运行不同类别的多个卡方检验

Question

我有二进制数据取决于个人pass/failed是否测试，以及特征信息（例如性别）和他们属于哪个部门（例如x，y，z）在df（数据）中

head(data,9)
department  gender   pass 
x           Male     1               
y           Female   1             
y           Male     0         
y           Male     1              
x           Female   1              
z           Female   0            
z           Male     1
x           Male     0
z           Female   0

我可以轻松地运行对性别与通过之间的关系进行卡方检验：

chisq.test(data$gender, data$pass)

但是有没有一种方法可以将运行分别用于 'department' (x,y,z) 中的值，而不必每次都手动对数据进行子集化？

我可以使用 tapply 创建一个新的数据框来分解每个部门的总体通过率：

as.data.frame(tapply(data$pass, data$department,mean))

但是我有没有办法添加一个新变量来指示上述测试的结果（假设 p 值）？

Answer 1

使用 broom 和 dplyr 是一种优雅的方法。首先，我们按部门变量分组并嵌套我们的数据框。然后我们运行 chisq.test 对每个 "subset"。最后，为了获得相关统计数据（例如 p.value），我们利用 broom::tidy。由于这些都嵌套在每个子集中，我们取消嵌套我们最终想要看到的任何组件。

有关详细信息，请参阅 this vignette

library(tidyverse)
library(broom)

df <- data.frame(
  stringsAsFactors = FALSE,
        department = c("x", "y", "y", "y", "x", "z", "z", "x", "z"),
            gender = c("Male","Female","Male",
                       "Male","Female","Female","Male","Male","Female"),
              pass = c(1L, 1L, 0L, 1L, 1L, 0L, 1L, 0L, 0L)
)


df %>%
  group_by(department) %>%
  nest() %>% 
  mutate(
    chi_test = map(data, ~ chisq.test(.$gender, .$pass)),
    tidied = map(chi_test, tidy)
  ) %>% 
  unnest(tidied)

#> # A tibble: 3 x 7
#> # Groups:   department [3]
#>   department data      chi_test statistic p.value parameter method              
#>   <chr>      <list>    <list>       <dbl>   <dbl>     <int> <chr>               
#> 1 x          <tibble ~ <htest>   4.62e-32   1.00          1 Pearson's Chi-squar~
#> 2 y          <tibble ~ <htest>   4.62e-32   1.00          1 Pearson's Chi-squar~
#> 3 z          <tibble ~ <htest>   1.88e- 1   0.665         1 Pearson's Chi-squar~

^{由 reprex package (v0.3.0)}

于 2020-05-20 创建

如果你想使用基础 R，你可以利用 split 和 lapply 这样的东西：

lapply(split(df, df$department), function(x) { chisq.test(x$gender, x$pass)$p.value })

Answer 2

是的！使用 by.

res <- do.call(rbind, by(dat, dat$department, function(x) {
  c(M=unname(tapply(x$pass, x$department, mean)),
    p=chisq.test(x$gender, x$pass)$p.value)
}))
res
#           M            p
# x 0.6788732 1.484695e-18
# y 0.6516517 3.045009e-22
# z 0.3205128 7.945768e-69

数据：

dat <- read.table(text="department  gender   pass 
x           Male     1               
y           Female   1             
y           Male     0         
y           Male     1              
x           Female   1              
z           Female   0            
z           Male     1
x           Male     0
z           Female   0", header=T)
set.seed(42)
dat <- dat[sample(1:nrow(dat), 1000, replace=T), ]

Answer 3

不是完全不同的问题答案，而是如果您尝试回答不同问题的答案。 @JasonAizkalns 为每个部门提供了一个优雅的答案，但如果您有兴趣将部门相互比较，则需要针对多重比较进行调整。所以它可能看起来像这样。

library(dplyr)
library(rcompanion)

df <- data.frame(
  stringsAsFactors = FALSE,
  department = c("x", "y", "y", "y", "x", "z", "z", "x", "z"),
  gender = c("Male","Female","Male",
             "Male","Female","Female","Male","Male","Female"),
  pass = c(1L, 1L, 0L, 1L, 1L, 0L, 1L, 0L, 0L)
)

df %>%
  group_by(department, gender) %>%
  summarise(Freq = n()) %>%
  xtabs(formula = Freq ~ ., data = .) %>% 
  pairwiseNominalIndependence(x = ., method = "holm", gtest = FALSE)

#> Warning in chisq.test(Dataz, ...): Chi-squared approximation may be incorrect

#> Warning in chisq.test(Dataz, ...): Chi-squared approximation may be incorrect

#> Warning in chisq.test(Dataz, ...): Chi-squared approximation may be incorrect
#>   Comparison p.Fisher p.adj.Fisher p.Chisq p.adj.Chisq
#> 1      x : y        1            1       1           1
#> 2      x : z        1            1       1           1
#> 3      y : z        1            1       1           1

运行不同类别的多个卡方检验

Running multiple chi-squared tests for different categories

for-loop

r

chi-squared

运行 不同类别的多个卡方检验

Running multiple chi-squared tests for different categories

for-loop

r

chi-squared

运行不同类别的多个卡方检验