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Index an ordered vector with random number of repeating values [duplicate](为具有随机数量重复值的有序向量编制索引[重复])

转载 作者：bug小助手 更新时间：2023-10-25 22:18:04 25 4
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df<-data.frame(old=c(1,1,1,5,7,7,7,11,13,13,16,18,20,20,20,20,25,25,25,29),
               new=c(1,1,1,2,3,3,3,4,5,5,6,7,8,8,8,8,9,9,9,10))

   old new
1    1   1
2    1   1
3    1   1
4    5   2
5    7   3
6    7   3
7    7   3
8   11   4
9   13   5
10  13   5
11  16   6
12  18   7
13  20   8
14  20   8
15  20   8
16  20   8
17  25   9
18  25   9
19  25   9
20  29  10

How do I turn old into new easily? Basically it's the order of values repeated the same number of times. The values are always increasing.

如何将旧的轻松变成新的？基本上，它是重复相同次数的值的顺序。价值总是在增加。


Thanks in advance. I don't even know how to Google something this simple.

先谢谢你。我甚至不知道怎么用谷歌搜索这么简单的东西。


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You could use dplyr::consecutive_id or data.table::rleid to get an identifier for each run of identical values:

您可以使用dplyr：：Continucative_id或data.table：：rleid来获取每次运行相同值的标识符：


df <- data.frame(
  old = c(1, 1, 1, 5, 7, 7, 7, 11, 13, 13, 16, 18, 20, 20, 20, 20, 25, 25, 25, 29)
)

library(dplyr, warn.conflicts = FALSE)
library(data.table)

df |>
  mutate(new = consecutive_id(old))
#>    old new
#> 1    1   1
#> 2    1   1
#> 3    1   1
#> 4    5   2
#> 5    7   3
#> 6    7   3
#> 7    7   3
#> 8   11   4
#> 9   13   5
#> 10  13   5
#> 11  16   6
#> 12  18   7
#> 13  20   8
#> 14  20   8
#> 15  20   8
#> 16  20   8
#> 17  25   9
#> 18  25   9
#> 19  25   9
#> 20  29  10

df |>
  mutate(new = rleid(old))
#>    old new
#> 1    1   1
#> 2    1   1
#> 3    1   1
#> 4    5   2
#> 5    7   3
#> 6    7   3
#> 7    7   3
#> 8   11   4
#> 9   13   5
#> 10  13   5
#> 11  16   6
#> 12  18   7
#> 13  20   8
#> 14  20   8
#> 15  20   8
#> 16  20   8
#> 17  25   9
#> 18  25   9
#> 19  25   9
#> 20  29  10


You're essentially just checking when the old column changes. You can do this in a variety of ways, but the easiest is checking if the previous value in df$old matches the current one, then find the cumulative sum of that:

本质上，您只是检查旧列何时发生更改。您可以使用多种方法来完成此操作，但最简单的方法是检查df$old中的前一个值是否与当前值匹配，然后求出其累计和：


df$new <- cumsum(df$old != dplyr::lag(df$old, default = 0))

or

或


df$new <- cumsum(df$old != c(0, df$old[-nrow(df)]))

or

或


library(dplyr)
df |> mutate(new = cumsum(old != lag(df$old, default = 0)))

or

或


df$new <- cumsum(diff(c(0, df$old)) > 0)

or

或


df |> mutate(new = cumsum(diff(c(0, old)) > 0)

更多回答

Thank you! I should have known it was in dplyr but had no idea how to phrase a search for it.

谢谢!我应该知道它在dplyr，但我不知道如何措辞搜索它。

Thanks. Yeah, these all work. I had a clunky ifelse method along the same lines but i figured there must be a specific function that does this, and of course there is in dplyr and also data.table. See the answer from @stefan

谢谢。是的，这些都起作用了。我使用了一个类似的笨重的ifElse方法，但我认为一定有一个特定的函数来完成这项工作，当然dplyr和data.table中也有这个函数。请看@stefan的答案

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