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tibble，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？
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)
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