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add_row with group_by nest tibble

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我正在尝试使用do将add_row（）分组数据 without .

library(dplyr)
library(tidyr)
library(purrr)
library(tibble)


my.data <- data.frame(

  supplier = c("a","a","a","a","a","a","b","b","b","b","b","b"),
  date = rep(c("2017-06-01","2017-03-01","2017-02-01","2017-01-12",
               "2017-05-01","2017-04-01"), 2), 
  order = c(1,0,0,1,1,0,0,1,0,0,1,0)

)

Solution with do

my.data %>%
  group_by(supplier) %>% 
  do(add_row(.,.before=0))

这使

# A tibble: 14 x 3
# Groups:   supplier [3]
   supplier       date order
      <chr>      <chr> <dbl>
 1     <NA>       <NA>    NA
 2        a 2017-06-01     1
 3        a 2017-03-01     0
 4        a 2017-02-01     0
 5        a 2017-01-12     1
 6        a 2017-05-01     1
 7        a 2017-04-01     0
 8     <NA>       <NA>    NA
 9        b 2017-06-01     0
10        b 2017-03-01     1
11        b 2017-02-01     0
12        b 2017-01-12     0
13        b 2017-05-01     1
14        b 2017-04-01     0

Attempt with nest and mutate or purrr::map

my.data %>%
  group_by(supplier) %>%
  nest() %>%
  mutate(extra.row = add_row(data, .before = 0))

mutate_impl（.data，dots）出错：评估错误：不支持的索引类型：NULL .

有什么建议 . 缩放时做的很慢 .

r dplyr purrr tibble

1 回答

  • 3

    您可以使用 bind_rows 将汇总数据集绑定到原始数据集上 .

    您也可以使用 complete ，虽然现在您的每组日期相同，但可能无法按照每组不同日期编写的日期 . 此外，我相信当你扩大规模时， complete 往往会很慢 .

    这两个解决方案都依赖于 date ，它是原始数据集中的实际 date 变量 .

    my.data = mutate(my.data, date = as.Date(date) )

    用 summarize 和 bind_rows 汇总和绑定 . arrange 是为了使事情井井有条，在实际案例中很可能不需要 .

    my.data %>%
    group_by(supplier) %>%
    summarize(date = min(date) - 30) %>%
    bind_rows(., my.data) %>%  
    arrange(supplier, date)

    如果组之间的日期相同，则使用 complete .

    my.data %>%
    group_by(supplier) %>%
    complete(date = c(min(.$date) - 30, .$date ) )

    两者的结果：

    # A tibble: 14 x 3
# Groups:   supplier [2]
   supplier       date order
     <fctr>     <date> <dbl>
 1        a 2016-12-13    NA
 2        a 2017-01-12     1
 3        a 2017-02-01     0
 4        a 2017-03-01     0
 5        a 2017-04-01     0
 6        a 2017-05-01     1
 7        a 2017-06-01     1
 8        b 2016-12-13    NA
 9        b 2017-01-12     0
10        b 2017-02-01     0
11        b 2017-03-01     1
12        b 2017-04-01     0
13        b 2017-05-01     1
14        b 2017-06-01     0
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