Question

我有来自多个团队的团队成员如何评价的数据。每个人都有自己的身份证号码，但团队中也有团队和评分人员，如下所示：

  StudyID TeamID CATMERater Rated   Rating
    (int)  (int)      (int) (dbl)    (dbl)
1    2930    551          1     1 5.000000 #How rater 1 rated 1 (themselves)
2    2938    551          2     1 3.800000 #How rater 2 rated 1
3    2939    551          3     1 5.000000 #How rater 3 rated 1
4    2930    551          1     2 3.666667 #How rater 1 rated 2
5    2938    551          2     2 4.000000 #...
6    2939    551          3     2 3.866667
...

等等。我使用tidyr获取此格式，并且我正在尝试获取StudyID的新列，其中TeamID和被评级的人员是相同的。这是我尝试过的，但是不起作用，因为我不确定如何引用同一个表：

edges %>% mutate(RatedStudyID = filter(edges, TeamID == TeamID & Rated == CATMERater))

希望这是有道理的，但我很欣赏建议朝着正确的方向前进。如果是left_join，我该怎么说TeamID == TeamID？

这是我最后想要看到的内容（尽管大多数是最后一栏）：

  StudyID TeamID CATMERater Rated   Rating   RatedStudyID
    (int)  (int)      (int) (dbl)    (dbl)   
1    2930    551          1     1 5.000000   2930
2    2938    551          2     1 3.800000   2930
3    2939    551          3     1 5.000000   2930
4    2930    551          1     2 3.666667   2938
5    2938    551          2     2 4.000000   2938
6    2939    551          3     2 3.866667   2938
...

每个@akron的输出结果会产生错误：

structure(list(StudyID = c(2930L, 2938L, 2939L, 2930L, 2938L, 
2939L, 2930L, 2938L, 2939L, 2930L, 2938L, 2939L, 2930L, 2938L, 
2939L, 2930L, 2938L, 2939L, 2920L, 2941L, 2989L, 2920L, 2941L, 
2989L, 2920L, 2941L, 2989L, 2920L, 2941L, 2989L, 2920L, 2941L, 
2989L, 2920L, 2941L, 2989L, 2922L, 2924L, 2943L, 2922L, 2924L, 
2943L, 2922L, 2924L, 2943L, 2922L, 2924L, 2943L, 2922L, 2924L
), TeamID = c(551L, 551L, 551L, 551L, 551L, 551L, 551L, 551L, 
551L, 551L, 551L, 551L, 551L, 551L, 551L, 551L, 551L, 551L, 552L, 
552L, 552L, 552L, 552L, 552L, 552L, 552L, 552L, 552L, 552L, 552L, 
552L, 552L, 552L, 552L, 552L, 552L, 553L, 553L, 553L, 553L, 553L, 
553L, 553L, 553L, 553L, 553L, 553L, 553L, 553L, 553L), CATMERater = c(1L, 
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 
3L, 2L, 1L, 3L, 2L, 1L, 3L, 2L, 1L, 3L, 2L, 1L, 3L, 2L, 1L, 3L, 
2L, 1L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 
2L), Rated = c(1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 
6, 6, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 1, 
1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5), Rating = c(5, 3.8, 5, 
3.66666666666667, 4, 3.86666666666667, 4.53333333333333, 4, 4.8, 
NaN, NaN, NaN, NaN, NaN, NaN, NA, NA, NA, 3.93333333333333, 5, 
5, 5, 5, 5, 5, 5, 5, NaN, NaN, NaN, NaN, NaN, NaN, NA, NA, NA, 
4, 4, 4, 4, 4, 4, 4, 3.86666666666667, 4, NaN, NaN, NaN, NaN, 
NaN)), .Names = c("StudyID", "TeamID", "CATMERater", "Rated", 
"Rating"), class = c("tbl_df", "data.frame"), row.names = c(NA, 
-50L))

Answer 1

来自评论：

library(dplyr)
x %>%
   group_by(Rated, TeamID) %>% #group by each team/rated individual
   filter(any(CATMERater == Rated)) %>% #filter out any groups with unrated individuals
   mutate(new = StudyID[CATMERater == Rated]) #make the new column

通过对每个组进行子集化来创建新列 - 它与整个数据帧上的x$StudyID[x$CATMERater == x$Rated]相同。只要我们有一个地方，这是真的（即自我评级），该值将设置为该组的每个成员。

Answer 2

使用map() transformations are lazy so they don’t like to work until asked for. I was expecting messages in a map() transformation. put this expectation in rdd.take() which in non-lazy inside foreachRDD() It worked.

data.table

在新数据集中，有些组在同一行中没有CATMERater和Rated的任何类似值。因此，我们可以使用异常来返回NA。

library(data.table)
setDT(edges)[ , RatedStudyID := StudyID[CATMERater == Rated] , .(Rated, TeamID)]
edges
#   StudyID TeamID CATMERater Rated   Rating RatedStudyID
#1:    2930    551          1     1 5.000000         2930
#2:    2938    551          2     1 3.800000         2930
#3:    2939    551          3     1 5.000000         2930
#4:    2930    551          1     2 3.666667         2938
#5:    2938    551          2     2 4.000000         2938
#6:    2939    551          3     2 3.866667         2938

Answer 3

我认为您可以通过加入来解决此问题

edges %>%
  select(TeamID, Rated = CATMERater, RaterStudyID = StudyID) %>%
  inner_join(edges, by = c("TeamID", "Rated"))

dplyr从同一个表

3 个答案: