test in the spirit of the Raup and Crick (1979) coefficient. The p-values act as
dissimilarities: they have very small values for highly co-occurring species. We
provide an R function called test.a() to compute this coefficient. Readers are
invited to apply it to the species of the Doubs fish data transformed to presenceabsence form. A critical point is to specify enough permutations for the probabilities
to survive a correction for multiple testing (see Sect. 7.2.6). There are 27 species, and
thus 27 Â 26/2 ¼ 351 tests will be run. A Bonferroni correction requires a p-value of
0.05/351 ¼ 0.0001425 to remain significant at the 0.05 level. This requires at least
9999 permutations, since the smallest p-value would be 1/(9999 + 1) ¼ 0.0001.
99,999 permutations provide a finer estimation of the p-value but beware: computation can take several minutes.
# Transform the data to presence-absence
spe.pa <- decostand(spe, "pa")
# Test the co-occurrence of species
res <- test.a(spe.pa, nperm = 99999)
summary(res)
In the output object, res$p.a.dist contains a matrix of p-values of class dist.
The next step is to compute a Holm correction (see Sect. 7.2.6) on the matrix of pvalues unfolded as a vector.
# Compute a Holm correction on the matrix of p-values
res.p.vec <- as.vector(res$p.a.dist)
adjust.res <- p.adjust(res.p.vec, method = "holm")
range(adjust.res)
Among the corrected Holm p-values, find 0.05 or the closest value smaller than
0.05:
(adj.sigth <- max(adjust.res[adjust.res <= 0.05]))
Now find the uncorrected p-value corresponding to adj.sigth:
(sigth <- max(res.p.vec[adjust.res <= 0.05]))
In this run it is 0.00017. The significant values thus have an uncorrected
probability of 0.00017 or less. Replace all larger values in the probability matrix
by 1:
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4 Cluster Analysis
dissimilarities: they have very small values for highly co-occurring species. We
provide an R function called test.a() to compute this coefficient. Readers are
invited to apply it to the species of the Doubs fish data transformed to presenceabsence form. A critical point is to specify enough permutations for the probabilities
to survive a correction for multiple testing (see Sect. 7.2.6). There are 27 species, and
thus 27 Â 26/2 ¼ 351 tests will be run. A Bonferroni correction requires a p-value of
0.05/351 ¼ 0.0001425 to remain significant at the 0.05 level. This requires at least
9999 permutations, since the smallest p-value would be 1/(9999 + 1) ¼ 0.0001.
99,999 permutations provide a finer estimation of the p-value but beware: computation can take several minutes.
# Transform the data to presence-absence
spe.pa <- decostand(spe, "pa")
# Test the co-occurrence of species
res <- test.a(spe.pa, nperm = 99999)
summary(res)
In the output object, res$p.a.dist contains a matrix of p-values of class dist.
The next step is to compute a Holm correction (see Sect. 7.2.6) on the matrix of pvalues unfolded as a vector.
# Compute a Holm correction on the matrix of p-values
res.p.vec <- as.vector(res$p.a.dist)
adjust.res <- p.adjust(res.p.vec, method = "holm")
range(adjust.res)
Among the corrected Holm p-values, find 0.05 or the closest value smaller than
0.05:
(adj.sigth <- max(adjust.res[adjust.res <= 0.05]))
Now find the uncorrected p-value corresponding to adj.sigth:
(sigth <- max(res.p.vec[adjust.res <= 0.05]))
In this run it is 0.00017. The significant values thus have an uncorrected
probability of 0.00017 or less. Replace all larger values in the probability matrix
by 1:
116
4 Cluster Analysis