I am trying to replicate Mathematica's LocalClusteringCoefficient function (as well as compute $C_i^{out}$) for a directed graph. Which was presumed that Mathematica calculates as the cyclic clustering coefficient, $C_i^{cyc}$ in this post: What exactly does LocalClusteringCoefficient compute for directed graphs? The equation comes from page 13 of this paper, also linked in Szabolcs's post:

$$C_i^{cyl}=$$ $$(A)_{ii}^3\over d_i^{in}d_i^{out}-d_i^{\leftarrow\rightarrow}$$ Where:
$$d_i^{in} = (A^T)_i 1$$ $$d_i^{out} = (A)_i 1$$ $$d_i^{\leftarrow\rightarrow}=(A)_{ii}^2$$

A = Adjacency Matrix, $(A)_i$ = $i$-th row of A, 1 = N-dimensional column vector $(1,1,....1,)^T$

For a graph g:

g = Graph[{1, 2, 3, 4}, {1 -> 2, 2 -> 3, 3 -> 1, 1 -> 4}, VertexLabels ->"Name"]

Here is what I've tried:

s = Range[VertexCount[g]];
one = Table[1, Max[VertexCount[g]]];

din = Transpose[AdjacencyMatrix[g]][[#]] & /@ s;
dout = AdjacencyMatrix[g][[#]] & /@ s;
dlink = b2.b2.b2

numer = dout.dout.dout
denom = (din.one).(dout.one) - dlink;

With these results:

(* {1/3, 2/3, 1/3, 0} *)

However, LocalClusteringCoefficient[g] yields:

(*{1/2, 1, 1, 0}*)

1 Answer 1


Your code is incorrect. Here is a direct implementation:

This is the adjacency matrix:

am = AdjacencyMatrix[g];

$(A^3)_{ii}$ in the formula refers to the $i$th element of the diagonal of its cube:

numerator = Normal@Diagonal@MatrixPower[am, 3]

You can get the in- and out-degrees of the vertices using VertexInDegree and VertexOutDegree.

We also need the degrees of reciprocal connections. A simple way to get these is to first construct an adjacency matrix of reciprocal connections, then sum up its rows:

Total[am Transpose[am]]

Then we are ready to compute the denominator of the formula:

denominator = VertexOutDegree[g] VertexInDegree[g] - Total[am Transpose[am]]

Finally just take the ratio:

(* {1/2, 1, 1, Indeterminate} *)

The indeterminate comes from 0/0. One way to avoid it is to use

div[0,0] = 0;
div[x_,y_] := Divide[x,y]

MapThread[div, {numerator, denominator}]

You can do it in many other ways of course.

Your code:

It would help if next time you commented the code, and indicated what each line is meant to do, like I did above with my code.

s = Range[VertexCount[g]];

one = Table[1, Max[VertexCount[g]]];

I do not understand Max[VertexCount[g]]. VertexCount[g] is a number. What is the point of Max?

din = Transpose[AdjacencyMatrix[g]][[#]] & /@ s

Why are you defining a function and mapping it? The result is simply `Transpose@Adja

dout = AdjacencyMatrix[g][[#]] & /@ s;

Same comment as above. Also, it is confusing to me to see the adjacency matrix and its transpose be called dout and din. These names suggest degree vectors.

dlink = b2.b2.b2

b2 is not defined, so I can't comment on this.

numer = dout.dout.dout

The result is a matrix. The formula asks for its diagonal. You are not taking the diagonal.

denom = (din.one).(dout.one) - dlink;

din.one is a weird way to obtain in-degrees. If you need to do summation, use Total on the adjacency matrix.

(din.one).(dout.one) is the dot product of the degree vectors, and the result is a number. The formula has an element-wise product and the result should be a vector, not a number.

(In case this is what confused you: there is no indication of Einstein summation convention being used in this paper.)

  • $\begingroup$ Thanks. I was working on calculating the in & out clustering coefficients and bumped into a complex infinity from "1/0", I can define div[1,0]=0; but it seems that I should define it as div[any numerical value, 0]=0; Thoughts? $\endgroup$
    – E3labs
    Apr 12, 2017 at 1:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.