1

尝试从 data.frame 执行 ttest（并获取 p.value），有一列包含组（好与坏），其余列是数字。

我在这里生成了一个玩具数据集：

W <- rep(letters[seq( from = 1, to = 2)], 25)
X <- rnorm(n=50, mean = 10, sd = 5)
Y <- rnorm(n=50, mean = 15, sd = 6)
Z <- rnorm(n=50, mean = 20, sd = 5)
test_data <- data.frame(W, X, Y, Z)

然后我将数据转换为长格式：

melt_testdata <- melt(test_data)

并进行了 t.test

lapply(unique(melt_testdata$variable),function(x){
  Good <- subset(melt_testdata, W  == 'a' & variable ==x)$variable
  Bad <- subset(melt_testdata, W == 'b' & variable ==x)$variable
  t.test(Good,Bad)$p.value
})

但我没有得到 t.test 结果，而是收到以下错误消息：

Error in if (stderr < 10 * .Machine$double.eps * max(abs(mx), abs(my))) stop("data are essentially constant") : 
  missing value where TRUE/FALSE needed In addition: Warning messages:
1: In mean.default(x) : argument is not numeric or logical: returning NA
2: In var(x) :
  Calling var(x) on a factor x is deprecated and will become an error.
  Use something like 'all(duplicated(x)[-1L])' to test for a constant vector.
3: In mean.default(y) : argument is not numeric or logical: returning NA
4: In var(y) :
  Calling var(x) on a factor x is deprecated and will become an error.
  Use something like 'all(duplicated(x)[-1L])' to test for a constant vector.

然后我尝试编写循环（第一次..）

good <- matrix(,50)
bad <- matrix(,50)
cnt=3
out <- rep(0,cnt)


for (i in 2:4){
  good[i] <- subset(test_data, W == 'a', select= test_data[,i])
  bad[i] <- subset(test_data, W == 'b', select= test_data[,i])
  out[i] <- print(t.test(good[[i]], bad[[i]])$p.value)
}

仍然没有得到 p.values .......这是错误消息

Error in x[j] : only 0's may be mixed with negative subscripts

我感谢任何方法的帮助，谢谢！

4

2 回答 2

2

我认为formula使用t.test. 尝试

library(broom)
library(magrittr)
library(dplyr)

W <- rep(letters[seq( from = 1, to = 2)], 25)
X <- rnorm(n=50, mean = 10, sd = 5)
Y <- rnorm(n=50, mean = 15, sd = 6)
Z <- rnorm(n=50, mean = 20, sd = 5)
test_data <- data.frame(W, X, Y, Z)

lapply(test_data[c("X", "Y", "Z")],
       function(x, y) t.test(x ~ y),
       y = test_data[["W"]]) %>% 
  lapply(tidy) %>% 
  do.call("rbind", .) %>% 
  mutate(variable = rownames(.))

编辑：

更严格地遵守这一dplyr理念，您可以使用以下内容：实际上看起来更干净一些。

library(broom)
library(dplyr)
library(tidyr)

W <- rep(letters[seq( from = 1, to = 2)], 25)
X <- rnorm(n=50, mean = 10, sd = 5)
Y <- rnorm(n=50, mean = 15, sd = 6)
Z <- rnorm(n=50, mean = 20, sd = 5)

test_data <- data.frame(W, X, Y, Z) 

test_data %>% 
  gather(variable, value, X:Z) %>% 
  group_by(variable) %>% 
  do(., tidy(t.test(value ~ W, data = .)))
于 2017-08-30T20:13:00.330 回答
0

这是一个解决方案，使用dplyr和 的公式参数t.test。作用于由从输出中提取值do定义的每个组，并将它们变成.group_by. glancet.testdata.frame

library(tidyverse)
library(broom)

melt_testdata %>% 
  group_by(variable) %>% 
  do(glance(t.test(value ~ W, data = .)))
于 2017-08-30T20:17:38.460 回答