R：用于nnet multinom多项式的Tukey posthoc测试适合测试多项分布的整体差异

Question

我使用nnet的{​​{1}}函数拟合了一个mutinomial模型（在这种情况下，数据给出了男性和女性的饮食偏好以及不同湖泊中不同大小的鳄鱼类型）：

multinom

我可以使用的因素的总体意义

data=read.csv("https://www.dropbox.com/s/y9elunsbv74p2h6/alligator.csv?dl=1")
head(data)
  id size  sex    lake food
1  1 <2.3 male hancock fish
2  2 <2.3 male hancock fish
3  3 <2.3 male hancock fish
4  4 <2.3 male hancock fish
5  5 <2.3 male hancock fish
6  6 <2.3 male hancock fish

library(nnet)
fit=multinom(food~lake+sex+size, data = data, Hess = TRUE)

我得到的影响情节，例如因子“湖”使用

library(car)
Anova(fit, type="III")  # type III tests
Analysis of Deviance Table (Type III tests)

Response: food
     LR Chisq Df Pr(>Chisq)    
lake   50.318 12  1.228e-06 ***
sex     2.215  4   0.696321    
size   17.600  4   0.001477 ** 

除了整体的Anova测试之外，我还想进行成对的Tukey posthoc测试，以测试多食物分布中吃掉猎物的总体差异，例如：穿过不同的湖泊。

我首先考虑在包library(effects) plot(effect(fit,term="lake"),ylab="Food",type="probability",style="stacked",colors=rainbow(5)) 中使用函数glht，但这似乎不起作用，例如因子multcomp：

lake

替代方法是使用包library(multcomp) summary(glht(fit, mcp(lake = "Tukey"))) Error in summary(glht(fit, mcp(lake = "Tukey"))) : error in evaluating the argument 'object' in selecting a method for function 'summary': Error in glht.matrix(model = list(n = c(6, 0, 5), nunits = 12L, nconn = c(0, : ‘ncol(linfct)’ is not equal to ‘length(coef(model))’ 来实现此目的，我尝试了

lsmeans

这会对每种特定类型食品的比例进行测试。

我想知道是否也可以通过这种或那种方式获得Tukey posthoc测试，其中在不同的湖泊中比较整体多项分布，即在任何猎物的比例中测试差异吃过？ 我试过

lsmeans(fit, pairwise ~ lake | food, adjust="tukey", mode = "prob")
$contrasts
food = bird:
 contrast               estimate         SE df t.ratio p.value
 george - hancock    -0.04397388 0.05451515 24  -0.807  0.8507
 george - oklawaha    0.03680712 0.03849268 24   0.956  0.7751
 george - trafford   -0.02123255 0.05159049 24  -0.412  0.9760
 hancock - oklawaha   0.08078100 0.04983303 24   1.621  0.3863
 hancock - trafford   0.02274133 0.06242724 24   0.364  0.9831
 oklawaha - trafford -0.05803967 0.04503128 24  -1.289  0.5786

food = fish:
 contrast               estimate         SE df t.ratio p.value
 george - hancock    -0.02311955 0.09310322 24  -0.248  0.9945
 george - oklawaha    0.19874095 0.09273047 24   2.143  0.1683
 george - trafford    0.32066789 0.08342262 24   3.844  0.0041
 hancock - oklawaha   0.22186050 0.09879102 24   2.246  0.1396
 hancock - trafford   0.34378744 0.09088119 24   3.783  0.0047
 oklawaha - trafford  0.12192695 0.08577365 24   1.421  0.4987

food = invert:
 contrast               estimate         SE df t.ratio p.value
 george - hancock     0.23202865 0.06111726 24   3.796  0.0046
 george - oklawaha   -0.13967425 0.08808698 24  -1.586  0.4053
 george - trafford   -0.07193252 0.08346283 24  -0.862  0.8242
 hancock - oklawaha  -0.37170290 0.07492749 24  -4.961  0.0003
 hancock - trafford  -0.30396117 0.07129577 24  -4.263  0.0014
 oklawaha - trafford  0.06774173 0.09384594 24   0.722  0.8874

food = other:
 contrast               estimate         SE df t.ratio p.value
 george - hancock    -0.12522495 0.06811177 24  -1.839  0.2806
 george - oklawaha    0.03499241 0.05141930 24   0.681  0.9035
 george - trafford   -0.08643898 0.06612383 24  -1.307  0.5674
 hancock - oklawaha   0.16021736 0.06759887 24   2.370  0.1103
 hancock - trafford   0.03878598 0.08135810 24   0.477  0.9635
 oklawaha - trafford -0.12143138 0.06402725 24  -1.897  0.2560

food = rep:
 contrast               estimate         SE df t.ratio p.value
 george - hancock    -0.03971026 0.03810819 24  -1.042  0.7269
 george - oklawaha   -0.13086622 0.05735022 24  -2.282  0.1305
 george - trafford   -0.14106384 0.06037257 24  -2.337  0.1177
 hancock - oklawaha  -0.09115595 0.06462624 24  -1.411  0.5052
 hancock - trafford  -0.10135358 0.06752424 24  -1.501  0.4525
 oklawaha - trafford -0.01019762 0.07161794 24  -0.142  0.9989

Results are averaged over the levels of: sex, size 
P value adjustment: tukey method for comparing a family of 4 estimates 

但这似乎不起作用：

lsmeans(fit, pairwise ~ lake, adjust="tukey", mode = "prob")

有什么想法？

或者有人知道如何$contrasts contrast estimate SE df t.ratio p.value george - hancock 3.252607e-19 1.879395e-10 24 0 1.0000 george - oklawaha -8.131516e-19 1.861245e-10 24 0 1.0000 george - trafford -1.843144e-18 2.504062e-10 24 0 1.0000 hancock - oklawaha -1.138412e-18 NaN 24 NaN NaN hancock - trafford -2.168404e-18 NaN 24 NaN NaN oklawaha - trafford -1.029992e-18 NaN 24 NaN NaN 为glht模型工作吗？

Answer 1

刚刚收到了Russ Lenth的一条消息，他认为在湖泊中进行这些成对比较的语法是为了测试鳄鱼食用的食物项目的多项分布差异

lsm = lsmeans(fit, ~ lake|food, mode = "latent")
cmp = contrast(lsm, method="pairwise", ref=1) 
test = test(cmp, joint=TRUE, by="contrast") 
There are linearly dependent rows - df are reduced accordingly
test
 contrast            df1 df2     F p.value
 george - hancock      4  24 3.430  0.0236
 george - oklawaha     4  24 2.128  0.1084
 george - trafford     4  24 3.319  0.0268
 hancock - oklawaha    4  24 5.820  0.0020
 hancock - trafford    4  24 5.084  0.0041
 oklawaha - trafford   4  24 1.484  0.2383

谢谢Russ！