如何在 R 中以一系列日期为条件"grepl"

Question

假设我有两个数据帧df1和df2:

df1 = structure(list(surname = c("Duisenberg", "Trichet", "Draghi"), 
`start term` = structure(c(896659200, 1067644800, 1320105600
), class = c("POSIXct", "POSIXt"), tzone = "UTC"), `end term` = structure(c(1067558400, 
1320019200, 1572480000), class = c("POSIXct", "POSIXt"), tzone = "UTC")), row.names = c(1L, 
9L, 15L), class = "data.frame") %>% data.frame(stringsAsFactors = F)
surname start.term   end.term
1  Duisenberg 1998-06-01 2003-10-31
9     Trichet 2003-11-01 2011-10-31
15     Draghi 2011-11-01 2019-10-31

df2= data.frame(Date = c("2010-01-01","1997-01-01","2020-01-01","2004-01-01","2012-01-01","1999-01-01","2000-01-01","2020-01-01","2022-01-01","1996-01-01"), speaker = c("Mario Draghi","W.L. Duisenberg","Ciao","Jean-Claude Trichet","M. Draghi","W.L. Duisenberg","Jean-Claude Trichet","Bye","Ciao","Mario Draghi"), stringsAsFactors = F)
Date             speaker
1  2010-01-01        Mario Draghi
2  1997-01-01     W.L. Duisenberg
3  2020-01-01                Ciao
4  2004-01-01 Jean-Claude Trichet
5  2012-01-01           M. Draghi
6  1999-01-01     W.L. Duisenberg
7  2000-01-01 Jean-Claude Trichet
8  2020-01-01                 Bye
9  2022-01-01                Ciao
10 1996-01-01        Mario Draghi

当df1中的名称出现在df2中时，我可以很容易地找到：

which(grepl(paste0(df1$surname, collapse = "|"), df2$speaker, ignore.case = TRUE))
[1]  1  2  4  5  6  7 10

相反，更棘手的是：只有当df2中的日期超出df1(start.term和end.term(的边界时，df1中的名称才会出现在df2中。

结果应该是：

[1] 1 2 10

我该怎么做？有人能帮我吗？

谢谢！

Answer 1

我认为本质上您希望在这里对匹配的名称执行联接操作。因此，第一步是找出它们是什么：

library(dplyr)
surnames_regex <- paste0(df1$surname, collapse = "|")
df2$matching_name <- strsplit(df2$speaker, split = "\s") |>
lapply(
(name) {
matching_name <- grep(surnames_regex, name, v = T)
matching_name <- ifelse(
length(matching_name) > 0,
matching_name[1],
NA_character_
)
matching_name
}
) |>
unlist()

df2
#          Date             speaker matching_name
# 1  2010-01-01        Mario Draghi        Draghi
# 2  1997-01-01     W.L. Duisenberg    Duisenberg
# 3  2020-01-01                Ciao          <NA>
# 4  2004-01-01 Jean-Claude Trichet       Trichet
# 5  2012-01-01           M. Draghi        Draghi
# 6  1999-01-01     W.L. Duisenberg    Duisenberg
# 7  2000-01-01 Jean-Claude Trichet       Trichet
# 8  2020-01-01                 Bye          <NA>
# 9  2022-01-01                Ciao          <NA>
# 10 1996-01-01        Mario Draghi        Draghi

然后，这只是一个加入这些名称并根据您定义的条件进行过滤的例子：

df2 |>
inner_join(
df1,
by = c("matching_name" = "surname")
) |>
filter(
Date < start.term |
Date > end.term
)
#         Date             speaker matching_name start.term   end.term
# 1 2010-01-01        Mario Draghi        Draghi 2011-11-01 2019-10-31
# 2 1997-01-01     W.L. Duisenberg    Duisenberg 1998-06-01 2003-10-31
# 3 2000-01-01 Jean-Claude Trichet       Trichet 2003-11-01 2011-10-31
# 4 1996-01-01        Mario Draghi        Draghi 2011-11-01 2019-10-31

Answer 2

确保所有日期的格式一致，即使所有日期都属于Date类。然后这可以在一个SQL语句中完成。

library(sqldf)
df1a <- transform(df1, start.term = as.Date(start.term), 
end.term = as.Date(end.term))
df2a <- transform(df2, Date = as.Date(Date))
sqldf("select distinct b.rowid, *
from df1a a
join df2a b on 
(b.Date < a.[start.term] or b.Date > a.[end.term]) and
b.speaker like '%' || a.surname || '%'")

给予：

rowid    surname start.term   end.term       Date             speaker
1     1     Draghi 2011-11-01 2019-10-31 2010-01-01        Mario Draghi
2     2 Duisenberg 1998-06-01 2003-10-31 1997-01-01     W.L. Duisenberg
3     7    Trichet 2003-11-01 2011-10-31 2000-01-01 Jean-Claude Trichet
4    10     Draghi 2011-11-01 2019-10-31 1996-01-01        Mario Draghi