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我有一个看起来像这样的数据框

> head(data)
               LH3003     LH3004     LH3005     LH3006     LH3007     LH3008     LH3009     LH3010     LH3011
cg18478105 0.02329879 0.08103364 0.01611778 0.01691191 0.01886975 0.01885553 0.01647439 0.02120779 0.01168622
cg14361672 0.09479536 0.07821380 0.02522833 0.06467310 0.05387729 0.05866673 0.08121820 0.10920162 0.04413263
cg01763666 0.03625680 0.04633759 0.04401555 0.08371531 0.09866403 0.17611284 0.07306743 0.12422579 0.11125146
cg02115394 0.10014794 0.09274320 0.08743445 0.08906313 0.09934032 0.18164115 0.06526380 0.08158144 0.08862067
cg13417420 0.01811630 0.02221060 0.01314041 0.01964530 0.02367295 0.01209913 0.01612864 0.01306061 0.04421938
cg26724186 0.32776266 0.31386294 0.24167480 0.29036142 0.24751268 0.26894756 0.20927278 0.28070790 0.33188921
               LH3012     LH3013     LH3014
cg18478105 0.02466508 0.01909706 0.02054417
cg14361672 0.09172160 0.06170230 0.07752691
cg01763666 0.04328518 0.13693868 0.04288165
cg02115394 0.08682942 0.08601880 0.12413149
cg13417420 0.01980470 0.02241745 0.02038114
cg26724186 0.30832389 0.27644816 0.37630038

有近 850000 行，以及包含样本名称背后信息的不同数据框：

> variables
   Sample_ID     Name Group01
3     LH3003     pair1       0
4     LH3004     pair1       1
5     LH3005   pair2       0
6     LH3006   pair2       1
7     LH3007    pair3       0
8     LH3008    pair3       1
9     LH3009 pair4       0
10    LH3010 pair4       1
11    LH3011 pair5       0
12    LH3012 pair5       1
13    LH3013  pair6       0
14    LH3014  pair6       1

是否可以通过基于另一个数据框定义样本的对和组注释来进行配对 t 检验？

谢谢您的帮助！

4

3 回答 3

0

您需要堆叠数据并定义一对列，然后运行 ​​t.test，这是 6 个测试中的 1 个：

data2 <- data.frame(x = c(data$LH3003, data$LH3004), pair = c(rep(0, nrow(data)), rep(1, nrow(data))))
t.test(x ~ pair, data2)
于 2016-11-15T13:55:23.733 回答
0

这是一种lapply将每个测试的结果存储在列表中的方法。这假设每对在第二个 data.frame,df2 中是相邻的，并且第一个 data.frame 被命名为 df1。

myTestList <- lapply(seq(1, nrow(df2), 2),  function(i) 
                     t.test(df1[[df2$Sample_ID[i]]], df1[[df2$Sample_ID[i+1]]], paired=TRUE))

返回

myTestList
[[1]]

    Paired t-test

data:  df1[[df2$Sample_ID[i]]] and df1[[df2$Sample_ID[i + 1]]]
t = -0.50507, df = 5, p-value = 0.635
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.03453201  0.02319070
sample estimates:
mean of the differences 
           -0.005670653 


[[2]]

    Paired t-test

data:  df1[[df2$Sample_ID[i]]] and df1[[df2$Sample_ID[i + 1]]]
t = -2.5322, df = 5, p-value = 0.05239
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.0459320947  0.0003458114
sample estimates:
mean of the differences 
            -0.02279314

数据

df1 <- read.table(header=TRUE, text="LH3003  LH3004  LH3005   LH3006  LH3007  LH3008  LH3009  LH3010   LH3011
cg18478105 0.02329879 0.08103364 0.01611778 0.01691191 0.01886975 0.01885553 0.01647439 0.02120779 0.01168622
cg14361672 0.09479536 0.07821380 0.02522833 0.06467310 0.05387729 0.05866673 0.08121820 0.10920162 0.04413263
cg01763666 0.03625680 0.04633759 0.04401555 0.08371531 0.09866403 0.17611284 0.07306743 0.12422579 0.11125146
cg02115394 0.10014794 0.09274320 0.08743445 0.08906313 0.09934032 0.18164115 0.06526380 0.08158144 0.08862067
cg13417420 0.01811630 0.02221060 0.01314041 0.01964530 0.02367295 0.01209913 0.01612864 0.01306061 0.04421938
cg26724186 0.32776266 0.31386294 0.24167480 0.29036142 0.24751268 0.26894756 0.20927278 0.28070790 0.33188921")[1:4]

df2 <- read.table(header=TRUE, text="   Sample_ID     Name Group01
3     LH3003     pair1       0
4     LH3004     pair1       1
5     LH3005   pair2       0
6     LH3006   pair2       1")
于 2016-11-15T13:41:58.647 回答
0

这是@Imo 的一个变体：

lapply(unique(df2$Name), function(x){
  samples <- df2[df2$Name==x,1]
  t.test(df1[,samples[1]], df1[,samples[2]], paired=T)
})
于 2016-11-15T15:16:16.083 回答