.

/**
* The Jaccard Similarity Coefficient or Jaccard Index is used to compare the
* similarity/diversity of sample sets. It is defined as the size of the
* intersection of the sets divided by the size of the union of the sets. Here,
* it is used to determine how similar to clustering assignments are.
*
* INPUTS:
*    L1: A list. Each element of the list is a number indicating the cluster
*        assignment of that number.
*    L2: The same as L1. Must be the same length as L1.
*
* RETURNS:
*    The Jaccard Similarity Index
*
* SIDE-EFFECTS:
*    None
*
* COMPLEXITY:
*    Time:  O(K^2+n), where K = number of clusters
*    Space: O(K^2)
*
* SOURCES:
*    Asa Ben-Hur, Andre Elisseeff, and Isabelle Guyon (2001) A stability based
*    method for discovering structure in clustered data. Biocomputing 2002: pp.
*    6-17.
*/
// [[Rcpp::export]]
NumericVector JaccardIndex(const NumericVector L1, const NumericVector L2){
int n = L1.size();
int K = max(L1);

int overlaps[K][K];
int cluster_sizes1[K], cluster_sizes2[K];

for(int i = 0; i < K; i++){    // We can use NumericMatrix (default 0)
cluster_sizes1[i] = 0;
cluster_sizes2[i] = 0;
for(int j = 0; j < K; j++)
overlaps[i][j] = 0;
}

//O(n) time. O(K^2) space. Determine the size of each cluster as well as the
//size of the overlaps between the clusters.
for(int i = 0; i < n; i++){
cluster_sizes1[(int)L1[i] - 1]++; // -1's account for zero-based indexing
cluster_sizes2[(int)L2[i] - 1]++;
overlaps[(int)L1[i] - 1][(int)L2[i] - 1]++;
}

// O(K^2) time. O(1) space. Square the overlap values.
int C1dotC2 = 0;
for(int j = 0; j < K; j++){
for(int k = 0; k < K; k++){
C1dotC2 += pow(overlaps[j][k], 2);
}
}

// O(K) time. O(1) space. Square the cluster sizes
int C1dotC1 = 0, C2dotC2 = 0;
for(int i = 0; i < K; i++){
C1dotC1 += pow(cluster_sizes1[i], 2);
C2dotC2 += pow(cluster_sizes2[i], 2);
}

return NumericVector::create((double)C1dotC2/(double)(C1dotC1+C2dotC2-C1dotC2));
}