从长到宽重塑时保留所有匹配的行

Question

我正在尝试将数据从长格式重塑为宽格式。我想保留与 sn3=1 和 sn3=2 匹配的所有行，而不是仅采用第一个值。我怎样才能实现这个目标？

所需输出：

sn2	休闲.1	h月.1	hyear.1	休闲.2	h月.2	hyear.2
227	230	6	2000	540	6	2000
227	130	6	2000	170	6	2000
250	370	6	2000	380	6	2000
250	380	6	2000	190	6	2000
294	120	6	2000	210	6	2000
294	200	6	2000	310	6	2000
307	130	7	2000	220	7	2000
307	480	7	2000	270	7	2000

> dput(df)
structure(list(sn2 = structure(c(227, 227, 227, 227, 249, 249, 
250, 250, 250, 250, 294, 294, 294, 294, 307, 307, 307, 307), label = "household number", format.stata = "%8.0g"), 
    sn3 = structure(c(1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 2, 
    2, 1, 1, 2, 2), label = "person number", format.stata = "%8.0g"), 
    leisure = c(230, 130, 540, 170, 430, 480, 370, 380, 380, 
    190, 120, 200, 210, 310, 130, 480, 220, 270), hmonth = structure(c(6, 
    6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7), label = "month of household interview", format.stata = "%8.0g", labels = c(january = 1, 
    february = 2, march = 3, april = 4, may = 5, june = 6, july = 7, 
    august = 8, september = 9, october = 10, november = 11, december = 12
    ), class = c("haven_labelled", "vctrs_vctr", "double")), 
    hyear = structure(c(2000, 2000, 2000, 2000, 2000, 2000, 2000, 
    2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 2000, 
    2000), label = "year of household interview", format.stata = "%8.0g")), row.names = c(NA, 
-18L), class = c("tbl_df", "tbl", "data.frame"), na.action = structure(c(`3492` = 3492L, 
`3493` = 3493L, `3494` = 3494L, `3495` = 3495L, `3496` = 3496L, 
`3497` = 3497L, `3498` = 3498L, `3499` = 3499L, `3500` = 3500L, 
`3501` = 3501L, `3508` = 3508L, `3509` = 3509L, `3510` = 3510L, 
`3511` = 3511L, `3512` = 3512L, `3513` = 3513L, `3518` = 3518L, 
`3519` = 3519L, `3520` = 3520L, `3521` = 3521L, `3522` = 3522L, 
`3523` = 3523L, `3524` = 3524L, `3525` = 3525L), class = "omit"))

Answer 1

您可以使用

pivot_wider

中的

tidyr

。此外，您还必须确保每一行都有一个唯一的 ID，因为它们不是像这样唯一标识的：

library(dplyr)
library(tidyr)
df %>%
  group_by(sn3) %>%
  mutate(id = row_number()) %>%
  pivot_wider(names_from = sn3, 
              values_from = leisure:hyear, 
              names_glue = "{.value}.{sn3}") %>%
  select(-id)
#> # A tibble: 16 × 7
#>      sn2 leisure.1 leisure.2   hmonth.1   hmonth.2 hyear.1 hyear.2
#>    <dbl>     <dbl>     <dbl> <hvn_lbll> <hvn_lbll>   <dbl>   <dbl>
#>  1   227       230       540          6          6    2000    2000
#>  2   227       130       170          6          6    2000    2000
#>  3   249       430        NA          6         NA    2000      NA
#>  4   249       480        NA          6         NA    2000      NA
#>  5   250       370        NA          6         NA    2000      NA
#>  6   250       380        NA          6         NA    2000      NA
#>  7   250        NA       380         NA          6      NA    2000
#>  8   250        NA       190         NA          6      NA    2000
#>  9   294       120        NA          6         NA    2000      NA
#> 10   294       200        NA          6         NA    2000      NA
#> 11   294        NA       210         NA          6      NA    2000
#> 12   294        NA       310         NA          6      NA    2000
#> 13   307       130        NA          7         NA    2000      NA
#> 14   307       480        NA          7         NA    2000      NA
#> 15   307        NA       220         NA          7      NA    2000
#> 16   307        NA       270         NA          7      NA    2000

^{创建于 2024-03-18，使用 reprex v2.0.2}