Question

我有一个示例数据框：

x <- data.frame(x = c(1, 1, 2, 2, 3, 3, 4, 4, 1), 
                y = c("a", "a", "b", "b", "c", "c", "d", "d", "z"))

我可以轻松地按小组获取row_number()：

x %>%
    group_by(x) %>%
    mutate(id = row_number())


x y id
1 a  1
1 a  2
2 b  1
2 b  2
3 c  1
3 c  2
4 d  1
4 d  2
1 z  3

我想要的是将x$x和x$y的相同组合编号为相同的数字，例如

以便c(x$x[1], x&y[1]) == c(x$x[2], x$y[2]) == c(x$x[n], x$y[n]在新列中获得相同的值。

我如何在dplyr中执行此操作？

Answer 1

另一种可能的选择：

library(dplyr)

x <- data_frame(x = c(1, 1, 2, 2, 3, 3, 4, 4, 1), 
                y = c("a", "a", "b", "b", "c", "c", "d", "d", "z"))

x %>% 
  group_by(x,y) %>% 
  summarise(y_list = list(y)) %>% 
  group_by(x) %>% 
  mutate(id = row_number()) %>% 
  tidyr::unnest() %>% 
  select(-y_list)

#output
      x     y    id
  <dbl> <chr> <int>
1     1     a     1
2     1     a     1
3     1     z     2
4     2     b     1
5     2     b     1
6     3     c     1
7     3     c     1
8     4     d     1
9     4     d     1

Answer 2

以下是使用factor的另一种解决方案：

## levels=unique(y) is so that levels of y are numbered according to their order of appearance and not alphabetical order
df %>% group_by(x) %>% mutate(id=as.numeric(factor(y,levels=unique(y))))

它返回：

  x     y     id
  <dbl> <chr> <dbl>
1     1     a     1
2     1     a     1
3     2     b     1
4     2     b     1
5     3     c     1
6     3     c     1
7     4     d     1
8     4     d     1
9     1     z     2

Answer 3

我们可以使用的另一个选项是match

library(dplyr)
x %>% 
   group_by(x) %>% 
   mutate(id = match(y, unique(y)))
# A tibble: 9 x 3
# Groups: x [4]
#      x      y    id
#  <dbl> <fctr> <int>
#1     1      a     1
#2     1      a     1
#3     2      b     1
#4     2      b     1
#5     3      c     1
#6     3      c     1
#7     4      d     1
#8     4      d     1
#9     1      z     2

Answer 4

x          %>% 
arrange(x) %>% 
mutate(xid = lag(x$x, default = 1), 
       yid = lag(as.character(x$y), default = "a")) %>% 
group_by(x) %>%
mutate(id = cumsum(x != xid | as.character(y) != yid) + 1) %>% 
mutate(xid = NULL, yid = NULL)

这会产生：

      x      y    id
  <dbl> <fctr> <dbl>
1     1      a     1
2     1      a     1
3     1      z     2
4     2      b     1
5     2      b     1
6     3      c     1
7     3      c     1
8     4      d     1
9     4      d     1

在dplyr中使用相同的编号对相同的分组值进行编号

4 个答案: