加入两个间隔不当的数据帧？

Question

编辑（2019-06）：此问题不再存在，因为 this issue 已关闭并实现了相关功能。如果您现在运行包含更新包的代码，它将起作用。

我试图找到重叠的间隔，并决定将间隔数据与 dplyr::left_join() 连接起来，这样我就可以将具有 lubridate::int_overlaps() 的间隔与具有相同 ID 的每个其他间隔进行比较。

这是我期望 left_join() 的行为方式。三行的两个小标题交叉形成 9 行的 tibble:

library(tidyverse)

tibble(a = rep("a", 3), b = rep(1, 3)) %>% 
  left_join(tibble(a = rep("a", 3), c = rep(2, 3)))
Joining, by = "a"
# A tibble: 9 x 3
      a     b     c
  <chr> <dbl> <dbl>
1     a     1     2
2     a     1     2
3     a     1     2
4     a     1     2
5     a     1     2
6     a     1     2
7     a     1     2
8     a     1     2
9     a     1     2

下面是相同代码在不同时间间隔内的行为方式。我有九行，但行不像上面那样交叉：

tibble(a = rep("a", 3), b = rep(make_date(2001) %--% make_date(2002), 3)) %>% 
  left_join(tibble(a = rep("a", 3), c = rep(make_date(2002) %--% make_date(2003))))
Joining, by = "a"
# A tibble: 9 x 3
      a                              b                              c
  <chr>                 <S4: Interval>                 <S4: Interval>
1     a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
2     a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
3     a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
4     a                         NA--NA                         NA--NA
5     a                         NA--NA                         NA--NA
6     a                         NA--NA                         NA--NA
7     a                         NA--NA                         NA--NA
8     a                         NA--NA                         NA--NA
9     a                         NA--NA                         NA--NA

我认为这是出乎意料的，但我可能遗漏了什么？或者这是一个错误？

我正在使用 lubridate 1.7.1、tibble 1.3.4 和 dplyr 0.7.4。

Answer 1

看起来像是 tibble() 中的错误：

> AA <- tibble(a = rep("a", 3), b = rep(make_date(2001) %--% make_date(2002), 3))
> class(AA$b)
[1] "Interval"
attr(,"package")
[1] "lubridate"
> AA
Error in round_x - lhs :
  Arithmetic operators undefined for 'Interval' and 'Interval' classes:
  convert one to numeric or a matching time-span class.

但是：

> AA <- as.data.frame(AA)
class(AA$b)
> class(AA$b)
[1] "Interval"
attr(,"package")
[1] "lubridate"
> AA
  a                              b
1 a 2001-01-01 UTC--2002-01-01 UTC
2 a 2001-01-01 UTC--2002-01-01 UTC
3 a 2001-01-01 UTC--2002-01-01 UTC

因此，这有效：

> AA <- tibble(a = rep("a", 3), b = rep(make_date(2001) %--% make_date(2002), 3))
> BB <- tibble(a = rep("a", 3), c = rep(make_date(2002) %--% make_date(2003)))
> AA %>% as.data.frame %>% left_join(BB)
Joining, by = "a"
  a                              b                              c
1 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
2 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
3 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
4 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
5 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
6 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
7 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
8 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
9 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC

虽然这不是：

> AA %>% left_join(BB)
Joining, by = "a"
Error in round_x - lhs :
  Arithmetic operators undefined for 'Interval' and 'Interval' classes:
  convert one to numeric or a matching time-span class.

注意：我在 x86_64-pc-linux-gnu

的 R 3.4.3 上使用 tibble_1.4.1（与您相同版本的 lubridate 和 dplyr）

Answer 2

错误

该对象还包含相关信息：

res <- tibble(a = rep("a", 3), b = rep(make_date(2001) %--% make_date(2002), 3)) %>% 
  left_join(tibble(a = rep("a", 3), c = rep(make_date(2002) %--% make_date(2003)))) 

print.data.frame(res)
# a                              b                              c
# 1 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 2 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 3 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 4 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 5 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 6 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 7 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 8 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 9 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC

res$c    
# [1] 2002-01-01 UTC--2003-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# [5] 2002-01-01 UTC--2003-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# [9] 2002-01-01 UTC--2003-01-01 UTC

但是当按索引进行子集化时，它不再起作用了:

res_df <- as.data.frame(res)

head(res_df)
  a                              b                              c
1 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
2 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
3 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
4 a                         NA--NA                         NA--NA
5 a                         NA--NA                         NA--NA
6 a                         NA--NA                         NA--NA

res_df[4,"c"]
[1] NA--NA

和tibble:::print.tbl利用了head。这就是为什么使用 tibbles 而不是 data.frames.

时问题会立即可见的原因

键入 str(res$b) 我们看到我们只有 3 start 个值对应 9 个 data 个值。

如果我们这样做：

res_df$b@start <- rep(res_df$b@start,3)
res_df$c@start <- rep(res_df$c@start,3)

现在打印一切正常：

  a                              b                              c
1 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
2 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
3 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
4 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
5 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
6 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
7 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
8 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
9 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC

解决方案

我们已经看到 as.data.frame 是不够的，left_join 是把事情搞砸的函数，请改用 merge：

res <- tibble(a = rep("a", 3), b = rep(make_date(2001) %--% make_date(2002), 3)) %>% 
  merge(tibble(a = rep("a", 3), c = rep(make_date(2002) %--% make_date(2003))),
        all.x=TRUE) 

head(res)
# a                              b                              c
# 1 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 2 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 3 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 4 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 5 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
# 6 a 2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC

res[4,"c"]
#[1] 2002-01-01 UTC--2003-01-01 UTC

我已经报告了这个问题here

Answer 3

此问题已不存在，因为 this issue 已关闭并实现了相关功能。如果您现在运行包含更新包的代码，它将起作用。

library(lubridate)
library(tidyverse)

tibble(a = rep("a", 3), b = rep(make_date(2001) %--% make_date(2002), 3)) %>% 
  left_join(tibble(a = rep("a", 3), c = rep(make_date(2002) %--% make_date(2003))))
#> Joining, by = "a"
#> # A tibble: 9 x 3
#>   a     b                              c                             
#>   <chr> <Interval>                     <Interval>                    
#> 1 a     2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
#> 2 a     2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
#> 3 a     2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
#> 4 a     2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
#> 5 a     2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
#> 6 a     2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
#> 7 a     2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
#> 8 a     2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC
#> 9 a     2001-01-01 UTC--2002-01-01 UTC 2002-01-01 UTC--2003-01-01 UTC

^{由 reprex package (v0.3.0)}

于 2019-06-07 创建

加入两个间隔不当的数据帧？

Joining two data frames with intervals misbehaves?

r

lubridate

dplyr

tidyverse

tibble