我认为这个标题解释了一切。我想在两个数据集之间进行t.test。我想逐行比较。
让我们使用mtcars并略微修改mtcars_mod。
structure(list(mpg = c(21, 25, 22.8, 21.4, 18.7, 18.1, 14.3,
24.4, 24.8, 19.2, 17.8, 16.4, 17.3, 15.2, 10.4, 10.4, 14.7, 32.4,
36.4, 31.9, 21.5, 15.5, 15.2, 13.3, 19.2, 27.3, 26, 30.4, 15.8,
29.7, 15, 21.4), cyl = c(6, 6, 4, 6, 8, 6, 8, 4, 4, 6, 6, 8,
8, 8, 8, 8, 7, 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, 6, 460, 440, 78.7, 75.7,
71.1, 120.1, 318, 304, 350, 400, 79, 15, 97, 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, 7, 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, 6, 2.78), qsec = c(16.46, 17.02, 18.61,
114, 17.02, 20.22, 15.84, 12, 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,
32, 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, 1, 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(4, 4, 1, 1, 2, 1, 4,
2, 2, 4, 4, 3, 3, 3, 4, 4, 4, 1, 2, 1, 1, 2, 2, 4, 2, 1,
2, 2, 4, 6, 8, 2)), .Names = c("mpg", "cyl", "disp", "hp",
"drat", "wt", "qsec", "vs", "am", "gear", "carb"), row.names = c("Mazda RX4",
"Mazda RX4 Wag", "Datsun 710", "Hornet 4 Drive", "Hornet Sportabout",
"Valiant", "Duster 360", "Merc 240D", "Merc 230", "Merc 280",
"Merc 280C", "Merc 450SE", "Merc 450SL", "Merc 450SLC", "Cadillac Fleetwood",
"Lincoln Continental", "Chrysler Imperial", "Fiat 128", "Honda Civic",
"Toyota Corolla", "Toyota Corona", "Dodge Challenger", "AMC Javelin",
"Camaro Z28", "Pontiac Firebird", "Fiat X1-9", "Porsche 914-2",
"Lotus Europa", "Ford Pantera L", "Ferrari Dino", "Maserati Bora",
"Volvo 142E"), class = "data.frame"
我尝试在循环中执行此操作，但我不知道如何存储结果。我只得到最后一个值...
for(z in 1:nrow(mtcars)){
vec_1 <- mtcars[z,1:7]
vec_2 <- mtcars_mod[z,1:7]
vec_results <- unlist(t.test(vec_1, vec_2)[3])
}
有人可以告诉我如何纠正我的循环吗？我更喜欢使用apply函数，但仍想知道我的循环错误...
答案 0 :(得分：3)
（我只会使用我自己修改的mtcarsmod ...抱歉，你的至少缺少一个paren，而且 - 尽管我确切地知道发生了什么 - 它是丑陋的在那个SO窗口！）
set.seed(42)
mtcarsmod <- as.data.frame(lapply(mtcars, jitter, factor = 5))
head(mtcarsmod)
# mpg cyl disp hp drat wt qsec vs am gear carb
# 1 21.1 5.55 160 109.7 3.89 2.62 16.5 -0.373 0.221 3.68 3.861
# 2 21.1 6.74 160 110.0 3.90 2.88 17.0 0.641 1.080 3.06 3.788
# 3 22.8 2.02 108 93.5 3.86 2.32 18.6 0.614 1.142 4.73 0.284
# 4 21.5 7.33 258 110.2 3.08 3.21 19.4 0.371 0.238 3.46 0.560
# 5 18.7 6.03 360 175.3 3.15 3.44 17.0 -0.903 0.430 2.63 2.130
# 6 18.1 4.83 225 104.4 2.77 3.46 20.2 0.491 -0.753 2.77 1.870
您应该使用sapply或其中一个亲属来代替循环。
sapply(seq_len(nrow(mtcars)),
function(r) unlist(t.test(mtcars[r,1:7], mtcarsmod[r,1:7])[3]))
# p.value p.value p.value p.value p.value p.value p.value p.value p.value p.value p.value
# 0.998 0.998 0.992 0.996 0.998 0.995 0.999 1.000 0.999 0.998 0.995
# p.value p.value p.value p.value p.value p.value p.value p.value p.value p.value p.value
# 0.995 0.999 0.999 0.998 0.999 0.997 0.999 0.995 0.997 0.995 0.999
# p.value p.value p.value p.value p.value p.value p.value p.value p.value p.value
# 0.997 0.998 1.000 0.990 0.997 0.999 0.999 0.995 0.997 0.995
使用lapply的一个好处可能是使用了更多的测试结果。例如：
ret <- lapply(seq_len(nrow(mtcars)),
function(r) t.test(mtcars[r,1:7], mtcarsmod[r,1:7]))
str(head(ret, n = 2))
# List of 2
# $ :List of 9
# ..$ statistic : Named num 0.0024
# .. ..- attr(*, "names")= chr "t"
# ..$ parameter : Named num 12
# .. ..- attr(*, "names")= chr "df"
# ..$ p.value : num 0.998
# ..$ conf.int : atomic [1:2] -73.4 73.5
# .. ..- attr(*, "conf.level")= num 0.95
# ..$ estimate : Named num [1:2] 45.7 45.6
# .. ..- attr(*, "names")= chr [1:2] "mean of x" "mean of y"
# ..$ null.value : Named num 0
# .. ..- attr(*, "names")= chr "difference in means"
# ..$ alternative: chr "two.sided"
# ..$ method : chr "Welch Two Sample t-test"
# ..$ data.name : chr "mtcars[r, 1:7] and mtcarsmod[r, 1:7]"
# ..- attr(*, "class")= chr "htest"
# $ :List of 9
# ..$ statistic : Named num -0.00311
# .. ..- attr(*, "names")= chr "t"
# ..$ parameter : Named num 12
# .. ..- attr(*, "names")= chr "df"
# ..$ p.value : num 0.998
# ..$ conf.int : atomic [1:2] -73.4 73.2
# .. ..- attr(*, "conf.level")= num 0.95
# ..$ estimate : Named num [1:2] 45.8 45.9
# .. ..- attr(*, "names")= chr [1:2] "mean of x" "mean of y"
# ..$ null.value : Named num 0
# .. ..- attr(*, "names")= chr "difference in means"
# ..$ alternative: chr "two.sided"
# ..$ method : chr "Welch Two Sample t-test"
# ..$ data.name : chr "mtcars[r, 1:7] and mtcarsmod[r, 1:7]"
# ..- attr(*, "class")= chr "htest"
ret[[1]]$p.value
# [1] 0.998
你仍然可以从结果中轻松获得p值的向量：
sapply(ret, `[[`, "p.value")
# [1] 0.998 0.998 0.992 0.996 0.998 0.995 0.999 1.000 0.999 0.998 0.995 0.995 0.999 0.999
# [15] 0.998 0.999 0.997 0.999 0.995 0.997 0.995 0.999 0.997 0.998 1.000 0.990 0.997 0.999
# [29] 0.999 0.995 0.997 0.995