删除R中具有聚合组的重复项

以下是我的数据示例：

kod <- structure(list(ID_WORKES = c(28029571L, 28029571L, 28029571L, 
28029571L, 28029571L, 28029571L, 28029571L, 28029571L, 28029571L
), TABL_NOM = c(9716L, 9716L, 9716L, 9716L, 9716L, 9716L, 9716L, 
9716L, 9716L), NAME = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L), .Label = "Dim", class = "factor"), ID_SP_NAR = c(20L, 
20L, 20L, 30L, 30L, 30L, 30L, 30L, 30L), KOD_DOR = c(28L, 28L, 
28L, 28L, 28L, 28L, 28L, 28L, 28L), KOD_DEPO = c(9167L, 9167L, 
9167L, 9167L, 9167L, 9167L, 9167L, 9167L, 9167L), COLUMN_MASH = c(13L, 
13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L), prop_violations = c(0.00561797752808989, 
0.00293255131964809, 0.00495049504950495, 0.00215982721382289, 
0.0120481927710843, 0.00561797752808989, 0.00293255131964809, 
0.00591715976331361, 0.00495049504950495), mash_score = c(0L, 
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), row.names = c(NA, -9L), class = "data.frame")
W

但还有一件事：如果对于ID_SP_NAR的prop_违规中的一些重复值，mash_分数的值大于0，那么最后一个mash_分数的值大于0

比如说

  ID_WORKES TABL_NOM NAME ID_SP_NAR KOD_DOR KOD_DEPO COLUMN_MASH prop_violations mash_score
1  28029571     9716  Dim        30      28     9167          13          0,0056          0
2  28029571     9716  Dim        30      28     9167          13     0,012048193          0
3  28029571     9716  Dim        30      28     9167          13     0,005617978          0
4  28029571     9716  Dim        30      28     9167          13     0,002932551          1
5  28029571     9716  Dim        30      28     9167          13      0,00591716          0
6  28029571     9716  Dim        30      28     9167          13     0,004950495          0

在这种情况下，ID_SP_NAR=30的prop_违规只会留下值0002932551，导致mash_得分>0

---

如何达到此条件？

以下是使用

tidyverse

软件包的解决方案：

kod %>% 
  group_by(ID_WORKES, TABL_NOM, NAME, KOD_DOR, KOD_DEPO, ID_SP_NAR) %>%
  summarise(prop_violations = if (all(mash_score == 0)) mean(prop_violations) else last(prop_violations[mash_score > 0]))

---

如果特定组的所有

mash_得分

均等于零，则返回平均值（使用

mean

）。如果至少有一个

mash_得分

大于零，则返回

mash_得分>0

的

prop_违规

的最后一个值（使用

dplyr:：last

）。

使用

数据的选项。表

：

setDT(kod)
kod[, {
        if(any(mash_score)>0) {
            i <- which(mash_score>0)[1L]
            .(prop_violations=prop_violations[i], mash_score=mash_score[i])
        } else 
            .(prop_violations=mean(prop_violations), mash_score=mash_score[1L])
    }, 
    .(ID_WORKES, TABL_NOM, NAME, KOD_DOR, KOD_DEPO, ID_SP_NAR)]

数据：

kod
setDT(kod)
kod[, {
        if(any(mash_score)>0) {
            i <- which(mash_score>0)[1L]
            .(prop_violations=prop_violations[i], mash_score=mash_score[i])
        } else 
            .(prop_violations=mean(prop_violations), mash_score=mash_score[1L])
    }, 
    .(ID_WORKES, TABL_NOM, NAME, KOD_DOR, KOD_DEPO, ID_SP_NAR)]

   ID_WORKES TABL_NOM NAME KOD_DOR KOD_DEPO ID_SP_NAR prop_violations mash_score
1:  28029571     9716  Dim      28     9167        20     0.004500341          0
2:  28029571     9716  Dim      28     9167        30     0.002932551          1

kod <- structure(list(ID_WORKES = c(28029571L, 28029571L, 28029571L, 
    28029571L, 28029571L, 28029571L, 28029571L, 28029571L, 28029571L
), TABL_NOM = c(9716L, 9716L, 9716L, 9716L, 9716L, 9716L, 9716L, 
    9716L, 9716L), NAME = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
        1L, 1L), .Label = "Dim", class = "factor"), ID_SP_NAR = c(20L, 
            20L, 20L, 30L, 30L, 30L, 30L, 30L, 30L), KOD_DOR = c(28L, 28L, 
                28L, 28L, 28L, 28L, 28L, 28L, 28L), KOD_DEPO = c(9167L, 9167L, 
                    9167L, 9167L, 9167L, 9167L, 9167L, 9167L, 9167L), COLUMN_MASH = c(13L, 
                        13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L), prop_violations = c(0.00561797752808989, 
                            0.00293255131964809, 0.00495049504950495, 0.00215982721382289, 
                            0.0120481927710843, 0.00561797752808989, 0.00293255131964809, 
                            0.00591715976331361, 0.00495049504950495), mash_score = c(0L, 
                                0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L)), row.names = c(NA, -9L), class = "data.frame")