Question

我有以下表格：

x = data.table(Id=c(1,1,2,3,3,4), Name=c("A", "A", "B", "C", "C", "D"), TxId=c(10, 11, 20, 30, 31, 40))
#Id Name TxId
#1:  1    A   10
#2:  1    A   11
#3:  2    B   20
#4:  3    C   30
#5:  3    C   31
#6:  4    D   40

y = data.table(Name=c("A", "B", "B", "C"), Family=c("A-alpha", "B-beta", "B-gamma", "C-delta"))
#   Name  Family
#1:    A A-alpha
#2:    B  B-beta
#3:    B B-gamma
#4:    C C-delta

我可以进行左连接和连接，但我只想为X中的每一行输出一行。

# Left join X to Y on Name column
xy = y[x, on="Name"]
#   Name  Family Id TxId
#1:    A A-alpha  1   10
#2:    A A-alpha  1   11
#3:    B  B-beta  2   20
#4:    B B-gamma  2   20
#5:    C C-delta  3   30
#6:    C C-delta  3   31
#7:    D      NA  4   40

# Concatenate Family column
xy[, Family:=paste0(Family, collapse=", "), by=c("Name", "TxId")]
#   Name          Family Id TxId
#1:    A         A-alpha  1   10
#2:    A         A-alpha  1   11
#3:    B B-beta, B-gamma  2   20
#4:    B B-beta, B-gamma  2   20
#5:    C         C-delta  3   30
#6:    C         C-delta  3   31
#7:    D              NA  4   40

如何摆脱B的额外行？我希望它在Id / TxId上是唯一的。即。

#   Name          Family Id TxId
#1:    A         A-alpha  1   10
#2:    A         A-alpha  1   11
#3:    B B-beta, B-gamma  2   20
#5:    C         C-delta  3   30
#6:    C         C-delta  3   31
#7:    D              NA  4   40

如果我做eddi评论：

xy[, .(Family=paste0(Family, collapse=", "), by=c("Name", "TxId")])

我得到了正确的结果。但是如果我尝试添加其他列，它就不起作用（我得到的结果与我完成:=版本的结果相同）：

xy[, .(Id, Family=paste0(Family, collapse=", ")), by=c("Name", "TxId")]

Answer 1

请尝试

xy[, .(Family = paste0(Family, collapse = ", "), by = c("Id", "Name", "TxId")]

我尝试解释：
如果Id是该群组的一部分，那么对于Id的每个唯一值，它只会出现一次（确切地说，对于Id，Name的每个唯一组合， TxId）。如果Id - 表达式中包含j，即.(Id, Family = paste0(Family, collapse = ", ")，那么Id的每一行都将包含在结果集中，尽管正在汇总Family

Data.table left join和aggregate / concatenate / group_concat

1 个答案: