R 两条曲线之间的统计差异

我有两个物种的猎物消耗数据，我试图确定它们的功能反应曲线的不同程度

以下是我的数据的一个小例子（每个物种有一个单独的Excel文件，非本地物种每种密度有6次重复，而不是5次）：

库（frair）
specie1请发布一个可复制的示例。@jmuhlenkamp我已经更新了数据和代码。我建议使用
dput
并将结果粘贴到此处，以便其他人可以轻松复制您的数据。您问题的标题没有描述您的问题，或者确实如此@使用dput更新jmuhlenkamp。
library(frair)

specie1 <- structure(list(density = c(2L, 2L, 2L, 2L, 2L, 4L, 4L, 4L, 4L, 
4L, 8L, 8L, 8L, 8L, 8L, 12L, 12L, 12L, 12L, 12L, 16L, 16L, 16L, 
16L, 16L, 20L, 20L, 20L, 20L, 20L, 24L, 24L, 24L, 24L, 24L, 28L, 
28L, 28L, 28L, 28L, 32L, 32L, 32L, 32L, 32L, 42L, 42L, 42L, 42L, 
42L, 54L, 54L, 54L, 54L, 54L), eaten = c(2L, 2L, 2L, 2L, 2L, 
4L, 0L, 4L, 4L, 4L, 8L, 8L, 8L, 6L, 8L, 12L, 12L, 12L, 12L, 12L, 
16L, 16L, 16L, 9L, 16L, 20L, 20L, 20L, 18L, 20L, 24L, 24L, 24L, 
24L, 24L, 28L, 28L, 28L, 28L, 28L, 30L, 22L, 32L, 32L, 32L, 27L, 
33L, 36L, 28L, 36L, 37L, 33L, 27L, 42L, 42L)), .Names = c("density", 
"eaten"), class = "data.frame", row.names = c(NA, -55L))

specie2 <- structure(list(density = c(2L, 2L, 2L, 2L, 2L, 2L, 4L, 4L, 4L, 
4L, 4L, 4L, 8L, 8L, 8L, 8L, 8L, 8L, 12L, 12L, 12L, 12L, 12L, 
12L, 16L, 16L, 16L, 16L, 16L, 16L, 20L, 20L, 20L, 20L, 20L, 20L, 
24L, 24L, 24L, 24L, 24L, 24L, 28L, 28L, 28L, 28L, 28L, 28L, 32L, 
32L, 32L, 32L, 32L, 32L, 42L, 42L, 42L, 42L, 42L, 42L, 54L, 54L, 
54L, 54L, 54L, 54L), eaten = c(0L, 0L, 2L, 0L, 0L, 0L, 0L, 0L, 
4L, 0L, 3L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 7L, 0L, 4L, 4L, 4L, 
9L, 15L, 8L, 15L, 16L, 0L, 0L, 1L, 4L, 0L, 0L, 13L, 2L, 2L, 4L, 
18L, 5L, 10L, 4L, 7L, 14L, 13L, 0L, 5L, 18L, 0L, 18L, 20L, 15L, 
21L, 5L, 9L, 15L, 5L, 1L, 2L, 31L, 20L, 7L, 13L, 10L, 1L)), .Names = c("density", 
"eaten"), class = "data.frame", row.names = c(NA, -66L))

specie1_fit <- frair_fit(eaten~density, data=specie1, response='rogersII', start=list(a=1.2, h=0.015), fixed=list(T=24))
specie2_fit <- frair_fit(eaten~density, data=specie2, response='rogersII', start=list(a=1.2, h=0.015), fixed=list(T=24))

frair_compare(specie1_fit, specie2_fit)

#FUNCTIONAL RESPONSE COEFFICIENT TEST

#Response:            rogersII
#Optimised variables: a,h
#Fixed variables:     T

#Original coefficients: 
#                 a       h
#specie1_fit 0.22065 0.48170
#specie2_fit 0.01706 0.54308

#Test: specie1_fit - specie2_fit

#   Estimate Std. Error z value Pr(z)
#Da  0.20631         NA      NA    NA
#Dh  0.03813         NA      NA    NA
#Warning message:
#In bbmle::mle2(minuslogl = fr_nll_difffunc, start = start, fixed = fixed,  :
#  couldn't invert Hessian