R-合并和融化列表到数据帧

Question

情景与数据

我有一个员工df_employees的数据框架

df_employees <- structure(list(empNo = c(1001, 1002, 1003)), .Names = "empNo", row.names = c(NA, 
  -3L), class = "data.frame")

> df_employees
  empNo
1  1001
2  1002
3  1003

和一张技能清单l_skills

l_skills <- list(c("skill1", "skill2", "skill3"), c("skill1", "skill2"), 
             "skill1")


> l_skills
[[1]]
[1] "skill1" "skill2" "skill3"

[[2]]
[1] "skill1" "skill2"

[[3]]
[1] "skill1"

问题

如何合并和融化数据，以给出结果数据df_result

> df_result
  empNo skills
1  1001 skill1
2  1001 skill2
3  1001 skill3
4  1002 skill1
5  1002 skill2
6  1003 skill1

尝试

我想我可以使用类似的方法来实现this cSplit function，但是当我试图安装cSplit时会出错。

> install.packages("cSplit")
Installing package into ‘C:/Users/<username>/Documents/R/win-library/3.1’
(as ‘lib’ is unspecified)
Warning in install.packages :
  package ‘cSplit’ is not available (for R version 3.1.2)

Answer 1

您可以使用来自melt包的reshape2函数：

library(reshape2)

L <- l_skills
names(L) <- df_employees$empNo

result <- melt(L)
colnames(result) <- c('skills','empNo')

result
#   skills empNo
# 1 skill1  1001
# 2 skill2  1001
# 3 skill3  1001
# 4 skill1  1002
# 5 skill2  1002
# 6 skill1  1003

R基解决方案：

do.call(rbind,mapply(cbind,df_employees$empNo,l_skills))
#       [,1]   [,2]    
#[1,] "1001" "skill1"
#[2,] "1001" "skill2"
#[3,] "1001" "skill3"
#[4,] "1002" "skill1"
#[5,] "1002" "skill2"
#[6,] "1003" "skill1"

Answer 2

您还可以使用来自stack的base R

 setNames(stack(setNames(L, df_employees$empNo)), c('skills', 'empNo'))
 #   skills empNo
 #1 skill1  1001
 #2 skill2  1001
 #3 skill3  1001
 #4 skill1  1002
 #5 skill2  1002
 #6 skill1  1003

或者使用splitstackshape

 library(splitstackshape)
 listCol_l(transform(df_employees, skills=I(L)), 'skills')[]
 #   empNo skills_ul
 #1:  1001    skill1
 #2:  1001    skill2
 #3:  1001    skill3
 #4:  1002    skill1
 #5:  1002    skill2
 #6:  1003    skill1

哪里

 L <- l_skills