匹配成对评级的行数据

Question

我有一对数据框，其中包含自我评分（1,0）并评定其合作伙伴（1,0）的个人对。

have <- data.frame(group=c(1, 1, 2, 2, 3, 3),
                   person=c(1, 2, 1, 2, 1, 2),
                   rateSelf=c(1, 0, 1, 0, 1, 0),
                   rateOther=c(1, 1, 1, 0, 0, 0))
have

#  group person rateSelf rateOther
#1     1      1        1         1
#2     1      2        0         1
#3     2      1        1         1
#4     2      2        0         0
#5     3      1        1         0
#6     3      2        0         0

我想比较某人如何评价自己（rateSelf）对他们如何评价他人（ratedByOther）。

want <- data.frame(group=c(1, 1, 2, 2, 3, 3),
                   person=c(1, 2, 1, 2, 1, 2),
                   rateSelf=c(1, 0, 1, 0, 1, 0),
                   rateOther=c(1, 1, 1, 0, 0, 0),
                   ratedByOther=c(1, 1, 0, 1, 0, 0))
want

#  group person rateSelf rateOther ratedByOther
#1     1      1        1         1            1
#2     1      2        0         1            1
#3     2      1        1         1            0
#4     2      2        0         0            1
#5     3      1        1         0            0
#6     3      2        0         0            0

获得ratedByOther的明智之举是什么？

Answer 1

我认为这可以做你想要的：

library(dplyr)
want <- have %>% 
  group_by(group) %>% 
  mutate(ratedByOther = rev(rateOther))

  group person rateSelf rateOther ratedByOther
  <dbl>  <dbl>    <dbl>     <dbl>        <dbl>
1     1      1        1         1            1
2     1      2        0         1            1
3     2      1        1         1            0
4     2      2        0         0            1
5     3      1        1         0            0
6     3      2        0         0            0

Answer 2

我们可以使用data.table

library(data.table)
setDT(have)[, ratedByOther := rev(rateOther), group]