如何“grepl”以R中的日期范围为条件答案

【问题标题】：How to "grepl" conditional on a range of dates in R如何“grepl”以R中的日期范围为条件
【发布时间】：2022-10-04 22:26:47
【问题描述】：

假设我有两个数据框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 的边界（start.term 和end.term）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？有人可以帮我弄这个吗？

谢谢！

【问题讨论】：

标签： r dataframe for-loop if-statement

【解决方案1】：

我认为基本上你想在这里对匹配的名称进行连接操作。所以第一步是找出那些是什么：

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，这只是加入这些名称并根据您定义的条件进行过滤的情况：

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

【讨论】：