我想执行t.test以获取指定向量之间的pvalue。我们以下面的数据为例：
structure(list(mpg = c(21, 21, 22.8, 21.4, 18.7, 18.1, 14.3,
24.4, 22.8, 19.2, 17.8, 16.4, 17.3, 15.2, 10.4, 10.4, 14.7, 32.4,
30.4, 33.9, 21.5, 15.5, 15.2, 13.3, 19.2, 27.3, 26, 30.4, 15.8,
19.7, 15, 21.4), cyl = c(6, 6, 4, 6, 8, 6, 8, 4, 4, 6, 6, 8,
8, 8, 8, 8, 8, 4, 4, 4, 4, 8, 8, 8, 8, 4, 4, 4, 8, 6, 8, 4),
disp = c(160, 160, 108, 258, 360, 225, 360, 146.7, 140.8,
167.6, 167.6, 275.8, 275.8, 275.8, 472, 460, 440, 78.7, 75.7,
71.1, 120.1, 318, 304, 350, 400, 79, 120.3, 95.1, 351, 145,
301, 121), hp = c(110, 110, 93, 110, 175, 105, 245, 62, 95,
123, 123, 180, 180, 180, 205, 215, 230, 66, 52, 65, 97, 150,
150, 245, 175, 66, 91, 113, 264, 175, 335, 109), drat = c(3.9,
3.9, 3.85, 3.08, 3.15, 2.76, 3.21, 3.69, 3.92, 3.92, 3.92,
3.07, 3.07, 3.07, 2.93, 3, 3.23, 4.08, 4.93, 4.22, 3.7, 2.76,
3.15, 3.73, 3.08, 4.08, 4.43, 3.77, 4.22, 3.62, 3.54, 4.11
), wt = c(2.62, 2.875, 2.32, 3.215, 3.44, 3.46, 3.57, 3.19,
3.15, 3.44, 3.44, 4.07, 3.73, 3.78, 5.25, 5.424, 5.345, 2.2,
1.615, 1.835, 2.465, 3.52, 3.435, 3.84, 3.845, 1.935, 2.14,
1.513, 3.17, 2.77, 3.57, 2.78), qsec = c(16.46, 17.02, 18.61,
19.44, 17.02, 20.22, 15.84, 20, 22.9, 18.3, 18.9, 17.4, 17.6,
18, 17.98, 17.82, 17.42, 19.47, 18.52, 19.9, 20.01, 16.87,
17.3, 15.41, 17.05, 18.9, 16.7, 16.9, 14.5, 15.5, 14.6, 18.6
), vs = c(0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0,
0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1), am = c(1,
1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1,
0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1), gear = c(4, 4, 4, 3,
3, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 4, 3, 3, 3,
3, 3, 4, 5, 5, 5, 5, 5, 4), carb = c("M_PP", "O_PP", "C_PP", "K_MM",
"T_MM", "C_MM", "R_PP", "E_PP", "W_PP", "Q_PP", "R_MM", "T_MM",
"V_MM", "Q_MM", "F_PP", "D_PP", "S_PP", "Z_PP", "K_PP", "G_PP", "F_MM",
"D_MM", "S_MM", "Z_MM", "K_MM", "F_MM", "A_PP", "D_PP", "T_PP",
"R_MM", "D_MM", "T_MM"), Name = c("Mark", "Mark", "Mark", "Mark",
"Mark", "Mark", "Tom", "Tom", "Tom", "Tom", "Tom", "Tom",
"Tom", "Tom", "Tim", "Tim", "Tim", "Tim", "Tim", "Tim", "Tim",
"Tim", "Tim", "Tim", "Tim", "Tim", "Greg", "Greg", "Greg",
"Greg", "Greg", "Greg")), .Names = c("mpg", "cyl", "disp",
"hp", "drat", "wt", "qsec", "vs", "am", "gear", "carb", "Name"
), row.names = c(NA, -32L), class = "data.frame")
您可以在下面看到一个可以与此数据框区分开的组：
mpg cyl disp hp drat wt qsec vs am gear carb Name
1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 M_PP Mark
2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 O_PP Mark
3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 C_PP Mark
4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 K_MM Mark
5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 T_MM Mark
6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 C_MM Mark
所以，我想在Mark {t.test列的PP和MM - 子组之间执行carb。我感兴趣的专栏是gear。我想知道齿轮数的差异对于这些小组来说具有统计学意义。
应该从这些数据中为Mark等所有组执行此类分析。
结果（pvalues）可以存储在附加列的同一数据框中。这意味着pvalues将在属于同一组的所有行中重复。
答案 0 :(得分：1)
我将使用cyl而不是gear，因为t.test在使用gear时会抛出“数据基本上不变”的错误。假设数据是名为d的对象。
我们分几步进行。
Name将数据框拆分为split(d, d$Name)。 group后缀创建一个新变量carb。我们在子集中为t.test执行了cyl。
D <- lapply(split(d, d$Name), function(x) {
x$group <- factor(gsub(".*(PP|MM)", "\\1", x$carb))
t.test(x$cyl[x$group=="PP"], x$cyl[x$group=="MM"])$p.value
})
到目前为止的输出：
D
# $Greg
# [1] 0.7250302
# $Mark
# [1] 0.2301996
# $Tim
# [1] 0.5995106
# $Tom
# [1] 0.1294094
我们将D重新整形为数据框，然后将其与原始数据框合并完成：
D <- data.frame(Name = names(D),
pvalue = unlist(D))
merge(d, D)
# Name mpg cyl disp hp drat wt qsec vs am gear carb pvalue
# 1 Greg 26.0 4 120.3 91 4.43 2.140 16.70 0 1 0.37495820 A_PP 0.7250302
# 2 Greg 30.4 4 95.1 113 3.77 1.513 16.90 1 1 -2.07140903 D_PP 0.7250302
# 3 Greg 15.8 8 351.0 264 4.22 3.170 14.50 0 1 -0.73900855 T_PP 0.7250302
# 4 Greg 19.7 6 145.0 175 3.62 2.770 15.50 0 1 -0.09174744 R_MM 0.7250302
# 5 Greg 15.0 8 301.0 335 3.54 3.570 14.60 0 1 -1.55889142 D_MM 0.7250302
# 6 Greg 21.4 4 121.0 109 4.11 2.780 18.60 1 1 0.78601261 T_MM 0.7250302
# 7 Mark 21.0 6 160.0 110 3.90 2.620 16.46 0 1 1.60209096 M_PP 0.2301996
# 8 Mark 21.0 6 160.0 110 3.90 2.875 17.02 0 1 0.25393125 O_PP 0.2301996
# 9 Mark 22.8 4 108.0 93 3.85 2.320 18.61 1 1 -1.14837484 C_PP 0.2301996
# 10 Mark 21.4 6 258.0 110 3.08 3.215 19.44 1 0 0.68440881 K_MM 0.2301996
# 11 Mark 18.7 8 360.0 175 3.15 3.440 17.02 0 0 -1.04994050 T_MM 0.2301996
# 12 Mark 18.1 6 225.0 105 2.76 3.460 20.22 1 0 -2.18665934 C_MM 0.2301996
# ..snip..
答案 1 :(得分：1)
使用dplyr，
library(dplyr)
df %>%
group_by(Name) %>%
mutate(carb1 = gsub('.*_', '', carb), p_values = t.test(cyl[carb1 == 'PP'], cyl[carb1 == 'MM'])$p.value) %>%
select(-carb1)
#Source: local data frame [32 x 13]
#Groups: Name [4]
# mpg cyl disp hp drat wt qsec vs am gear carb Name p_values
# <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <chr> <dbl>
#1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 M_PP Mark 0.2301996
#2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 O_PP Mark 0.2301996
#3 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 C_PP Mark 0.2301996
#4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 K_MM Mark 0.2301996
#5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 T_MM Mark 0.2301996
#6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 C_MM Mark 0.2301996
#7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 R_PP Tom 0.1294094
#8 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 E_PP Tom 0.1294094
#9 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 W_PP Tom 0.1294094
#10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 Q_PP Tom 0.1294094
注意：我使用cyl作为gear会引发错误
错误：数据基本上是常数