如何循环进行R中所有可能的因子水平比较
考虑以下数据帧:如何循环进行R中所有可能的因子水平比较,r,for-loop,categorical-data,R,For Loop,Categorical Data,考虑以下数据帧: type = c('A', 'A', 'A', 'B', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D') val1 = c(.35, .36, .35, .22, .27, .25, .88, .9, .87, .35, .35, .36) val2 = c(.35, .35, .37, .40, .42, .46, .9, .91, .82, .36, .36, .36) df = data.frame (type, val1, val2)
type = c('A', 'A', 'A', 'B', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D')
val1 = c(.35, .36, .35, .22, .27, .25, .88, .9, .87, .35, .35, .36)
val2 = c(.35, .35, .37, .40, .42, .46, .9, .91, .82, .36, .36, .36)
df = data.frame (type, val1, val2)
我有四个类别(称为类型;A、B、C和D)。每种类型的三个观察值可以平均,以创建一种类型的多元平均值(由val1和val2的平均值组成)。我想使用霍特林测试比较所有可能的类型组合(AB、AC、AD、BC、BD、CD),以确定哪种类型意味着(如果有的话)是相同的。我可以将其硬编码为:
a = filter (df, type == "A") [,2:3]
b = filter (df, type == "B") [,2:3]
c = filter (df, type == "C") [,2:3]
d = filter (df, type == "D") [,2:3]
然后对每对指定类型运行Hotelling的T2测试:
library('Hotelling')
hotelling.test(a, b, shrinkage=FALSE)
hotelling.test(b, c, shrinkage=FALSE)
hotelling.test(a, c, shrinkage=FALSE)
#And so on
这显然是非常低效和不切实际的,因为我的实际数据集有55种不同的类型。我知道答案在于for循环,但我很难找出如何告诉hotelling.test来比较所有可能类型组合的val1/val2多元平均数。我对创建for循环非常陌生,希望有人能给我指出正确的方向
在比较了所有类型之后,理想情况下,我能够得到一个输出,显示Hotelling测试p值大于0.05的类型对,这意味着这两种类型可能是重复的。在示例数据帧中,类型A和D返回的p值大于0.05,而其他比较具有p我们可以使用
combn
创建成对组合,对数据集进行子集划分并应用函数
library(Hotelling)
outlst <- combn(as.character(unique(df$type)), 2,
FUN = function(x) hotelling.test(subset(df, type == x[1], select = -1),
subset(df, type == x[2], select = -1)), simplify = FALSE)
names(outlst) <- combn(as.character(unique(df$type)), 2, FUN = paste, collapse = "_")
outlst[1]
#$A_B
#Test stat: 36.013
#Numerator df: 2
#Denominator df: 3
#P-value: 0.007996
库(霍特林)
outlst如果要用于循环:
type = c('A', 'A', 'A', 'B', 'B', 'B', 'C', 'C', 'C', 'D', 'D', 'D')
val1 = c(.35, .36, .35, .22, .27, .25, .88, .9, .87, .35, .35, .36)
val2 = c(.35, .35, .37, .40, .42, .46, .9, .91, .82, .36, .36, .36)
df = data.frame (type, val1, val2)
for (first in unique(df$type)) {
for (second in unique(df$type)) {
if (first != second) {
print(c(first, second))
}
}
}
[1] "A" "B"
[1] "A" "C"
[1] "A" "D"
[1] "B" "A"
[1] "B" "C"
[1] "B" "D"
[1] "C" "A"
[1] "C" "B"
[1] "C" "D"
[1] "D" "A"
[1] "D" "B"
[1] "D" "C"
筛选的第一步与拆分
相同,这将为您节省大量工作,并能跟上如此多的对象sp不敢相信我以前从未遇到过“拆分”,这是一个很棒的工具。谢谢