根据其他列的最旧组成员的值重新编码整个组的列

Question

我想根据每个组中最老成员的另一列的值重新编码指示整个组的状态（x1 或 x2 = 3 或 0）的两列。

在下面的示例中，x1(x2) 是每个组内 key1(key2) 的总和（每人总是有三个值/插补）。但是，我只想为每个组设置 x1>0 或 x2>0。在那些有一个 key1=1 的人和一个 key2=1 的人（因此 x1=3 AND x2=3）的组中，应该由最年长的人决定。如果最年长的人有 key1=1 和 key2=0，就像在 A 组中，x1 应该是 3，x2 应该是整个组的 0，依此类推。

可重现的例子：

id <- c("A11", "A12", "A13", "A21", "A22", "A23", "B11", "B12", "B13", "C11", "C12", "C13", "C21", "C22", "C23", "D11", "D12", "D13", "D21", "D22", "D23", "E11", "E12", "E13", "E21", "E22", "E23")
group <- c("A","A","A","A","A","A","B","B","B","C","C","C","C","C","C","D","D","D","D","D","D","E","E","E","E","E","E")
imputation <- c(rep(1:3, 9))
age <- c(45,45,45,17,17,17,20,20,20,70,70,70,60,60,60,25,25,25,30,30,30,28,28,28,34,34,34)
key1 <- c(1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,1,1,1,0,0,0)
key2 <- c(0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0)
x1 <- c(3,3,3,3,3,3,0,0,0,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3)
x2 <- c(3,3,3,3,3,3,0,0,0,3,3,3,3,3,3,3,3,3,3,3,3,0,0,0,0,0,0)
test <- data.frame(id, group, imputation, age, key1, key2, x1, x2)

应重新编码 x1 和 x2 的子集：

 > test %>% group_by(group) %>% filter(x1==x2 & x1>0 | x1==x2 & x2>0)
 # A tibble: 18 x 8
 # Groups:   group [3]
   id    group imputation   age  key1  key2    x1    x2
   <fct> <fct>      <int> <dbl> <dbl> <dbl> <dbl> <dbl>
 1 A11   A              1    45     1     0     3     3
 2 A12   A              2    45     1     0     3     3
 3 A13   A              3    45     1     0     3     3
 4 A21   A              1    17     0     1     3     3
 5 A22   A              2    17     0     1     3     3
 6 A23   A              3    17     0     1     3     3
 7 C11   C              1    70     0     1     3     3
 8 C12   C              2    70     0     1     3     3
 9 C13   C              3    70     0     1     3     3
10 C21   C              1    60     1     0     3     3
11 C22   C              2    60     1     0     3     3
12 C23   C              3    60     1     0     3     3
13 D11   D              1    25     1     0     3     3
14 D12   D              2    25     1     0     3     3
15 D13   D              3    25     1     0     3     3
16 D21   D              1    30     0     1     3     3
17 D22   D              2    30     0     1     3     3
18 D23   D              3    30     0     1     3     3

输出应该是：

    id group imputation age key1 key2 x1 x2
1  A11     A          1  45    1    0  3  0
2  A12     A          2  45    1    0  3  0
3  A13     A          3  45    1    0  3  0
4  A21     A          1  17    0    1  3  0
5  A22     A          2  17    0    1  3  0
6  A23     A          3  17    0    1  3  0
7  C11     C          1  70    0    1  0  3
8  C12     C          2  70    0    1  0  3
9  C13     C          3  70    0    1  0  3
10 C21     C          1  60    1    0  0  3
11 C22     C          2  60    1    0  0  3
12 C23     C          3  60    1    0  0  3
13 D11     D          1  25    1    0  0  3
14 D12     D          2  25    1    0  0  3
15 D13     D          3  25    1    0  0  3
16 D21     D          1  30    0    1  0  3
17 D22     D          2  30    0    1  0  3
18 D23     D          3  30    0    1  0  3

我猜它可以通过 group_by、filter、mutate 和 ifelse 的组合来完成，但我还没有弄清楚。然而，重要的是它包含过滤器或类似的东西，因为x1==x2 & x1>0 | x1==x2 & x2>0 的观察只是我数据框的一个子集。

Answer 1

在每个group 中，您可以比较age 的unique 值（其中key1 为1）与unique 值age（其中key2 为1）并更新x1 和@987654329 @ 相应地：

id <- c("A11", "A12", "A13", "A21", "A22", "A23", "B11", "B12", "B13", "C11", "C12", "C13", "C21", "C22", "C23", "D11", "D12", "D13", "D21", "D22", "D23", "E11", "E12", "E13", "E21", "E22", "E23")
group <- c("A","A","A","A","A","A","B","B","B","C","C","C","C","C","C","D","D","D","D","D","D","E","E","E","E","E","E")
imputation <- c(rep(1:3, 9))
age <- c(45,45,45,17,17,17,20,20,20,70,70,70,60,60,60,25,25,25,30,30,30,28,28,28,34,34,34)
key1 <- c(1,1,1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,1,1,1,0,0,0)
key2 <- c(0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0)
x1 <- c(3,3,3,3,3,3,0,0,0,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3)
x2 <- c(3,3,3,3,3,3,0,0,0,3,3,3,3,3,3,3,3,3,3,3,3,0,0,0,0,0,0)
test <- data.frame(id, group, imputation, age, key1, key2, x1, x2)

library(dplyr)

test %>% 
  group_by(group) %>% 
  filter(x1==x2 & x1>0 | x1==x2 & x2>0) %>%
  mutate(x1 = ifelse(unique(age[key1==1]) > unique(age[key2==1]), 3, 0),
         x2 = ifelse(unique(age[key1==1]) > unique(age[key2==1]), 0, 3)) %>%
  ungroup()

# # A tibble: 18 x 8
#   id    group imputation   age  key1  key2    x1    x2
#   <fct> <fct>      <int> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 A11   A              1    45     1     0     3     0
# 2 A12   A              2    45     1     0     3     0
# 3 A13   A              3    45     1     0     3     0
# 4 A21   A              1    17     0     1     3     0
# 5 A22   A              2    17     0     1     3     0
# 6 A23   A              3    17     0     1     3     0
# 7 C11   C              1    70     0     1     0     3
# 8 C12   C              2    70     0     1     0     3
# 9 C13   C              3    70     0     1     0     3
#10 C21   C              1    60     1     0     0     3
#11 C22   C              2    60     1     0     0     3
#12 C23   C              3    60     1     0     0     3
#13 D11   D              1    25     1     0     0     3
#14 D12   D              2    25     1     0     0     3
#15 D13   D              3    25     1     0     0     3
#16 D21   D              1    30     0     1     0     3
#17 D22   D              2    30     0     1     0     3
#18 D23   D              3    30     0     1     0     3