找到物种积累曲线到达渐近线的位置
Find where species accumulation curve reaches asymptote
我已经使用 specaccum()
命令为我的样本制定了物种积累曲线。
这是一些示例数据:
site1<-c(0,8,9,7,0,0,0,8,0,7,8,0)
site2<-c(5,0,9,0,5,0,0,0,0,0,0,0)
site3<-c(5,0,9,0,0,0,0,0,0,6,0,0)
site4<-c(5,0,9,0,0,0,0,0,0,0,0,0)
site5<-c(5,0,9,0,0,6,6,0,0,0,0,0)
site6<-c(5,0,9,0,0,0,6,6,0,0,0,0)
site7<-c(5,0,9,0,0,0,0,0,7,0,0,3)
site8<-c(5,0,9,0,0,0,0,0,0,0,1,0)
site9<-c(5,0,9,0,0,0,0,0,0,0,1,0)
site10<-c(5,0,9,0,0,0,0,0,0,0,1,6)
site11<-c(5,0,9,0,0,0,5,0,0,0,0,0)
site12<-c(5,0,9,0,0,0,0,0,0,0,0,0)
site13<-c(5,1,9,0,0,0,0,0,0,0,0,0)
species_counts<-rbind(site1,site2,site3,site4,site5,site6,site7,site8,site9,site10,site11,site12,site13)
accum <- specaccum(species_counts, method="random", permutations=100)
plot(accum)
为了确保我有足够的采样,我需要确保物种积累图的曲线达到渐近线,定义为最后两点之间的斜率 <0.3(站点 12 和 13 之间的 ei) .
results <- with(accum, data.frame(sites, richness, sd))
产生这个:
sites richness sd
1 1 3.46 0.9991916
2 2 4.94 1.6625403
3 3 5.94 1.7513054
4 4 7.05 1.6779918
5 5 8.03 1.6542263
6 6 8.74 1.6794660
7 7 9.32 1.5497149
8 8 9.92 1.3534841
9 9 10.51 1.0492422
10 10 11.00 0.8408750
11 11 11.35 0.7017295
12 12 11.67 0.4725816
13 13 12.00 0.0000000
我觉得我快到了。我可以生成一个具有站点 vs 丰富度的 lm,并提取站点 12 和 13 之间的确切斜率(切线?)。要在这里搜索更长的时间。
稍微简化您的数据生成过程:
species_counts <- matrix(c(0,8,9,7,0,0,0,8,0,7,8,0,
5,0,9,0,5,0,0,0,0,0,0,0, 5,0,9,0,0,0,0,0,0,6,0,0,
5,0,9,0,0,0,0,0,0,0,0,0, 5,0,9,0,0,6,6,0,0,0,0,0,
5,0,9,0,0,0,6,6,0,0,0,0, 5,0,9,0,0,0,0,0,7,0,0,3,
5,0,9,0,0,0,0,0,0,0,1,0, 5,0,9,0,0,0,0,0,0,0,1,0,
5,0,9,0,0,0,0,0,0,0,1,6, 5,0,9,0,0,0,5,0,0,0,0,0,
5,0,9,0,0,0,0,0,0,0,0,0, 5,1,9,0,0,0,0,0,0,0,0,0),
byrow=TRUE,nrow=13)
在 运行 随机化测试之前 set.seed()
总是一个好主意(让我们知道 specaccum
在 vegan
包中):
set.seed(101)
library(vegan)
accum <- specaccum(species_counts, method="random", permutations=100)
从返回的对象中提取 richness
和 sites
分量并计算 d(richness)/d(sites)
(请注意,斜率向量比原点 site/richness 向量短一个元素: 如果您尝试将斜坡与特定数量的站点匹配,请小心)
(slopes <- with(accum,diff(richness)/diff(sites)))
## [1] 1.45 1.07 0.93 0.91 0.86 0.66 0.65 0.45 0.54 0.39 0.32 0.31
在这种情况下,斜率实际上从未低于 0.3,因此此代码用于查找斜率第一次低于 0.3 的时间:
which(slopes<0.3)[1]
returns NA
.
我已经使用 specaccum()
命令为我的样本制定了物种积累曲线。
这是一些示例数据:
site1<-c(0,8,9,7,0,0,0,8,0,7,8,0)
site2<-c(5,0,9,0,5,0,0,0,0,0,0,0)
site3<-c(5,0,9,0,0,0,0,0,0,6,0,0)
site4<-c(5,0,9,0,0,0,0,0,0,0,0,0)
site5<-c(5,0,9,0,0,6,6,0,0,0,0,0)
site6<-c(5,0,9,0,0,0,6,6,0,0,0,0)
site7<-c(5,0,9,0,0,0,0,0,7,0,0,3)
site8<-c(5,0,9,0,0,0,0,0,0,0,1,0)
site9<-c(5,0,9,0,0,0,0,0,0,0,1,0)
site10<-c(5,0,9,0,0,0,0,0,0,0,1,6)
site11<-c(5,0,9,0,0,0,5,0,0,0,0,0)
site12<-c(5,0,9,0,0,0,0,0,0,0,0,0)
site13<-c(5,1,9,0,0,0,0,0,0,0,0,0)
species_counts<-rbind(site1,site2,site3,site4,site5,site6,site7,site8,site9,site10,site11,site12,site13)
accum <- specaccum(species_counts, method="random", permutations=100)
plot(accum)
为了确保我有足够的采样,我需要确保物种积累图的曲线达到渐近线,定义为最后两点之间的斜率 <0.3(站点 12 和 13 之间的 ei) .
results <- with(accum, data.frame(sites, richness, sd))
产生这个:
sites richness sd
1 1 3.46 0.9991916
2 2 4.94 1.6625403
3 3 5.94 1.7513054
4 4 7.05 1.6779918
5 5 8.03 1.6542263
6 6 8.74 1.6794660
7 7 9.32 1.5497149
8 8 9.92 1.3534841
9 9 10.51 1.0492422
10 10 11.00 0.8408750
11 11 11.35 0.7017295
12 12 11.67 0.4725816
13 13 12.00 0.0000000
我觉得我快到了。我可以生成一个具有站点 vs 丰富度的 lm,并提取站点 12 和 13 之间的确切斜率(切线?)。要在这里搜索更长的时间。
稍微简化您的数据生成过程:
species_counts <- matrix(c(0,8,9,7,0,0,0,8,0,7,8,0,
5,0,9,0,5,0,0,0,0,0,0,0, 5,0,9,0,0,0,0,0,0,6,0,0,
5,0,9,0,0,0,0,0,0,0,0,0, 5,0,9,0,0,6,6,0,0,0,0,0,
5,0,9,0,0,0,6,6,0,0,0,0, 5,0,9,0,0,0,0,0,7,0,0,3,
5,0,9,0,0,0,0,0,0,0,1,0, 5,0,9,0,0,0,0,0,0,0,1,0,
5,0,9,0,0,0,0,0,0,0,1,6, 5,0,9,0,0,0,5,0,0,0,0,0,
5,0,9,0,0,0,0,0,0,0,0,0, 5,1,9,0,0,0,0,0,0,0,0,0),
byrow=TRUE,nrow=13)
在 运行 随机化测试之前 set.seed()
总是一个好主意(让我们知道 specaccum
在 vegan
包中):
set.seed(101)
library(vegan)
accum <- specaccum(species_counts, method="random", permutations=100)
从返回的对象中提取 richness
和 sites
分量并计算 d(richness)/d(sites)
(请注意,斜率向量比原点 site/richness 向量短一个元素: 如果您尝试将斜坡与特定数量的站点匹配,请小心)
(slopes <- with(accum,diff(richness)/diff(sites)))
## [1] 1.45 1.07 0.93 0.91 0.86 0.66 0.65 0.45 0.54 0.39 0.32 0.31
在这种情况下,斜率实际上从未低于 0.3,因此此代码用于查找斜率第一次低于 0.3 的时间:
which(slopes<0.3)[1]
returns NA
.