一列串联字符串中的唯一字符

Question

我有一个带有字符串列“city”的 data.frame，它由由;分隔的连接字母组成;

dt = data.frame(id = letters[1:6],
                city = c("A;B","B;D","A;D;G","A;C","F;G","C;D"))
dt
#   id    city
# 1  a     A;B
# 2  b     B;D
# 3  c   A;D;G
# 4  d     A;C
# 5  e     F;G
# 6  f     C;D`

我希望从“城市”列中获得独特的个人字母：

city = c("A","B","C","D","F","G")`

如何做到这一点？

Answer 1

更清洁的解决方案是：

dt= data.frame(id=letters[1:6],city = c("A;B","B;D","A;D;G","A;C","F;G","C;D"))

city=strsplit(as.character(dt$city), ";")

city=sort(unique(unlist(city)))

[1] "A" "B" "C" "D" "F" "G"

Answer 2

数据：

dt=  data.frame(id=letters[1:6],city = c("A;B","B;D","A;D;G","A;C","F;G","C;D"))

> dt
  id  city
1  a   A;B
2  b   B;D
3  c A;D;G
4  d   A;C
5  e   F;G
6  f   C;D

拆分列city ，使用as.character转换为字符串：

city <- unlist(strsplit(as.character(dt$city), ";", fixed = T))

> city
 [1] "A" "B" "B" "D" "A" "D" "G" "A" "C" "F" "G" "C" "D"

现在使用unique和order来获取输出：

city <- unique(city)

> city
[1] "A" "B" "D" "G" "C" "F"

city <- city[order(city)]

> city
[1] "A" "B" "C" "D" "F" "G"

> dput(city)
c("A", "B", "C", "D", "F", "G")

---

编辑：更新了 OP 的新数据。

Edit2：更新以省略sapply ，因为显然strsplit是矢量化的。 谢谢@Cris！