我是生态学家,主要使用纯素R包.

我有2个矩阵(样本x丰度)(见下面的数据):

矩阵1/nrow = 6replicates*24sites,ncol = 15种丰度(鱼)矩阵2/nrow = 3replicates*24sites,ncol = 10种丰度(无脊椎动物)

两个矩阵中的站点相同.我想在两对网站之间得到整体的bray-curtis不同(考虑两个矩阵).我看到两个选项:

选项1,平均重复(在场地规模)鱼类和大型无脊椎动物丰度,cbind两个平均丰度矩阵(nrow = 24sites,ncol = 15 + 10平均丰度)和计算bray-curtis.

选项2,对于每个组合,计算站点对之间的bray-curtis不相似性,计算站点质心之间的距离.然后总结2个距离矩阵.

如果我不清楚,我在下面的R代码中做了这两个操作.

请问,您能告诉我选项2是否正确且比选项1更合适.

先感谢您.

皮埃尔

这里是R代码的例子

生成数据

library(plyr);library(vegan)

#assemblage 1: 15 fish species, 6 replicates per site
a1.env=data.frame(
  Habitat=paste("H",gl(2,12*6),sep=""),
  Site=paste("S",gl(24,6),sep=""),
  Replicate=rep(paste("R",1:6,sep=""),24))

summary(a1.env)

a1.bio=as.data.frame(replicate(15,rpois(144,sample(1:10,1))))

names(a1.bio)=paste("F",1:15,sep="")

a1.bio[1:72,]=2*a1.bio[1:72,]

#assemblage 2: 10 taxa of macro-invertebrates, 3 replicates per site

a2.env=a1.env[a1.env$Replicate%in%c("R1","R2","R3"),]

summary(a2.env)

a2.bio=as.data.frame(replicate(10,rpois(72,sample(10:100,1))))

names(a2.bio)=paste("I",1:10,sep="")

a2.bio[1:36,]=0.5*a2.bio[1:36,]


#environmental data at the sit scale

env=unique(a1.env[,c("Habitat","Site")])

env=env[order(env$Site),]

方案1,平均丰度和cbind

a1.bio.mean=ddply(cbind(a1.bio,a1.env),.(Habitat,Site),numcolwise(mean))

a1.bio.mean=a1.bio.mean[order(a1.bio.mean$Site),]

a2.bio.mean=ddply(cbind(a2.bio,a2.env),.(Habitat,Site),numcolwise(mean))

a2.bio.mean=a2.bio.mean[order(a2.bio.mean$Site),]

bio.mean=cbind(a1.bio.mean[,-c(1:2)],a2.bio.mean[,-c(1:2)])

dist.mean=vegdist(sqrt(bio.mean),"bray")

选项2,计算质心之间的每个组合距离并对2个距离矩阵求和

a1.dist=vegdist(sqrt(a1.bio),"bray")

a1.coord.centroid=betadisper(a1.dist,a1.env$Site)$centroids

a1.dist.centroid=vegdist(a1.coord.centroid,"eucl")

a2.dist=vegdist(sqrt(a2.bio),"bray")

a2.coord.centroid=betadisper(a2.dist,a2.env$Site)$centroids …