如何优雅地重新编码包含多个值的多个列

如何解决如何优雅地重新编码包含多个值的多个列

我有一个包含连续数据和分类数据的数据框。

df<- data.frame(gender=c("male","female","transgender"),education=c("high-school","grad-school","home-school"),smoke=c("yes","no","prefer not tell"))
> print(df)
       gender   education           smoke
1        male high-school             yes
2      female grad-school              no
3 transgender home-school prefer not tell
> str(df)
'data.frame':   3 obs. of  3 variables:
 $ gender   : chr  "male" "female" "transgender"
 $ education: chr  "high-school" "grad-school" "home-school"
 $ smoke    : chr  "yes" "no" "prefer not tell"

我正在尝试将分类列重新编码为名义格式。我当前的方法非常繁琐。 首先，我必须将所有字符变量转换为因子格式，

# Coerce all character formats to Factors
df<- data.frame(df[sapply(df,is.character)] <-
  lapply(df[sapply(df,is.character)],as.factor))

library(plyr)
df$gender<- revalue(df$gender,c("male"="1","female"="2","transgender"="3"))
df$education<- revalue(df$education,c("high-school"="1","grad-school"="2","home-school"="3"))
df$smoke<- revalue(df$smoke,c("yes"="1","no"="2","prefer not tell"="3"))
> print(df)
  gender education smoke
1      1         1     1
2      2         2     2
3      3         3     3

是否有更优雅的方法来解决此问题？遵循tidyverse风格的做法会有所帮助。我已经见过类似的问题，例如1，2，3。这些解决方案的问题是它们与我寻求的内容无关，或者它们使用诸如lapply()或sapply()之类的基本R方法，这对我来说很难解释。我也想知道是否有一种优雅的方法可以按照tidyverse方法将所有字符变量转换为因子格式。

解决方法

尝试一下。只需考虑一下我们正在使用mutate()和across()两次，以便首先将值转换为因数，然后按它们在每个变量（unique()）中的显示方式进行排序，然后将数字端与as.numeric()提取值。这里的代码：

library(tidyverse)
#Code
df %>% mutate(across(gender:smoke,~factor(.,levels = unique(.)))) %>%
  mutate(across(gender:smoke,~as.numeric(.)))

输出：

  gender education smoke
1      1         1     1
2      2         2     2
3      3         3     3

为了确定新值的分配方式，您可以使用以下方法：

#Code 2
df %>% summarise_all(.funs = unique) %>% pivot_longer(everything()) %>%
  arrange(name) %>%
  group_by(name) %>% mutate(Newval=1:n())

输出：

# A tibble: 9 x 3
# Groups:   name [3]
  name      value           Newval
  <chr>     <fct>            <int>
1 education high-school          1
2 education grad-school          2
3 education home-school          3
4 gender    male                 1
5 gender    female               2
6 gender    transgender          3
7 smoke     yes                  1
8 smoke     no                   2
9 smoke     prefer not tell      3

或者可能需要更多控制：

#Code 3
df %>% mutate(id=1:n()) %>% pivot_longer(-id) %>%
  left_join(df %>% summarise_all(.funs = unique) %>% pivot_longer(everything()) %>%
              arrange(name) %>%
              group_by(name) %>% mutate(Newval=1:n()) %>% ungroup()) %>%
  select(-value) %>%
  pivot_wider(names_from = name,values_from=Newval) %>%
  select(-id)

输出：

# A tibble: 3 x 3
  gender education smoke
   <int>     <int> <int>
1      1         1     1
2      2         2     2
3      3         3     3

如果您的变量属于character类，则可以使用此管道从字符转换为因数，然后重新组织因数，然后使其成为数字：

#Code 4
df %>% 
  mutate(across(gender:smoke,~as.factor(.))) %>%
  mutate(across(gender:smoke,~as.numeric(.)))

输出：

  gender education smoke
1      1         1     1
2      2         2     2
3      3         3     3

,

您可以将数据中的“字符和因子”列转换为数字，从而根据每个级别在数据中的出现为每个级别赋予唯一值。

library(dplyr)

df %>% 
  mutate(across(where(~is.character(.) | is.factor(.)),~match(.,unique(.))))

#  gender education smoke
#1      1         1     1
#2      2         2     2
#3      3         3     3

,

Base R解决方案：

lapply(df,function(x){
    if(is.character(x) | is.factor(x)){
      x <- as.integer(labels(as.factor(x)))
    }else{
      x
    }
  }
)