Question

我正在尝试在第三个值的组代码之后的两列上执行t.test或wilcox.test。这是我的数据

dput(data1)
structure(list(moda = structure(c(20L, 20L, 20L, 20L, 20L, 20L, 
20L, 20L, 20L, 20L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 10L, 
10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 
11L, 11L, 11L, 11L, 11L, 11L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 
5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 17L, 17L, 17L, 17L, 17L, 
17L, 17L, 17L, 17L, 17L, 18L, 18L, 18L, 18L, 18L, 18L, 18L, 18L, 
18L, 18L, 19L, 19L, 19L, 19L, 19L, 19L, 19L, 19L, 19L, 19L, 7L, 
7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 20L, 20L, 
20L, 20L, 20L, 20L, 20L, 20L, 20L, 20L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 8L, 8L, 8L, 8L, 
8L, 8L, 8L, 8L, 8L, 8L, 12L, 12L, 12L, 12L, 12L, 12L, 12L, 12L, 
12L, 12L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 14L, 
14L, 14L, 14L, 14L, 14L, 14L, 14L, 14L, 14L, 15L, 15L, 15L, 15L, 
15L, 15L, 15L, 15L, 15L, 15L, 16L, 16L, 16L, 16L, 16L, 16L, 16L, 
16L, 16L, 16L, 20L, 20L, 20L, 20L, 20L, 20L, 20L, 20L, 20L, 20L, 
21L, 21L, 21L, 21L, 21L, 21L, 21L, 21L, 21L, 22L, 22L, 22L, 22L, 
22L, 22L, 22L, 22L, 22L, 22L, 23L, 23L, 23L, 23L, 23L, 23L, 23L, 
23L, 23L), .Label = c("ACN1", "ACN2", "BA", "BM", "BS1", "BS2", 
"CN", "EK5", "HW1", "HW2", "HW3", "L27", "L5K", "LC", "M2K", 
"M630", "PB1", "PB2", "PB3", "PG", "RMB", "RMC", "RMM"), class =                     "factor"), 
epicotyle = c(1.5, 1.5, 2, 1, 1.5, 1.2, 1, 2.4, 1.3, 1.4, 
1.7, 2, 1.8, 2.3, 2.5, 2.5, 1.5, 1.5, 2, 1.3, 1.5, 1.8, 1.3, 
1.8, 1.7, 1.5, 2.3, 1.8, 2.2, 1.5, 1.5, 1.5, 1.3, 1.5, 1.5, 
1.5, 1.5, 1.8, 1.5, 2.1, 1.8, 1.3, 2, 1.5, 2, 3.5, 1.5, 1.7, 
1.7, 2, 1.7, 2, 1.5, 2, 1.5, 2, 2, 1.5, 2, 1.5, 1.8, 1, 2, 
3, 1.6, 1.5, 1.5, 1.3, 1.5, 1.5, 1.2, 1.5, 1.5, 1, 1.2, 1.5, 
1.5, 1.5, 1.5, 2, 1.1, 1.5, 1.5, 1.7, 1.8, 1.5, 1.3, 1.5, 
1.5, 2.5, 1.2, 1.4, 1, 1.5, 2, 1.5, 1.2, 1.5, 2, 2.3, 2.1, 
2, 2.4, 1.5, 1.7, 1.4, 2.4, 1, 1, 2, 1.5, 1.2, 2.4, 1.2, 
1, 0.8, 1.8, 1.5, 1.5, 2, 1, 1.5, 1.2, 1, 2.4, 1.3, 1.4, 
1.5, 1.5, 1.5, 2.1, 1.5, 1.4, 1.5, 1.3, 1.5, 3, 2.6, 1.5, 
2.2, 1.9, 1.5, 1.4, 1.4, 2.5, 2.1, 2, 1.5, 2, 2, 2, 1.5, 
2.1, 2, 1.5, 2.5, 2.5, 3, 3, 3.5, 3.5, 3, 2, 2.5, 3.5, 1, 
1.2, 1.5, 2.5, 1.5, 1.5, 1.5, 1.5, 1.5, 2.4, 1.5, 2, 3, 1.7, 
3, 2.5, 2, 2.5, 2.5, 2.5, 1.5, 1.5, 1.5, 1, 1.5, 2, 1.4, 
1.2, 1.7, 2.1, 1.5, 2, 1.5, 1.5, 2, 1.4, 2, 3, 2, 2, 1.5, 
1.5, 2, 1, 1.5, 1.2, 1, 2.4, 1.3, 1.4, 2, 2.5, 3, 3, 1.7, 
3, 1.8, 2, 1.8, 2.2, 2.3, 1.5, 2, 1.8, 1.8, 1.3, 2, 1.8, 
1.8, 2, 1.8, 1.5, 1.7, 2, 1.4, 1.5, 1.7, 1.5), hypocotyle = c(1.5, 
1.5, 2, 1, 1.5, 1.2, 1, 2.4, 1.3, 1.4, 5, 7, 2.5, 6.5, 5.4, 
5, 6, 5.7, 7, 5.5, 5.7, 5.5, 7, 6.5, 5.5, 5.5, 6.7, 4.9, 
5.3, 6.7, 5.8, 6.5, 6, 5.6, 5, 5.5, 6, 6, 6, 3.5, 4.7, 4.5, 
5.9, 5, 6, 7, 6, 5.5, 5, 5.8, 5.5, 5.5, 4.8, 5.7, 6, 7, 5.2, 
5, 5.2, 5.3, 5.6, 5, 5.3, 6, 5, 5.5, 4.5, 5.7, 6, 4.5, 4.4, 
5.2, 5.2, 4.1, 5.2, 5.2, 5.4, 6, 5.5, 6.5, 5, 6, 5.5, 7.5, 
5.2, 5.6, 5.4, 5.5, 5, 5, 6, 5.2, 6, 6.3, 6.3, 4.2, 5.1, 
3.5, 6, 6, 6, 6, 5, 5, 6, 5, 5.6, 5.5, 5, 5, 6, 5.2, 6, 6.3, 
6.3, 4.2, 5.1, 3.8, 4, 7, 5, 6, 4, 5.4, 3.5, 3.6, 5, 6, 4.8, 
4.7, 4.4, 5.5, 3.5, 5.3, 4.3, 5.5, 4.5, 5.5, 4.2, 6, 4.3, 
4, 4.7, 3.5, 3.7, 4.2, 5, 5, 5.1, 5.7, 5, 3.5, 4, 5.6, 3.9, 
3.5, 7, 6, 6, 6, 6.5, 5.5, 4.5, 6.5, 6.5, 3, 5, 5.5, 5.3, 
4, 5.5, 6, 4, 5.5, 6, 5, 4, 4.5, 4.5, 4, 3.5, 4.5, 5, 4, 
4.5, 5, 4.7, 6, 3.8, 4.5, 4.1, 4, 3.7, 4, 4.5, 5, 6, 4.5, 
6, 5.7, 3.7, 5.8, 6.2, 5.5, 5, 3.8, 4, 7, 5, 6, 4, 5.4, 3.5, 
3.6, 5, 7, 6.5, 8, 6.5, 5.7, 7.5, 7.3, 7.4, 7, 5.4, 6.5, 
6.5, 7.2, 7.4, 6, 6.5, 6, 7, 6, 7, 6.5, 6.5, 6.5, 8, 5.7, 
6.5, 6, 7)), class = "data.frame", row.names = c(NA, -243L
))
bartl <- function(p, yy=data1[,1]){barte <- bartlett.test(p,yy)}

aggre <- function (x, y=data1[,1]) aggregate(formula =x~data1[,1],data =         data1, FUN = t.test)
lshap <- lapply(data1[-1], FUN = aggre, y=data1[,1])
lshap

或者我也尝试过其他东西

result <- by(data1[-1], data1[,1], 
         function(x) t.test(data1[,2], data1[,3], mu=0, alt="two.sided",     paired = TRUE, conf.level = 0.95))
result$p-value

ttest.pval <- sapply(result, '[[', 'p.value')
ttest.pval

但是看起来它并不在乎我的组的代码，因此pvalue相同。

我的最终目标是创建一个脚本，该脚本在根据shapiro测试pvalue进行先前的均方差检验后，给出正确的t.test pvalue或wilcox。

Answer 1

只需使用传递给x函数的by变量，该变量是子集数据帧。您会忽略使用它，因此会得到重复的结果。为了清楚起见，下面为 subdata 重命名x并调整返回对象：

result_list <- by(data1, data1$moda, function(sub_data) {
  res <- t.test(sub_data$epicotyle, sub_data$hypocotyle, mu=0, alt="two.sided",     
                paired = TRUE, conf.level = 0.95)
  output <- c(t_stat=unname(res$estimate), p_value=res$p.value)
})

result_matrix <- do.call(rbind, result_list)

result_matrix
#         t_stat      p_value
# ACN1 -3.355556 1.562532e-06
# ACN2 -2.488889 1.711867e-05
# BA   -3.480000 8.156509e-08
# BM   -4.070000 4.481620e-08
# BS1  -3.411111 2.882650e-06
# BS2  -3.788889 6.311561e-08
# CN   -3.930000 1.464457e-07
# EK5  -2.830000 3.638520e-07
# HW1  -3.588889 5.753743e-05
# HW2  -4.090000 4.047017e-08
# HW3  -4.400000 4.571253e-10
# L27  -2.900000 4.274705e-06
# L5K  -3.370000 5.018480e-07
# LC   -2.030000 2.347509e-05
# M2K  -2.890000 4.196188e-07
# M630 -3.450000 2.544427e-08
# PB1  -3.500000 3.465025e-10
# PB2  -3.720000 2.646581e-09
# PB3  -4.280000 4.970850e-09
# PG   -2.166667 4.627951e-07
# RMB  -4.677778 4.395313e-08
# RMC  -4.600000 1.166993e-08
# RMM  -4.955556 3.320310e-09

在第三列中遵循特定代码的情况下，在两列之间进行t.test

1 个答案: