如何将相同的命令应用于变量列表

时间:2013-07-25 15:44:32

标签: r apply

我想对一堆变量应用t检验。下面是一些模拟数据

d <- data.frame(var1=rnorm(10), 
                var2=rnorm(10), 
                group=sample(c(0,1), 10, replace=TRUE))

# Is there a way to do this in some sort of loop?
with(d, t.test(var1~group))
with(d, t.test(var2~group))

# I tried this but the loop did not give a result!?
varnames <- c('var1', 'var2')
for (i in 1:2) {
  eval(substitute(with(d, t.test(variable~group)),
                  list(variable=as.name(varnames[i]))))  
}

此外,是否可以从t检验的结果中提取值(例如,两个组均值,p值),以便循环将在变量之间生成一个整齐的平衡表?换句话说,我想要的最终结果不是彼此之间的一堆t检验,而是一个像这样的表:

Varname   mean1   mean2   p-value
Var1        1.1    1.2     0.989
Var2        1.2    1.3     0.912

3 个答案:

答案 0 :(得分:6)

您可以像这样使用formulalapply

set.seed(1)
d <- data.frame(var1 = rnorm(10), 
                var2 = rnorm(10), 
                group = sample(c(0, 1), 10, replace = TRUE))


varnames <- c("var1", "var2")
formulas <- paste(varnames, "group", sep = " ~ ")
res <- lapply(formulas, function(f) t.test(as.formula(f), data = d))
names(res) <- varnames

如果要提取表格,可以像这样进行

t(sapply(res, function(x) c(x$estimate, pval = x$p.value)))
     mean in group 0 mean in group 1     pval
var1         0.61288        0.012034 0.098055
var2         0.46382        0.195100 0.702365

答案 1 :(得分:3)

这是一个重塑/ plyr解决方案: foo函数是主力,它运行t检验和提取均值和p值。

d <- data.frame(var1=rnorm(10), 
                var2=rnorm(10), 
                group=sample(c(0,1), 10, replace=TRUE))

require(reshape2)
require(plyr)

dfm <- melt(d, id = 'group')

foo <- function(x) {
  tt <- t.test(value ~ group, data = x)
  out <- data.frame(mean1 = tt$estimate[1], mean2 = tt$estimate[2], P = tt$p.value)
  return(out)
}

ddply(dfm, .(variable), .fun=foo)
#  variable      mean1      mean2         P
#1     var1 -0.2641942  0.3716034 0.4049852
#2     var2 -0.9186919 -0.2749101 0.5949053

答案 2 :(得分:0)

使用sapply对所有varnames应用t-test,并通过对“estimate”和“p.value”进行子集化来提取必要的数据。如果要提取其他信息,请检查names(with(d, t.test(var1~group)))

t(with(d, sapply(varnames, function(x) unlist(t.test(get(x)~group)[c("estimate", "p.value")]))))