计算变化是否大于 0.25

Question

考虑这个data.table

library(data.table)
dt1 <- data.table(id = c("01","01","01","01","01", "02","02","02",",02","02"),
                 change_total = c(.00,.90,-.10,.8,.3,.00,.90,-.10,.8,.3))

如何计算每个 ID 的行之间的变化是否大于 change_total 的 25% (change_greater_than_25_percent)，然后如果这些“是”大于每个 ID 的行的 25%。 看起来像这样

dt2 <- data.table(id = c("01","01","01","01", "01","02","02","02","02","02"),
                 change_total = c(.00,.90,-.10,.8,.3,.00,.90,-.10,.8,.3),
                 change_greater_than_25_percent = c("no","yes","no","no","no","no","yes","no","no","no"),
                 change_greater_than_25_percent_greaterthan_25_percent_ofthe_time_by_id = c("no","yes","yes","yes","no","no","yes","yes","yes","no"))

Answer 1

我能够重现列 change_greater_than_25_percent 的预期结果：

library(data.table)
library(magrittr) # piping used to improve readability
dt1[, change_greater_than_25_percent := 
      (change_total / shift(change_total, fill = Inf) - 1. > 0.25) %>%  
      fifelse("yes", "no"), by = id][]

    id change_total change_greater_than_25_percent
 1: 01          0.0                             no
 2: 01          0.9                            yes
 3: 01         -0.1                             no
 4: 01          0.8                             no
 5: 01          0.3                             no
 6: 02          0.0                             no
 7: 02          0.9                            yes
 8: 02         -0.1                             no
 9: 02          0.8                             no
10: 02          0.3                             no

很遗憾，我不明白列change_greater_than_25_percent_greaterthan_25_percent_ofthe_time_by_id的确切定义

然后如果这些“是”按每个 id 的行数大于 25%

编辑 1

如果我理解正确Pan Dora's comment，第二列表示change_greater_than_25_percent（取为0，"no"、"yes"的值为1）的累积平均值大于0.25 .

如果将"no"和"yes"分别替换为逻辑值FALSE和TRUE，这将更容易计算，因为根据help("TRUE")

在需要数值的上下文中，逻辑向量被强制转换为整数向量，TRUE 被映射到 1L、FALSE 到 0L 和 NA 到 NA_integer_。

library(data.table)
dt1[, change_greater_than_25_percent := 
      change_total / shift(change_total, fill = Inf) - 1. > 0.25, by = id][
        , change_greater_than_25_percent_greaterthan_25_percent_ofthe_time_by_id := 
          cumsum(change_greater_than_25_percent) / (1:.N) >= 0.25, by = id][]

    id change_total change_greater_than_25_percent change_greater_than_25_percent_greaterthan_25_percent_ofthe_time_by_id
 1: 01          0.0                          FALSE                                                                  FALSE
 2: 01          0.9                           TRUE                                                                   TRUE
 3: 01         -0.1                          FALSE                                                                   TRUE
 4: 01          0.8                          FALSE                                                                   TRUE
 5: 01          0.3                          FALSE                                                                  FALSE
 6: 02          0.0                          FALSE                                                                  FALSE
 7: 02          0.9                           TRUE                                                                   TRUE
 8: 02         -0.1                          FALSE                                                                   TRUE
 9: 02          0.8                          FALSE                                                                   TRUE
10: 02          0.3                          FALSE                                                                  FALSE

注意事项

1. 为了重现预期结果大于或等于 25% (>= 0.25) 必须使用而不是大于 25% (> 0.25 ）。我没有修改已经很长的列名。
2. 使用逻辑值代替字符简化了计算，因此我们不需要magrittr管道。
3. 使用了data.table 链接。
4. 为了计算累积平均值，dplyr::cummean() 或自适应frollmean() 可以交替使用。

编辑 2

以上代码要求dt1 属于data.table 类。为确保，请致电

setDT(dt1)

事先。