如何优雅地重新编码包含多个值的多个列
How to elegantly recode multiple columns containing multiple values
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。我目前的方法非常乏味。
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
有没有更优雅的方法来解决这个问题?类似于 tidyverse
风格的东西会有所帮助。我已经看到一些类似的问题,例如 1, ,3。这些解决方案的问题要么是它们与我寻求的内容无关,要么是它们使用基本的 R 方法,如 lapply()
或 sapply()
,这对我来说很难解释。我还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。
试试这个。只需考虑到我们使用 mutate()
和 across()
两次,以便首先将值转换为按它们在每个变量中的出现方式排序的因子 (unique()
),然后是数字端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:
#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
如果你的变量是 class character
你可以使用这个管道从字符转换为因子,然后重新组织因子,然后使它们成为数字:
#Code 4
df %>%
mutate(across(gender:smoke,~as.factor(.))) %>%
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
您可以将数据中的字符和因子列转换为数字,根据它们在数据中的出现为每个水平赋予唯一值。
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
基础 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
}
}
)
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。我目前的方法非常乏味。 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
有没有更优雅的方法来解决这个问题?类似于 tidyverse
风格的东西会有所帮助。我已经看到一些类似的问题,例如 1, lapply()
或 sapply()
,这对我来说很难解释。我还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。
试试这个。只需考虑到我们使用 mutate()
和 across()
两次,以便首先将值转换为按它们在每个变量中的出现方式排序的因子 (unique()
),然后是数字端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:
#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
如果你的变量是 class character
你可以使用这个管道从字符转换为因子,然后重新组织因子,然后使它们成为数字:
#Code 4
df %>%
mutate(across(gender:smoke,~as.factor(.))) %>%
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
您可以将数据中的字符和因子列转换为数字,根据它们在数据中的出现为每个水平赋予唯一值。
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
基础 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
}
}
)