我遇到了对存储在数据框中的某些数据运行t检验的问题。我知道怎么一个接一个地做，但根本没有效率。请问如何写一个循环来做呢？
例如，我在testData中有数据：
testData <- dput(testData)
structure(list(Label = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L
), .Label = c("Bad", "Good"), class = "factor"), F1 = c(0.647789237,
0.546087915, 0.461342005, 0.794212207, 0.569199511, 0.735685704,
0.650942066, 0.457497016, 0.808619288, 0.673100668, 0.68781739,
0.470094549, 0.958591821, 1, 0.46908343, 0.578755283, 0.289380462,
0.685117658, 0.296011479, 0.208821225, 0.461487258, 0.176144907,
0.325684001), F2 = c(0.634327378, 0.602685034, 0.70643658, 0.577336318,
0.61069332, 0.676176013, 0.685433524, 0.601847779, 0.641738937,
0.822097452, 0.549508092, 0.711380436, 0.605492874, 0.419354439,
0.654424433, 0.782191133, 0.826394651, 0.63269692, 0.835389099,
0.760279322, 0.711607982, 1, 0.858631893), F3 = c(0.881115444,
0.850553659, 0.855405201, 0.732706141, 0.816063806, 0.841134018,
0.899594853, 0.788591779, 0.767461265, 0.954481259, 0.840970764,
0.897785959, 0.789288481, 0.604922471, 0.865024811, 0.947356946,
0.96622214, 0.879623595, 0.953189022, 0.960153373, 0.868949632,
1, 0.945716439), F4 = c(0.96939781, 0.758302, 0.652984943, 0.803719964,
0.980135127, 0.945287339, 0.84045753, 0.926053105, 0.974856922,
0.829936068, 0.89662815, 0.823594767, 1, 0.886954348, 0.825638185,
0.798524271, 0.524755093, 0.844685467, 0.522120663, 0.388604114,
0.725126521, 0.46430556, 0.604943457), F5 = c(0.908895247, 0.614799496,
0.529111461, 0.726753028, 0.942601677, 0.86641298, 0.75771251,
0.88237302, 1, 0.817706498, 0.834060845, 0.813550164, 0.927107922,
0.827680764, 0.797814872, 0.768118872, 0.271122929, 0.790632558,
0.391325631, 0.257446927, 0.687042673, 0.239520504, 0.521753545
), F6 = c(0.589651031, 0.170481902, 0.137755423, 0.24453692,
0.505348067, 0.642589538, 0.308854104, 0.286913756, 0.60756673,
0.531315171, 0.389958915, 0.236113471, 1, 0.687877983, 0.305962183,
0.40469629, 0.08012222, 0.376774451, 0.098261016, 0.046544022,
0.201513755, 0.02085411, 0.113698232), F7 = c(0.460358642, 0.629499543,
0.598616653, 0.623674078, 0.526920757, 0.494086383, 0.504021253,
0.635105287, 0.558992452, 0.397770725, 0.543528957, 0.538542617,
0.646897446, 0.543646493, 0.47463817, 0.385081029, 0.555731206,
0.43769237, 0.501754893, 0.586155312, 0.496028109, 1, 0.522921361
), F8 = c(0.523850222, 0.448936418, 0.339311791, 0.487421437,
0.462073661, 0.493421514, 0.464091025, 0.496938844, 0.5817454,
0.474404602, 0.720114482, 0.493098785, 1, 0.528538582, 0.478233718,
0.2695123, 0.362377901, 0.462252858, 0.287725327, 0.335584366,
0.397324649, 0.469082387, 0.403397835), F9 = c(0.481230473, 0.349419856,
0.309729777, 0.410783763, 0.465172146, 0.520935471, 0.380916463,
0.422238573, 0.572283353, 0.434705384, 0.512705279, 0.358892539,
1, 0.606926979, 0.370574926, 0.319739889, 0.249984729, 0.381053882,
0.245597953, 0.22883148, 0.314061676, 0.233511631, 0.269890359
), F10 = c(0.592403628, 0.249811036, 0.256613757, 0.305839002,
0.497637944, 0.601946334, 0.401643991, 0.302626606, 0.623582766,
0.706254724, 0.435846561, 0.324357521, 1, 0.740362812, 0.402588813,
0.537414966, 0.216458806, 0.464852608, 0.251228269, 0.181500378,
0.31840514, 0.068594104, 0.253873772), F11 = c(0.490032261, 0.366486136,
0.336749996, 0.421899324, 0.479339762, 0.527364467, 0.398297911,
0.432190187, 0.584030586, 0.453666402, 0.526861753, 0.388880674,
1, 0.615835576, 0.39058525, 0.350811433, 0.290220147, 0.397424867,
0.288095106, 0.274852912, 0.340129804, 0.271099396, 0.305499273
)), .Names = c("Label", "F1", "F2", "F3", "F4", "F5", "F6", "F7",
"F8", "F9", "F10", "F11"), class = "data.frame", row.names = c(NA,
-23L))
我需要为两个独立的组运行每个列的t检验，即“好”与“坏”的几个功能“F1”到“F11”。我试着做类似的事情：
GoodF1 <- subset(testData, Label == 'Good', select=c("F1"))
BadF1 <- subset(testData, Label == 'Bad', select=c("F1"))
t.test(GoodF1$F1,BadF1$F1)
然后将“F2”的其余部分改为“F11”，但显然效率不高。如果你有更好的想法在循环中运行它，我真的很感激。非常感谢。
答案 0 :(得分：5)
这是一个简单的解决方案，不需要额外的包：
lapply(testData[-1], function(x) t.test(x ~ testData$Label))
此处testData[-1]引用testData的所有列，但第一列（包含标签）。负索引用于排除数据。
答案 1 :(得分：4)
您可以使用公式接口t.test并使用lapply迭代列名来构建公式：
lapply(names(testData)[-1],function(x)
t.test(as.formula(paste(x,"Label",sep="~")),
data=testData))
[[1]]
Welch Two Sample t-test
data: F1 by Label
t = -3.6391, df = 13.969, p-value = 0.002691
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.4519374 -0.1167204
sample estimates:
mean in group Bad mean in group Good
0.3776753 0.6620042
[[2]]
Welch Two Sample t-test
data: F2 by Label
t = 3.7358, df = 12.121, p-value = 0.002796
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.06997617 0.26529723
sample estimates:
mean in group Bad mean in group Good
0.8008989 0.6332622
...
答案 2 :(得分：3)
我使用reshape2将数据放入长格式，然后我在lapply内使用您的代码。
library(reshape2)
dat <- melt(testData)
lapply(unique(dat$variable),function(x){
Good <- subset(dat, Label == 'Good' & variable ==x)$value
Bad <- subset(dat, Label == 'Bad' & variable ==x)$value
t.test(Good,Bad)
})
答案 3 :(得分：2)
使用lapply包中的ldply或plyr之类的内容非常简单：
library(plyr)
cols_to_test <- c("F1", "F2", "F3")
results <- ldply(
cols_to_test,
function(colname) {
t_val = t.test(testData[[colname]] ~ testData$Label)$statistic
return(data.frame(colname=colname, t_value=t_val))
})
将结果整齐地打包在数据框中：
colname t_value
1 F1 -3.639136
2 F2 3.735834
3 F3 4.303688
答案 4 :(得分：1)
这似乎是一个很常见的场景，而且我自己一直在努力。为此目的，现在有一个名为matrixTests的包。因此，您可以在每个列上执行 t-test ，如下所示：
library(matrixTests)
goodMat <- testData[testData[,1]=="Good",-1]
badMat <- testData[testData[,1]=="Bad",-1]
result <- col_t_welch(goodMat, badMat)
每列的结果将按行显示：
> result
obs.x obs.y obs.tot mean.x mean.y mean.diff var.x var.y stderr df statistic pvalue conf.low conf.high alternative mean.null conf.level
F1 15 8 23 0.6620042 0.3776753 0.2843289 0.030422051 0.032610380 0.07813088 13.969401 3.639136 0.0026907550 0.11672039 0.45193741 two.sided 0 0.95
F2 15 8 23 0.6332622 0.8008989 -0.1676367 0.007950091 0.011868380 0.04487264 12.121463 -3.735834 0.0027964901 -0.26529723 -0.06997617 two.sided 0 0.95
F3 15 8 23 0.8256733 0.9401514 -0.1144781 0.006957733 0.001949685 0.02659999 20.988353 -4.303688 0.0003146478 -0.16979764 -0.05915849 two.sided 0 0.95
F4 15 8 23 0.8742631 0.6091331 0.2651299 0.009285928 0.027017832 0.06321622 9.639523 4.194017 0.0020007742 0.12355816 0.40670172 two.sided 0 0.95
F5 15 8 23 0.8164387 0.4908705 0.3255682 0.015196701 0.054574685 0.08851525 9.132819 3.678104 0.0049648236 0.12577586 0.52536063 two.sided 0 0.95
F6 15 8 23 0.4429950 0.1678080 0.2751870 0.055993325 0.021810423 0.08036909 20.281178 3.424040 0.0026474215 0.10768889 0.44268512 two.sided 0 0.95
F7 15 8 23 0.5450866 0.5606705 -0.0155839 0.005238405 0.035530607 0.06921382 8.119018 -0.225156 0.8274218437 -0.17478492 0.14361711 two.sided 0 0.95
F8 15 8 23 0.5328120 0.3734072 0.1594048 0.023064998 0.005458074 0.04711609 20.936316 3.383236 0.0028151348 0.06140341 0.25740625 two.sided 0 0.95
F9 15 8 23 0.4797677 0.2803339 0.1994337 0.027905214 0.002845209 0.04707440 18.511452 4.236565 0.0004696924 0.10072958 0.29813785 two.sided 0 0.95
F10 15 8 23 0.4961010 0.2865410 0.2095600 0.045493711 0.023072590 0.07692196 18.972832 2.724320 0.0134746988 0.04854491 0.37057514 two.sided 0 0.95
F11 15 8 23 0.4941480 0.3147666 0.1793814 0.025996108 0.001953517 0.04446643 17.527205 4.034086 0.0008157456 0.08577994 0.27298287 two.sided 0 0.95
对于p值，有一个名为 pvalue 的列：
> result$pvalue
[1] 0.0026907550 0.0027964901 0.0003146478 0.0020007742 0.0049648236 0.0026474215 0.8274218437 0.0028151348 0.0004696924 0.0134746988 0.0008157456