Question

我有一个配对数据的数据集（同一家庭的成员）。

ID 是个人标识符， householdid 是合作伙伴的标识符（反之亦然）。

我需要为其合作伙伴的每个 id 添加一个额外的列（职业）。

我的数据看起来像这样

dta = rbind( c(1013661,101366, 'Never worked'), 
   c(1013662, 101366, 'Intermediate occs'), 
   c(1037552, 103755, 'Managerial & professional occs'), 
   c(1037551, 103755, 'Intermediate occs')
)

colnames(dta) = c('idno', 'householdid', 'occup')
dta

 idno      householdid occup                           
"1013661" "101366"    "Never worked"                  
"1013662" "101366"    "Intermediate occs"             
"1037552" "103755"    "Managerial & professional occs"
"1037551" "103755"    "Intermediate occs"

我需要的应该是这样的

 idno      householdid occup                            occupPartner                    
"1013661" "101366"    "Never worked"                   "Intermediate occs"             
"1013662" "101366"    "Intermediate occs"              "Never worked"                  
"1037552" "103755"    "Managerial & professional occs" "Intermediate occs"             
"1037551" "103755"    "Intermediate occs"              "Managerial & professional occs"

我猜有一个mutate的解决方案，但我不确定group_by应该是什么。

有什么想法吗？

Answer 1

尝试

library(dplyr)
dta1 <-  as.data.frame(dta) %>% 
          group_by(householdid) %>% 
          mutate(occupPartner= rev(occup)) 
as.data.frame(dta1)
#     idno householdid                          occup
#1 1013661      101366                   Never worked
#2 1013662      101366              Intermediate occs
#3 1037552      103755 Managerial & professional occs
#4 1037551      103755              Intermediate occs
#                 occupPartner
#1              Intermediate occs
#2                   Never worked
#3              Intermediate occs
#4 Managerial & professional occs

如果数据已经订购，

 indx <- c(rbind(seq(2, nrow(dta), by=2), seq(1, nrow(dta), by=2)))
 cbind(dta, occupPartner=dta[,3][indx])

Answer 2

使用data.table

的另一个选项

library(data.table)
out = as.data.table(dta)[, occupPartner := rev(occup), by = householdid]

#> out
#      idno householdid                          occup
#1: 1013661      101366                   Never worked
#2: 1013662      101366              Intermediate occs
#3: 1037552      103755 Managerial & professional occs
#4: 1037551      103755              Intermediate occs
#                     occupPartner
#1:              Intermediate occs
#2:                   Never worked
#3:              Intermediate occs
#4: Managerial & professional occs

基于配对数据的额外列（变异）

2 个答案: