在 R 中按组创建连续年份的计数

Question

这里是R新手。我正在寻找一个 dplyr 解决方案（最好）来创建一个向量，该向量显示一个组内的连续年数。如果序列被任何间隙打断，即使是同一组，计数器也应该重新开始。

我的数据看起来与此类似：

library(lubridate)
#> 
#> Attaching package: 'lubridate'
#> The following objects are masked from 'package:base':
#> 
#>     date, intersect, setdiff, union
library(magrittr)
library(tidyverse)

df <- tribble(
    ~id, ~ref, ~branch, ~year, ~unit, ~client, ~group,
    1, 561, "LA", 2000, "x", "y", "z",  
    2, 561, "LA", 2001, "x", "y", "z",
    3, 561, "LA", 2002, "x", "y", "z",
    4, 561, "LA", 2003, "x", "y", "z",
    5, 561, "LA", 2004, "x", "y", "z",
    6, 561, "LA", 2005, "x", "y", "z",
    7, 561, "LA", 2007, "x", "y", "z",
    8, 561, "LA", 2008, "x", "y", "z",
    9, 561, "LA", 2009, "x", "y", "z",
    )

我的预期输出是这样的，其中添加了“seq_count”：

df_exp <- tribble(
    ~id, ~ref, ~branch, ~year, ~unit, ~client, ~group, ~seq_count,
    1, 561, "LA", 2000, "x", "y", "z", 6,
    2, 561, "LA", 2001, "x", "y", "z", 6,
    3, 561, "LA", 2002, "x", "y", "z", 6,
    4, 561, "LA", 2003, "x", "y", "z", 6,
    5, 561, "LA", 2004, "x", "y", "z", 6,
    6, 561, "LA", 2005, "x", "y", "z", 6,
    7, 561, "LA", 2007, "x", "y", "z", 3,
    8, 561, "LA", 2008, "x", "y", "z", 3,
    9, 561, "LA", 2009, "x", "y", "z", 3,
    )

我已尝试使用 dplyr::add_count，如下所示：

df1 <- df %>% 
    group_by(ref, branch, unit, client, group) %>% 
    add_count()

但是，这只添加了 group_by 命令指定的计数，并没有考虑 2005 年和 2007 年之间的差距。有没有一种方法可以在 R 中以简洁的方式做到这一点？

Answer 1

n() 将为您提供组中的观察次数。

df1 <- df %>% 
    group_by(ref, branch, unit, client, group) %>% 
    mutate(seq_count = n())

如果您只想要摘要，可以使用 summarise 而不是 mutate。

Answer 2

您可以创建另一个组，当年份之间有差距时会发生变化。

library(dplyr)
df %>% 
    add_count(group, grp = cumsum(year - lag(year, default = first(year)) > 1), 
               name = 'seq_count')

# A tibble: 9 x 9
#     id   ref branch  year unit  client group   grp seq_count
#  <dbl> <dbl> <chr>  <dbl> <chr> <chr>  <chr> <int>     <int>
#1     1   561 LA      2000 x     y      z         0         6
#2     2   561 LA      2001 x     y      z         0         6
#3     3   561 LA      2002 x     y      z         0         6
#4     4   561 LA      2003 x     y      z         0         6
#5     5   561 LA      2004 x     y      z         0         6
#6     6   561 LA      2005 x     y      z         0         6
#7     7   561 LA      2007 x     y      z         1         3
#8     8   561 LA      2008 x     y      z         1         3
#9     9   561 LA      2009 x     y      z         1         3

或 n()

df %>%
  group_by(group, grp = cumsum(year - lag(year, default = first(year)) > 1)) %>%
  mutate(seq_count = n())

在 R 中按组创建连续年份的计数

Create count of sequential years by groups in R

r

lubridate

dplyr

tidyr