创建从组中识别最小字符的列并标记关系

Question

我有 10 个主题的配对数据（有些缺失，有些并列）。我的目标是选择具有最佳disc_grade（A > B > C）的eye，并从下面的数据框中相应地标记关系。

我不知道如何使用 R 代码为每个主题选择具有最佳 disc_grade 的行。

df <- structure(list(patientID = c(1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 
6, 7, 7, 8, 8, 9, 9, 10, 10), eye = c("R", "L", "R", "L", "R", 
"L", "R", "L", "R", "L", "R", "L", "R", "L", "R", "L", "R", "L", 
"R", "L"), disc_grade = c(NA, "B", "C", "B", "B", "C", "B", "C", 
"B", "A", "B", "B", "C", "B", NA, NA, "B", "C", "B", "C")), .Names = c("patientID", "eye", "disc_grade"), class = c("tbl_df", "data.frame"), row.names = c(NA, -20L))

想要的输出是：

   patientID   eye disc_grade
2          1   L          B
4          2   L          B
5          3   R          B
7          4   R          B
10         5   L          A
11         6   Tie        B
14         7   L          B
17         9   R          B
19        10   R          B

Answer 1

这似乎有效：

df %>% 
  group_by(patientID) %>% 
  filter(disc_grade == min(disc_grade, na.rm=TRUE)) %>%
  summarise(eye = if (n()==1) eye else "Tie", disc_grade = first(disc_grade))

  patientID   eye disc_grade
      (dbl) (chr)      (chr)
1         1     L          B
2         2     L          B
3         3     R          B
4         4     R          B
5         5     L          A
6         6   Tie          B
7         7     L          B
8         9     R          B
9        10     R          B

第 8 组有一个警告，但由于filter 在NAs 上的工作方式，我们得到了预期的结果。

---

带data.table：

setDT(df)[, 
  .SD[ disc_grade == min(disc_grade, na.rm=TRUE) ][,
    .( eye = if (.N==1) eye else "Tie", disc_grade = disc_grade[1] )
  ]
, by=patientID]

同样，有一个警告，但现在我们确实得到了第 8 组的一行，因为 [ 不会忽略 NAs。为了解决这个问题，您可以在操作之前或之后过滤 NA（如其他答案所示）。我在主要操作期间执行此操作的最佳想法非常复杂：

setDT(df)[, 
  .SD[ which(disc_grade == min(disc_grade, na.rm=TRUE)) ][,
    if (.N >= 1) list( eye = if (.N==1) eye else "Tie", disc_grade = disc_grade[1] )
  ]
, by=patientID]

Answer 2

data.table 的一个选项

library(data.table)
na.omit(setDT(df))[, eye:=if(uniqueN(disc_grade)==1 & 
            .N >1)  'Tie'  else eye, patientID
         ][order(factor(disc_grade, levels=c('A', 'B', 'C'))),
              .SD[1L] ,patientID][order(patientID)]
#    patientID eye disc_grade
#1:         1   L          B
#2:         2   L          B
#3:         3   R          B
#4:         4   R          B
#5:         5   L          A
#6:         6 Tie          B
#7:         7   L          B
#8:         9   R          B
#9:        10   R          B

Answer 3

library(dplyr)

df <- structure(list(patientID = c(1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 
                                   6, 7, 7, 8, 8, 9, 9, 10, 10), eye = c("R", "L", "R", "L", "R", 
                                                                         "L", "R", "L", "R", "L", "R", "L", "R", "L", "R", "L", "R", "L", 
                                                                         "R", "L"), disc_grade = c(NA, "B", "C", "B", "B", "C", "B", "C", 
                                                                                                   "B", "A", "B", "B", "C", "B", NA, NA, "B", "C", "B", "C")), .Names = c("patientID", "eye", "disc_grade"), class = c("tbl_df", "data.frame"), row.names = c(NA, -20L))



df %>%
  filter(!is.na(disc_grade)) %>%                     ## remove rows with NAs
  group_by(patientID) %>%                            ## for each patient
  filter(disc_grade == min(disc_grade)) %>%          ## keep the row (his eye) that has the best score
  mutate(eye_upd = ifelse(n() > 1, "tie", eye)) %>%  ## if you kept both eyes you have a tie
  select(patientID,eye_upd,disc_grade) %>%
  distinct()

#    patientID eye_upd disc_grade
#        (dbl)   (chr)     (fctr)
# 1         1       L          B
# 2         2       L          B
# 3         3       R          B
# 4         4       R          B
# 5         5       L          A
# 6         6     tie          B
# 7         7       L          B
# 8         9       R          B
# 9        10       R          B

Answer 4

当然有更好的方法来做到这一点，但这可以完成工作......需要更多咖啡......

df_orig <- df

library(dplyr)

df %>%
  filter(!is.na(disc_grade)) %>%
  group_by(patientID) %>%
  summarise(best = min(disc_grade)) %>%
  left_join(., df_orig, by = c("patientID" = "patientID",
                               "best" = "disc_grade")) %>%
  group_by(patientID) %>%
  mutate(eye = ifelse(n() > 1, "tie", eye)) %>%
  distinct(patientID) %>% 
  select(patientID, eye, best)

注意：由于类型对话，我能够逃脱 min(disc_grade)。考虑看看as.numeric(as.factor(df$disc_grade))。