我想使用dplyr
创建7个虚拟变量 - 每天一个到目前为止，我已经设法使用sjmisc包和to_dummy函数来完成它，但是我分两步执行-1。创建一个df的dummies，2）追加到原来的df
#Sample dataframe
mydfdata.frame(x=rep(letters[1:9]),
day=c("Mon","Tues","Wed","Thurs","Fri","Sat","Sun","Fri","Mon"))
#1.Create the 7 dummy variables separately
daysdummy<-sjmisc::to_dummy(mydf$day,suffix="label")
#2. append to dataframe
mydf<-bind_cols(mydf,daysdummy)
> mydf
x day day_Fri day_Mon day_Sat day_Sun day_Thurs day_Tues day_Wed
1 a Mon 0 1 0 0 0 0 0
2 b Tues 0 0 0 0 0 1 0
3 c Wed 0 0 0 0 0 0 1
4 d Thurs 0 0 0 0 1 0 0
5 e Fri 1 0 0 0 0 0 0
6 f Sat 0 0 1 0 0 0 0
7 g Sun 0 0 0 1 0 0 0
8 h Fri 1 0 0 0 0 0 0
9 i Mon 0 1 0 0 0 0 0
我的问题是，我是否可以使用dplyr在单个工作流程中执行此操作，并将to_dummy添加到管道工作流程中 - 可能使用mutate？
* to_dummy documentation
答案 0 :(得分：5)
如果您想对管道执行此操作，您可以执行以下操作：
library(dplyr)
library(sjmisc)
mydf %>%
to_dummy(day, suffix = "label") %>%
bind_cols(mydf) %>%
select(x, day, everything())
返回：
# A tibble: 9 x 9 x day day_Fri day_Mon day_Sat day_Sun day_Thurs day_Tues day_Wed <fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> 1 a Mon 0. 1. 0. 0. 0. 0. 0. 2 b Tues 0. 0. 0. 0. 0. 1. 0. 3 c Wed 0. 0. 0. 0. 0. 0. 1. 4 d Thurs 0. 0. 0. 0. 1. 0. 0. 5 e Fri 1. 0. 0. 0. 0. 0. 0. 6 f Sat 0. 0. 1. 0. 0. 0. 0. 7 g Sun 0. 0. 0. 1. 0. 0. 0. 8 h Fri 1. 0. 0. 0. 0. 0. 0. 9 i Mon 0. 1. 0. 0. 0. 0. 0.
我们可以使用dplyr和tidyr：
library(dplyr)
library(tidyr)
mydf %>%
mutate(var = 1) %>%
spread(day, var, fill = 0, sep = "_") %>%
left_join(mydf) %>%
select(x, day, everything())
使用基数R我们可以做类似的事情：
as.data.frame.matrix(table(rep(mydf$x, lengths(mydf$day)), unlist(mydf$day)))
返回：
Fri Mon Sat Sun Thurs Tues Wed a 0 1 0 0 0 0 0 b 0 0 0 0 0 1 0 c 0 0 0 0 0 0 1 d 0 0 0 0 1 0 0 e 1 0 0 0 0 0 0 f 0 0 1 0 0 0 0 g 0 0 0 1 0 0 0 h 1 0 0 0 0 0 0 i 0 1 0 0 0 0 0
答案 1 :(得分：0)
使用dummies()的替代解决方案，我认为会更快
mydf = data.frame(x=rep(letters[1:9]),
day=c("Mon","Tues","Wed","Thurs","Fri","Sat","Sun","Fri","Mon"))
library(dummies)
mydf <- cbind(mydf, dummy(mydf$day, sep = "_"))
产生
x day mydf_Fri mydf_Mon mydf_Sat mydf_Sun mydf_Thurs mydf_Tues mydf_Wed
1 a Mon 0 1 0 0 0 0 0
2 b Tues 0 0 0 0 0 1 0
3 c Wed 0 0 0 0 0 0 1
4 d Thurs 0 0 0 0 1 0 0
5 e Fri 1 0 0 0 0 0 0
6 f Sat 0 0 1 0 0 0 0
7 g Sun 0 0 0 1 0 0 0
8 h Fri 1 0 0 0 0 0 0
9 i Mon 0 1 0 0 0 0 0
然后您可以使用gsub()来获得更清晰的名称
names(mydf) = gsub("mydf_", "", names(mydf))
head(mydf)
x day Fri Mon Sat Sun Thurs Tues Wed
1 a Mon 0 1 0 0 0 0 0
2 b Tues 0 0 0 0 0 1 0
3 c Wed 0 0 0 0 0 0 1
4 d Thurs 0 0 0 0 1 0 0
5 e Fri 1 0 0 0 0 0 0
6 f Sat 0 0 1 0 0 0 0
答案 2 :(得分：0)
您也可以使用基础R sjmisc::to_dummy代替model.matrix;一个dplyr解决方案是：
library(dplyr);
model.matrix(~ 0 + day, mydf) %>%
as.data.frame() %>%
bind_cols(mydf) %>%
select(x, day, everything());
# x day dayFri dayMon daySat daySun dayThurs dayTues dayWed
#1 a Mon 0 1 0 0 0 0 0
#2 b Tues 0 0 0 0 0 1 0
#3 c Wed 0 0 0 0 0 0 1
#4 d Thurs 0 0 0 0 1 0 0
#5 e Fri 1 0 0 0 0 0 0
#6 f Sat 0 0 1 0 0 0 0
#7 g Sun 0 0 0 1 0 0 0
#8 h Fri 1 0 0 0 0 0 0
#9 i Mon 0 1 0 0 0 0 0