Question

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。

Answer 1

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")

R：连接表，其中t1.key1 = t2.key1和t2.key2部分匹配t2.key2

1 个答案: