在dplyr中突变虚拟变量

Question

我想使用dplyr创建7个虚拟变量-每天创建一个

到目前为止，我已经设法使用sjmisc包和to_dummy函数来完成此操作，但是我分两个步骤进行了操作-1。创建虚拟变量的df，2）附加到原始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，是否可以在单个工作流程中使用dplyr并将其添加to_dummy到管道工作流程中（也许使用）mutate？

* to_dummy 文档

Answer 1

如果要使用管道执行此操作，则可以执行以下操作：

library(dplyr)
library(sjmisc)

mydf %>% 
  to_dummy(day, suffix = "label") %>% 
  bind_cols(mydf) %>% 
  select(x, day, everything())

返回值：

# 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.

通过dplyr和tidyr我们可以做到：

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，我们可以执行以下操作：

as.data.frame.matrix(table(rep(mydf$x, lengths(mydf$day)), unlist(mydf$day)))

返回值：

  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