我有一个数据集,其中有18个种群。每个种群中都有几个个体,每个个体都有一个"颜色"的调用。我只想在以群体为主要因素的单向方差分析中同时比较两个群体,以获得内部和中间的成对MS。
我知道如何使用以下代码从综合方差分析中提取MS:
mylm <- lm(Color ~ Pop, data=PopColor)
anova(mylm)[["Mean Sq"]]
首先产生受试者间MS(PopColor$Pop),然后分别产生受试对象间MS(残差):
[1] 3.7079911 0.4536985
- 有没有一种方法可以让我创建一个do循环,在所有人群之间进行所有成对的单向方差分析,然后提取MS之间和内部的方差分析
- 然后,我想将每次比较中的两个MS值移动到它们自己的对称矩阵中:一个在由群体标记的受试者MS矩阵中,另一个在受试者由人群标记的MS矩阵中。这些将具有与"总体"名称相同的列和行名称
以下是我的六个群体的数据子集:
dput(dat)
structure(list(Pop = structure(c(6L, 6L, 6L, 6L, 6L, 6L, 6L,
6L, 6L, 6L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("pop1001", "pop1026",
"pop252", "pop254a", "pop311", "pop317"), class = "factor"),
Color = c(4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 3L, 3L,
3L, 3L, 2L, 3L, 3L, 3L, 2L, 2L, 2L, 1L, 3L, 3L, 2L, 3L, 2L,
3L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 3L, 3L, 2L, 3L, 3L, 2L,
3L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 2L, 3L, 2L, 2L, 3L, 4L,
2L, 3L, 2L, 4L, 3L, 3L, 2L, 3L, 2L, 3L, 3L, 4L, 3L, 2L, 4L,
4L, 1L, 2L, 2L, 2L, 2L, 1L, 3L, 2L, 3L, 2L, 3L, 3L, 3L, 3L,
2L, 3L, 4L, 2L, 2L, 4L, 3L)), .Names = c("Pop", "Color"), class = "data.frame", row.names = c(NA,
-94L))
如有任何帮助,我们将不胜感激!谢谢
我不理解你的第2点(对于像我这样的统计学家来说,这有点技术性)。关于第一点,我理解它,因为你想在你的所有种群中应用lm/Anova。您可以使用combn
:
combn:生成每次取m的x元素的所有组合
pops <- unique(PopColor$Pop)
ll <- combn(pops,2,function(x){
dat.pair <- PopColor[PopColor$Pop %in% pops[x],]
mylm <- lm(Color ~ Pop, data=dat.pair)
c(as.character(pops[x]),anova(mylm)[["Mean Sq"]])
},simplify=FALSE)
do.call(rbind,ll)
[,1] [,2] [,3] [,4]
[1,] "pop1026" "pop254a" "0.210291858678956" "0.597865353037767"
[2,] "pop1026" "pop1001" "0.52409988385598" "0.486874236874237"
[3,] "pop1026" "pop317" "15.7296466973886" "0.456486042692939"
[4,] "pop1026" "pop311" "1.34392057218144" "0.631962930099576"
[5,] "pop1026" "pop252" "0.339324116743472" "0.528899835796388"
[6,] "pop254a" "pop1001" "0.0166666666666669" "0.351785714285714"
[7,] "pop254a" "pop317" "14.45" "0.227777777777778"
[8,] "pop254a" "pop311" "1.92898550724637" "0.561430575035063"
[9,] "pop254a" "pop252" "0.8" "0.344444444444444"
[10,] "pop1001" "pop317" "20.4166666666667" "0.205357142857143"
[11,] "pop1001" "pop311" "3.55030333670374" "0.46474019088017"
[12,] "pop1001" "pop252" "1.35" "0.280357142857143"
[13,] "pop317" "pop311" "9.60474308300398" "0.429172510518934"
[14,] "pop317" "pop252" "8.45" "0.116666666666667"
[15,] "pop311" "pop252" "0.110803689064557" "0.496914446002805"
正如您所看到的,我们有choose(6,2)=15
可能的对。