Ted Garland made the very good point to me that "positivization" is not
the same as taking the absolute value of contrasts when several traits are
being considered. If, for example, one is inclined to positivize the
contrasts for two traits with respect to trait x, then a pair of contrasts
(x,y) = (-1,2) will become (1,-2) - not (1,2). The same effect as
positivization can be achieved by rotating the descendant branches of a
node so that all contrasts in a given trait are positive (obviously, in
this case some or many contrasts for another trait can be negative).
To my knowledge, however, diagnostic tests on contrasts usually use
absolute values (or squares) of contrasts. In general, these should not
be used in statistical analyses such as correlation (although positivized
contrasts may be used with no ill effect).
- Liam
Liam J. Revell
NESCent, Duke University
web: http://anolis.oeb.harvard.edu/~liam/
NEW email: lrev...@nescent.org
On Wed, 16 Sep 2009, Liam J. Revell wrote:
Hi Manabu,
Yes, "positivized" is the same as taking the absolute value.
For some of Garland et al.'s diagnostic tests on contrasts you might
need the "expected variances" of the contrasts (actually, proportional
to their expected variance prior to standardization), or the
unstandardized contrasts. It is possible to obtain these using specific
settings in the pic() function. For example, to obtain the expected
variances you just run pic(var.contrasts=TRUE), e.g.,
> cont.X<-pic(x,tree,var.contrasts=TRUE)
where x is a vector of your phenotypic trait and tree is your phylogeny.
cont.X is a matrix with the standardized contrasts in the first column
and variances in the second column. Similarly:
> cont.x<-pic(x,tree,scaled=FALSE)
will return a vector (cont.x) containing the unstandardized contrasts
(although in most cases you don't want these).
If you compute:
> cont.X<-pic(x,tree,scaled=FALSE,var.contrasts=TRUE)
> cont.x<-cont.X[,1]/sqrt(cont.X[,2])
you will get the same vector of contrasts as you would obtain with:
> cont.x<-pic(x,tree)
- Liam
Liam J. Revell
NESCent, Duke University
web: http://anolis.oeb.harvard.edu/~liam/
NEW email: lrev...@nescent.org
Manabu Sakamoto wrote:
Hi Liam,
Thanks very much. Now, that makes sense that pic() contrasts are
already standardised, because I was getting strange results compared
to Mesquite when I tried to further standardise the contrasts...
As I am at it, I thought perhaps you could help me interpret what
Garland et al. (1992) referred to when they discuss "positivizing"
stardised contrasts. I thought perhaps it was just simply taking the
absolute values of the contrasts because the signs are arbitrary but I
was not sure because of the specific wording in that paper...
Many thanks,
Manabu
--------------------------------------
M. Sakamoto, PhD
Department of Earth Sciences,
University of Bristol,
Wills Memorial Building, Queen's Road,
Bristol BS8 1RJ, UK
m.sakam...@bristol.ac.uk
Liam J. Revell さんは書きました:
The contrasts returned by pic() should already be standardized.
Standardizing phylogenetic contrasts is performed by dividing each
contrast (calculated between internal or terminal nodes) by a sum
proportional to the square root of its expected variance. Assuming
Brownian motion, for two sister terminal nodes, this is just the
square root of the sum of their branch lengths to a common ancestor.
(This is because the variance of a difference between two random
variables is just the sum of their separate variances, and variance
is accumulates in direct proportion to elapsed time under BM.) For
contrasts that are deeper in the phylogeny, this sum has to be
adjusted to account for uncertainty in the reconstructed values at
internal nodes, which adds extra variance to the computed difference.
This is described in great detail in Felsenstein (1985; Am Nat
125:1-15). However, as noted above - the values returned by pic()
should have already been standardized in this way!
You should also not subtract the mean value of the contrasts. This is
stated explicitly in Garland et al. (1992; Syst Biol 41:18-32): "The
expected mean of any set of contrasts is zero because the direction
of subtraction is arbitrary..., so only standard deviations are needed."
- Liam
Liam J. Revell
NESCent, Duke University
web: http://anolis.oeb.harvard.edu/~liam/
NEW email: lrev...@nescent.org
On Tue, 15 Sep 2009, Manabu Sakamoto wrote:
Dear list,
I am trying to standardise independent contrasts following Garland et
al. (1992), but I am unsure just how you would extract the "sum of it's
[contrast's] branch lenghts" in R.
I computed independent contrasts using the "pic" function in "ape". I
then took the contrasts and subtracted the mean from each one and now I
need to divide them by their standard deviations or the square root of
the sum of their branch lengths (Garland et al., 1992). This is where I
am stuck. I assume the sum of the branch lengths here refer to the sum
of the two branch lengths that go into the computation of each
contrast?
I would greatly appreciate any kind support on this matter.
Many thanks in advance,
Manabu Sakamoto
--
--------------------------------------
M. Sakamoto, PhD
Department of Earth Sciences,
University of Bristol,
Wills Memorial Building, Queen's Road,
Bristol BS8 1RJ, UK
m.sakam...@bristol.ac.uk
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