Question

我们说data.frame名为countDF：

> countDF date count complete 1 20180124 16 FALSE 2 20180123 24 TRUE 3 20180122 24 TRUE 4 20180121 24 TRUE 5 20180120 23 FALSE 6 20180119 23 FALSE 7 20180118 24 TRUE

看起来像这样：

> dput(countDF)
structure(list(date = c("20180124", "20180123", "20180122", "20180121", 
"20180120", "20180119", "20180118"), count = c(16L, 24L, 24L, 
24L, 23L, 23L, 24L), complete = c(FALSE, TRUE, TRUE, TRUE, FALSE, 
FALSE, TRUE)), class = "data.frame", row.names = c(NA, -7L), .Names = c("date", 
"count", "complete"))

这个清单：

> last7D_missingHours
$`20180124`
[1]  3 17 18 19 20 21 22 23

$`20180120`
[1] 18

$`20180119`
[1] 7

看起来像这样：

> dput(last7D_missingHours)
structure(list(`20180124` = c(3L, 17L, 18L, 19L, 20L, 21L, 22L, 
23L), `20180120` = 18L, `20180119` = 7L), .Names = c("20180124", 
"20180120", "20180119"))

我想建一个data.frame（或者，data_frame）将left_join(countDF, last7D_missingHours, by = c('date' = names(last7D_missingHours)))加入前者NA并date > countDF date count complete missingHour 1 20180124 16 FALSE 3 17 18 19 20 21 22 23 2 20180123 24 TRUE NA 3 20180122 24 TRUE NA 4 20180121 24 TRUE NA 5 20180120 23 FALSE 18 6 20180119 23 FALSE 7 7 20180118 24 TRUE NA不匹配的行，如下所示：

tibbles

我可能会通过我猜测的递归子集来解决这个问题，但是想知道是否有人对更优化的方法有任何建议，因为我知道(db.query(TableA) .filter(TableA.id == TableB.table_a_id, TableA.record_id.is_(None)) .update({TableA.record_id: TableB.record_id}, synchronize_session=False))最近已经走了很长的路。 ..

Answer 1

将缺少的小时数放入tibble的列表列中，其他变量作为日期，然后只有left_join。

library(tidyverse)

countDF <- structure(list(date = c("20180124", "20180123", "20180122", "20180121", 
                                   "20180120", "20180119", "20180118"), 
                          count = c(16L, 24L, 24L, 24L, 23L, 23L, 24L), 
                          complete = c(FALSE, TRUE, TRUE, TRUE, FALSE, FALSE, TRUE)), 
                     class = "data.frame", row.names = c(NA, -7L), .Names = c("date", "count", "complete"))

last7D_missingHours <- structure(list(`20180124` = c(3L, 17L, 18L, 19L, 20L, 21L, 22L, 
                                                     23L), `20180120` = 18L, `20180119` = 7L), .Names = c("20180124", 
                                                                                                          "20180120", "20180119"))

lst_tbl <- tibble(date = c("20180124", "20180120", "20180119"),
                  missingHour = last7D_missingHours)

left_join(countDF, lst_tbl)
#> Joining, by = "date"
#>       date count complete                   missingHour
#> 1 20180124    16    FALSE 3, 17, 18, 19, 20, 21, 22, 23
#> 2 20180123    24     TRUE                          NULL
#> 3 20180122    24     TRUE                          NULL
#> 4 20180121    24     TRUE                          NULL
#> 5 20180120    23    FALSE                            18
#> 6 20180119    23    FALSE                             7
#> 7 20180118    24     TRUE                          NULL

我最终得到的是NULL而不是NA，我觉得这更有意义，所以我没有尝试改变它们只是为了得到你要求的东西。

左边根据列表名称将不规则列表列表连接到数据框

1 个答案: