Question

我正在使用R进行数据分析，但是在编码方面存在一些问题。我创建了自己的用于创建频率表的函数，并将其应用于数据中的变量，但是R显示错误消息。

任何人都可以给我任何解决方案，为什么它不起作用？

> str(diabetes)
Classes ‘tbl_df’, ‘tbl’ and 'data.frame':   56632 obs. of  30 variables:
 $ ID       : chr  "A308059801" "A308059802" "A308120201" "A308120202" ...
 $ year     : num  2010 2010 2010 2010 2010 2010 2010 2010 2010 2010 ...
 $ region   : num  1 1 1 1 1 1 1 1 1 1 ...
 $ sex      : num  1 2 1 2 2 1 2 1 2 1 ...
 $ age      : num  61 54 33 33 4 65 59 54 49 18 ...
 $ edu      : chr  "3.000000" "2.000000" "3.000000" "4.000000" ...
 $ occp     : chr  "5.000000" "3.000000" "4.000000" "1.000000" ...
 $ marri_1  : 'labelled' num  1 1 1 1 2 1 1 1 1 2 ...
  ..- attr(*, "label")= chr "Marriage Y/N"
 $ marri_2  : 'labelled' num  1 1 1 1 8 1 1 1 1 8 ...
  ..- attr(*, "label")= chr "Marriage status"
 $ tins     : 'labelled' num  10 20 10 10 10 20 20 10 10 10 ...
  ..- attr(*, "label")= chr "Insurance registration"
 $ D_1_1    : 'labelled' chr  "3.000000" "2.000000" "2.000000" "3.000000" ...
  ..- attr(*, "label")= chr "Self-report health status"
 $ DI1_dg   : 'labelled' chr  "1.000000" "8.000000" "8.000000" "8.000000" ...
  ..- attr(*, "label")= chr "HBP diagnosis"
 $ DI1_pr   : 'labelled' chr  "1.000000" "8.000000" "8.000000" "8.000000" ...
  ..- attr(*, "label")= chr "HBP current status"
 $ DI1_pt   : 'labelled' chr  "1.000000" "8.000000" "8.000000" "8.000000" ...
  ..- attr(*, "label")= chr "HBP care"
 $ DE1_dg   : 'labelled' chr  "8.000000" "8.000000" "8.000000" "8.000000" ...
  ..- attr(*, "label")= chr "Diabetes diagnosis"
 $ DE1_pr   : 'labelled' chr  "8.000000" "8.000000" "8.000000" "8.000000" ...
  ..- attr(*, "label")= chr "Diabetes status"
 $ DE1_pt   : 'labelled' chr  "8.000000" "8.000000" "8.000000" "8.000000" ...
  ..- attr(*, "label")= chr "Diabetes cure"
 $ HE_DMdg  : 'labelled' chr  "0.000000" "0.000000" "0.000000" "0.000000" ...
  ..- attr(*, "label")= chr "Diabetes doctor diagnosis"
 $ HE_BMI   : 'labelled' chr  "26.177198" "22.807647" "26.562865" "20.863743" ...
  ..- attr(*, "label")= chr "BMI"
 $ HE_DM    : 'labelled' chr  "2.000000" "3.000000" "1.000000" "1.000000" ...
  ..- attr(*, "label")= chr "With diagnosis(over 19 year-old)"
 $ LQ4_07   : 'labelled' chr  "8.000000" "8.000000" "8.000000" "8.000000" ...
  ..- attr(*, "label")= chr "Barries for physical activity - diabetes"
 $ HE_DMfh1 : 'labelled' chr  "0.000000" "0.000000" "9.000000" "1.000000" ...
  ..- attr(*, "label")= chr "Father with diagnosis"
 $ HE_DMfh2 : 'labelled' chr  "1.000000" "0.000000" "9.000000" "0.000000" ...
  ..- attr(*, "label")= chr "Mother with diagnosis"
 $ HE_DMfh3 : 'labelled' chr  "0.000000" "0.000000" "9.000000" "0.000000" ...
  ..- attr(*, "label")= chr "Sibling with diagnosis"
 $ HE_glu   : 'labelled' chr  "124.000000" "141.000000" "92.000000" "88.000000" ...
  ..- attr(*, "label")= chr "Diabetes indicator - glucose level"
 $ BE5_1    : 'labelled' chr  "1.000000" "1.000000" "1.000000" "1.000000" ...
  ..- attr(*, "label")= chr "Muscle training frequency"
 $ LQ4_04   : 'labelled' chr  "8.000000" "8.000000" "8.000000" "8.000000" ...
  ..- attr(*, "label")= chr "Barriers for physical activity - Have heart disease"
 $ DF2_dg   : 'labelled' chr  "8.000000" "8.000000" "8.000000" "8.000000" ...
  ..- attr(*, "label")= chr "Diagnosed with depression"
 $ HE_IHDfh1: 'labelled' chr  "0.000000" "0.000000" "9.000000" "0.000000" ...
  ..- attr(*, "label")= chr "Diagnosed with Ischaemic heart disease"
 $ HE_HP    : 'labelled' chr  "3.000000" "3.000000" "2.000000" "1.000000" ...
  ..- attr(*, "label")= chr "Hypertension Status (three levels)"


freq_table <- function (y) {
  d <-   select (y) %>% group_by (y) %>% summarise (n = n ()) %>% mutate (freq = n / sum (n))
}

lapply(diabetes$marri_1, freq_table)

Answer 1

select函数位于管道的开头，并且至少需要两个参数，您可以将数据框的名称添加到参数函数中另外，由于y存储在变量中，因此在使用dplyr动词时，必须在其前面加上!!来取消对y的引用。

library(tidyverse)
# add df as an argument and add it before the select
freq_table <- function (df,y) {
  d <-   df %>% select (!! y) %>% group_by (!! y) %>% summarise (n = n ()) %>% mutate (freq = n / sum (n))
}

freq_table(diabetes,"marri_1")

或者您可以通过更简单的方式完成

tab <- table(diabetes$marri_1)
tab <- as.data.frame(tab)
names(tab) <- c("marri_1","n")
tab$freq <- tab$n /sum(tab$n)

这是您要找的吗？

R：如何在我的数据上使用用户定义的函数

1 个答案: