我有一个有趣的连接，我需要在R中完成。所以这是我的两个表。
表1：
Date Name
2016-01-02 10:18:00 CARDOSO, RAMON
2016-01-02 15:02:00 HARRISON, KATHYANNE M
2016-01-02 15:02:00 PALEO, SHERI
2016-01-03 02:09:00 PHANOR, RENALDY
2016-01-03 09:42:00 GUAMAN, ANGEL
2016-01-03 18:47:00 AIME, MADELINE
2016-01-03 18:47:00 CADET, GARDY
2016-01-03 19:31:00 REID, ARTHUR D
2016-01-03 22:11:00 HERNANDEZ-JONES, FREDRICK JOSHUA
2016-01-04 12:32:00 AGUERO, RAUL
表2：
Date ID Name
2016-01-02 10:18:00 16-22-AR CARDOSO, RAMON
2016-01-02 15:02:00 16-24-AR HARRISON, KATHYANNE M", " PALEO, SHERI"
2016-01-02 15:02:00 16-25-AR HARRISON, KATHYANNE M", " PALEO, SHERI"
2016-01-03 02:09:00 16-31-AR PHANOR, RENALDY
2016-01-03 09:42:00 16-32-AR GUAMAN, ANGEL
2016-01-03 18:47:00 16-39-AR AIME, MADELINE", " CADET, GARDY"
2016-01-03 18:47:00 16-40-AR AIME, MADELINE", " CADET, GARDY"
2016-01-03 19:31:00 16-42-AR REID, ARTHUR D
2016-01-03 22:11:00 16-44-AR HERNANDEZ-JONES, FREDRICK JOSHUA
2016-01-04 12:32:00 16-49-AR AGUERO, RAUL
我的目标是将表1中的ID作为自己的列，但为了做到这一点，我需要在日期进行连接，并以某种方式在名称上进行匹配，我们在表2中查找表1中的名称
更新：
原始数据集看起来像这样
2016-01-02 10:18:00 16-22-AR CARDOSO, RAMON
2016-01-02 15:02:00 16-24-AR, 16-25-AR HARRISON, KATHYANNE M", " PALEO, SHERI"
2016-01-03 02:09:00 16-31-AR PHANOR, RENALDY
2016-01-03 09:42:00 16-32-AR GUAMAN, ANGEL
2016-01-03 18:47:00 16-39-AR, 16-40-AR AIME, MADELINE", " CADET, GARDY"
2016-01-03 19:31:00 16-42-AR REID, ARTHUR D
2016-01-03 22:11:00 16-44-AR HERNANDEZ-JONES, FREDRICK JOSHUA
2016-01-04 12:32:00 16-49-AR AGUERO, RAUL
目标是让每个名称在其自己的行上具有相应的ID。 ID与名称的顺序相同，第一个ID与第一个名称相同。
希望这个澄清有所帮助。
答案 0 :(得分：2)
我假设你有类似
的东西2016-01-02 15:02:00 16-24-AR HARRISON, KATHYANNE M", " PALEO, SHERI"
2016-01-02 15:02:00 16-25-AR HARRISON, KATHYANNE M", " PALEO, SHERI"
第一个ID对应第一个名称，第二个ID对应第二个名称。然后一种方法是创建一个具有正确名称的新列。
d$order <- unlist(sapply(rle(paste0(d$Date, d$Name))$lengths, seq_len))
split_names <- function(name, order = 1) {
names <- strsplit(name, '\\", \\"')[[1]] # Split
names <- gsub('^\\s|\\"', "", names) # Clean up leading space and trailing "
names[order]
}
d$Newname <- mapply(split_names, d$Name, d$order)
d[, c("Date", "ID", "Newname")]
# Date ID Newname
# 1 2016-01-02 10:18:00 16-22-AR CARDOSO, RAMON
# 2 2016-01-02 15:02:00 16-24-AR HARRISON, KATHYANNE M
# 3 2016-01-02 15:02:00 16-25-AR PALEO, SHERI
# 4 2016-01-03 02:09:00 16-31-AR PHANOR, RENALDY
# 5 2016-01-03 09:42:00 16-32-AR GUAMAN, ANGEL
# 6 2016-01-03 18:47:00 16-39-AR AIME, MADELINE
# 7 2016-01-03 18:47:00 16-40-AR CADET, GARDY
# 8 2016-01-03 19:31:00 16-42-AR REID, ARTHUR D
# 9 2016-01-03 22:11:00 16-44-AR HERNANDEZ-JONES, FREDRICK JOSHUA
# 10 2016-01-04 12:32:00 16-49-AR AGUERO, RAUL
数据：
structure(list(Date = c("2016-01-02 10:18:00", "2016-01-02 15:02:00",
"2016-01-02 15:02:00", "2016-01-03 02:09:00", "2016-01-03 09:42:00",
"2016-01-03 18:47:00", "2016-01-03 18:47:00", "2016-01-03 19:31:00",
"2016-01-03 22:11:00", "2016-01-04 12:32:00"), ID = c("16-22-AR",
"16-24-AR", "16-25-AR", "16-31-AR", "16-32-AR", "16-39-AR", "16-40-AR",
"16-42-AR", "16-44-AR", "16-49-AR"), Name = c("CARDOSO, RAMON",
"HARRISON, KATHYANNE M\", \" PALEO, SHERI\"", "HARRISON, KATHYANNE M\", \" PALEO, SHERI\"",
"PHANOR, RENALDY", "GUAMAN, ANGEL", "AIME, MADELINE\", \" CADET, GARDY\"",
"AIME, MADELINE\", \" CADET, GARDY\"", "REID, ARTHUR D", "HERNANDEZ-JONES, FREDRICK JOSHUA",
"AGUERO, RAUL")), .Names = c("Date", "ID", "Name"), row.names = c(NA,
-10L), class = "data.frame")