加入两个间隔不正确的数据帧？

时间：2017-12-20 11:17:36

标签： r dplyr tidyverse lubridate tibble

我正在尝试找到重叠的区间，并决定将区间数据与dplyr::left_join()连接起来，这样我就可以将lubridate::int_overlaps()的区间与每个其他区间的区间进行比较。

以下是我期望left_join()的行为方式。两个三行交叉形成一个，共有9行：

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

以下是相同代码的行为与间隔的关系。我得到九行，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，但我可能会遗漏一些东西？或者这是一个错误？

我正在使用 1.7.1， 1.3.4和 0.7.4。

3 个答案:

答案 0 :(得分：7)

错误

该对象仍包含相关信息：

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

但是当按指数进行子集化时，它不再有效：

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使用head。这就是tibbles而不是data.frames立即显示问题的原因。

键入str(res$b)，我们发现9个start值只有3 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)
现在打印好了：

  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还不够，left_join功能搞乱，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

我已报告问题here

答案 1 :(得分：4)

看起来像for(int i=0; i<FirstList.size()+SecondList.size(); i++) { HSSFRow row = sheet.createRow((short)i); row.createCell(0).setCellValue(FirstList.get(i)); row.createCell(1).setCellValue(SecondList.get(i)); }中的错误：

tibble()

然而：

> 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.

因此，这有效：

> 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

虽然没有：

> 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

注意：我正在使用tibble_1.4.1（相同版本的lubridate和dplyr），R 3.4.3 for x86_64-pc-linux-gnu

答案 2 :(得分：1)

此问题已不存在，因为this issue已关闭并且已实现相关功能。如果您现在使用更新的程序包运行代码，它将可以正常工作。

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

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