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## Re: [igraph] Kendall's tau of two graph ranking list

**From**: |
Tamas Nepusz |

**Subject**: |
Re: [igraph] Kendall's tau of two graph ranking list |

**Date**: |
Thu, 3 Jul 2008 00:10:02 +0200 |

Hi,

I uses cov(x, y, method="kendall") . But it did not work.

`Without any further detailed knowledge about R (I assume you are using
``that), I think that in order to use cov() or cor(), you should provide
``the rankings to cov(...). Example:
`
Let us assume that your vertices are called A, B, C and D.
Let your first ordering be A, B, D, C and the second one A, C, D, B.
Now, go according to your vertex names.

`A is the first one and it is placed first in both ordering, therefore
``the first element of your x and y vectors will be 1.
``B is the second vertex and it is placed second in the first ordering
``and fourth in the second ordering. Therefore, the next element of x
``will be 2 and the next element of y will be 4.
``Similarly, the elements corresponding to C in x and y will be 4 and 2,
``respectively, since C is the 4th in the first ordering and 2 in the
``second.
`Finally, the elements corresponding to D in x and y will be 3 and 3.
Now, your x is:
> x <- c(1, 2, 4, 3)
> y <- c(1, 4, 2, 3)
The Kendall correlation is then:
> cor(x, y, method="kendall")
[1] 0

`So, practically you'll have to transform your vectors before using
``them with cor(). For instance, if the 4th element of your vector is 2,
``then that means that the second vertex of the graph is ranked 4th, but
``cor() requires a vector for which the indices of the elements
``correspond to the vertex indices in the graph and the values of the
``elements correspond to the ranks. What you are doing is the other way
``round. Does it make sense?
`
--
Tamas