使用plyr的配对t检验

Question

我想使用R包plyr在一个非常大的数据框架上运行一个配对的t测试，但是我不知道怎么做。我最近学习了如何使用plyr进行关联，我非常喜欢您如何指定要比较哪些组，然后plyr为您细分数据。例如，您可以让plyr计算虹膜数据集中的每种虹膜的萼片长度与萼片宽度之间的相关性，如下所示：

Correlations <- ddply(iris, "Species", function(x) cor(x$Sepal.Length, x$Sepal.Width))

我可以自己分解数据框架，方法是指定刚毛虹膜物种的数据列在第1行:50行，等等，但是plyr不太可能像我那样搞砸，不小心说第1行:51行。

那么，我如何用配对t检验来做类似的事情呢？我怎样才能指定哪对观测是对的？下面是一些与我所做的工作类似的例子数据，我希望这两对成为研究对象，我想用农药来分解数据：

 Exposure <- data.frame("Subject" = rep(1:4, 6), 
                     "Season" = rep(c(rep("summer", 4), rep("winter", 4)),3),
                     "Pesticide" = rep(c("atrazine", "metolachlor", "chlorpyrifos"), each=8),
                      "Exposure" = sample(1:100, size=24))
 Exposure$Subject <- as.factor(Exposure$Subject)

换句话说，我想要评估的问题是，冬季和夏季，每个人的农药接触量是否存在差异，我想分别回答这三种杀虫剂中的每一种。

事先非常感谢！

编辑:为了澄清，这是如何在plyr中进行未配对t检验：

 TTests <- dlply(Exposure, "Pesticide", function(x) t.test(x$Exposure ~ x$Season))

如果我在里面加上"paired=T“，plyr会做一个配对的t检验，但它假设我总是有相同的顺序。虽然在上面的示例数据框架中，它们都是按相同的顺序排列的，但在实际数据中却没有，因为有时我会丢失数据。

Answer 1

您是想要这个吗？

library(data.table)

# convert to data.table in place
setDT(Exposure)

# make sure data is sorted correctly
setkey(Exposure, Pesticide, Season, Subject)

Exposure[, list(res = list(t.test(Exposure[Season == "summer"],
                                  Exposure[Season == "winter"],
                                  paired = T)))
         , by = Pesticide]$res
#[[1]]
#
#        Paired t-test
#
#data:  Exposure[Season == "summer"] and Exposure[Season == "winter"]
#t = -4.1295, df = 3, p-value = 0.02576
#alternative hypothesis: true difference in means is not equal to 0
#95 percent confidence interval:
# -31.871962  -4.128038
#sample estimates:
#mean of the differences 
#                    -18 
#
#
#[[2]]
#
#        Paired t-test
#
#data:  Exposure[Season == "summer"] and Exposure[Season == "winter"]
#t = -6.458, df = 3, p-value = 0.007532
#alternative hypothesis: true difference in means is not equal to 0
#95 percent confidence interval:
# -73.89299 -25.10701
#sample estimates:
#mean of the differences 
#                  -49.5 
#
#
#[[3]]
#
#        Paired t-test
#
#data:  Exposure[Season == "summer"] and Exposure[Season == "winter"]
#t = -2.5162, df = 3, p-value = 0.08646
#alternative hypothesis: true difference in means is not equal to 0
#95 percent confidence interval:
# -30.008282   3.508282
#sample estimates:
#mean of the differences 
#                 -13.25

Answer 2

我不知道ddply，但下面是使用一些base函数的方法。

by(data = Exposure, INDICES = Exposure$Pesticide, FUN = function(x) {
  t.test(Exposure ~ Season, data = x)
})

Exposure$Pesticide: atrazine

    Welch Two Sample t-test

data:  Exposure by Season
t = -0.1468, df = 5.494, p-value = 0.8885
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -49.63477  44.13477
sample estimates:
mean in group summer mean in group winter 
               60.50                63.25 

---------------------------------------------------------------------------------------------- 
Exposure$Pesticide: chlorpyrifos

    Welch Two Sample t-test

data:  Exposure by Season
t = -0.8932, df = 4.704, p-value = 0.4151
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -83.58274  41.08274
sample estimates:
mean in group summer mean in group winter 
               52.25                73.50 

---------------------------------------------------------------------------------------------- 
Exposure$Pesticide: metolachlor

    Welch Two Sample t-test

data:  Exposure by Season
t = 0.8602, df = 5.561, p-value = 0.4252
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -39.8993  81.8993
sample estimates:
mean in group summer mean in group winter 
                62.5                 41.5