Question

我有一个不同类型的人（pid）的数据集（type2=c("dad", "mom", "kid";并且为了方便，type=c("a", "b", "c")）嵌套在家庭（hid）中并重复测量（ time）。

向所有人询问某些变量，例如v1_，但这些值分布在三列中。例如，v1_a包含所有父亲（type==a）的值。
v2_之类的变量只询问爸爸和妈妈（a和b），而且这些值分布在两列中。
v3等变量也只向爸爸和妈妈询问，但这些值包含在一列中。
向所有人询问v4等变量，这些值包含在一列中。

有：

   hid pid type type2 time v1_a v1_b v1_c v2_a v2_b v3 v4
1    1   1    a   dad    1    6   NA   NA    2   NA  4  3
2    1   2    b   mom    1   NA    2   NA   NA    5  6  6
3    1   3    c   kid    1   NA   NA    1   NA   NA NA  5
4    2   4    a   dad    1    3   NA   NA    6   NA  2  6
5    2   5    b   mom    1   NA    5   NA   NA    2  4  3
6    2   6    c   kid    1   NA   NA    3   NA   NA NA  5
7    1   1    a   dad    2    3   NA   NA    2   NA  4  3
8    1   2    b   mom    2   NA    3   NA   NA    5  6  6
9    1   3    c   kid    2   NA   NA    2   NA   NA NA  5
10   2   4    a   dad    2    2   NA   NA    6   NA  2  6
11   2   5    b   mom    2   NA    3   NA   NA    2  4  3
12   2   6    c   kid    2   NA   NA    2   NA   NA NA  5

这是我想要的最终结果：

   hid pid type type2 time v1 v2 v3 v4
1    1   1    a   dad    1  6  2  4  3
2    1   2    b   mom    1  2  5  6  6
3    1   3    c   kid    1  1 NA NA  5
4    2   4    a   dad    1  3  6  2  6
5    2   5    b   mom    1  5  2  4  3
6    2   6    c   kid    1  3 NA NA  5
7    1   1    a   dad    2  3  2  4  3
8    1   2    b   mom    2  3  5  6  6
9    1   3    c   kid    2  2 NA NA  5
10   2   4    a   dad    2  2  6  2  6
11   2   5    b   mom    2  3  2  4  3
12   2   6    c   kid    2  2 NA NA  5

我正在寻找一种tidyverse方法来处理混合变量的更大实际用例，如下所示。变量命名是一致的。我在gather()之后去哪儿？

library(tidyverse)
df_have <- data.frame(hid=c(1, 1, 1, 2, 2, 2,
                            1, 1, 1, 2, 2, 2),
                      pid=c(1, 2, 3, 4, 5, 6,
                            1, 2, 3, 4, 5, 6),
                      type=c("a", "b", "c", "a", "b", "c",
                             "a", "b", "c", "a", "b", "c"),
                      type2=c("dad", "mom", "kid", "dad", "mom", "kid",
                              "dad", "mom", "kid", "dad", "mom", "kid"),
                      time=c(1, 1, 1, 1, 1, 1, 
                             2, 2, 2, 2, 2, 2),
                      v1_a=c(6, NA, NA, 3, NA, NA,
                             3, NA, NA, 2, NA, NA),
                      v1_b=c(NA, 2, NA, NA, 5, NA,
                             NA, 3, NA, NA, 3, NA),
                      v1_c=c(NA, NA, 1, NA, NA, 3,
                             NA, NA, 2, NA, NA, 2),
                      v2_a=c(2, NA, NA, 6, NA, NA,
                             2, NA, NA, 6, NA, NA),
                      v2_b=c(NA, 5, NA, NA, 2, NA,
                             NA, 5, NA, NA, 2, NA),
                      v3=c(4, 6, NA, 2, 4, NA,
                           4, 6, NA, 2, 4, NA),
                      v4=c(3, 6, 5, 6, 3, 5,
                           3, 6, 5, 6, 3, 5)
                      )
df_want <- data.frame(hid=c(1, 1, 1, 2, 2, 2,
                            1, 1, 1, 2, 2, 2),
                      pid=c(1, 2, 3, 4, 5, 6,
                            1, 2, 3, 4, 5, 6),
                      type=c("a", "b", "c", "a", "b", "c",
                             "a", "b", "c", "a", "b", "c"),
                      type2=c("dad", "mom", "kid", "dad", "mom", "kid",
                              "dad", "mom", "kid", "dad", "mom", "kid"),
                      time=c(1, 1, 1, 1, 1, 1, 
                             2, 2, 2, 2, 2, 2),
                      v1=c(6, 2, 1, 3, 5, 3,
                           3, 3, 2, 2, 3, 2),
                      v2=c(2, 5, NA, 6, 2, NA,
                           2, 5, NA, 6, 2, NA),
                      v3=c(4, 6, NA, 2, 4, NA,
                           4, 6, NA, 2, 4, NA),
                      v4=c(3, 6, 5, 6, 3, 5,
                           3, 6, 5, 6, 3, 5)
                      )

df_have %>%
  gather(key, value, -hid, -pid, -type, -type2, -time)

Answer 1

这让我在那里，但filter(!is.na(value))步骤似乎是一个黑客。更好的想法？

df_test <- 
df_have %>%
  gather(key, value, -hid, -pid, -type, -time, -type2) %>%
  mutate(key = str_replace(key, "_.*", "")) %>%
  filter(!is.na(value)) %>%
  spread(key, value) %>%
  arrange(time, hid, type, pid)

从@www更新：

df_test <- 
df_have %>%
  gather(key, value, -hid, -pid, -type, -time, -type2, na.rm=TRUE) %>%
  mutate(key = str_replace(key, "_.*", "")) %>%
  spread(key, value) %>%
  arrange(time, hid, type, pid)

Answer 2

使用来自coalesce的{{1}}和来自dplyr的{{1}}的{{1}}是另一个想法。

map

使用嵌套的重复度量收集多个列

2 个答案: