Question

我想使用R包plyr在一个非常大的数据框上运行成对t测试，但我不知道该怎么做。我最近学会了如何使用plyr进行相关，我真的很喜欢你如何指定你想要比较的组，然后plyr为你分解数据。例如，您可以让plyr计算虹膜数据集中每种虹膜的萼片长度和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))

我可以通过指定虹膜的setosa种类的数据在1:50行中等来自行打破数据框架，但是plyr比我更不容易陷入困境例如，不小心说出行1:51。

那么如何使用配对t检验做类似的事情呢？如何指定哪些观察对？这里的一些示例数据与我正在使用的数据类似，我希望这些对是主题，我希望通过Pesticide打破数据：< / p>

 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，我想评估的问题是，在冬季和夏季期间，每个人的农药暴露是否存在差异，我想分别回答这个问题。三种农药中的每一种。

提前多多谢谢！

编辑：为了澄清，这是如何在plyr中进行非配对t测试：

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

如果我添加＆＃34;配对= T＆＃34;在那里，plyr 将进行配对t检验，但它假设我总是以相同的顺序排列对。虽然我在上面的示例数据框中以相同的顺序将它们全部放在一起，但我不会在我的实际数据中，因为我有时会丢失数据。

Answer 1

你想要这个吗？

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，但这是我使用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

使用plyr进行成对t检验

2 个答案: