用dplyr和rle总结连续的故障 [英] Summarize consecutive failures with dplyr and rle

问题描述

我正在尝试构建一个流失模型，其中包括每个客户的连续数量的UX故障，并遇到困难。这是我的简化数据和所需的输出：

I'm trying to build a churn model that includes the maximum consecutive number of UX failures for each customer and having trouble. Here's my simplified data and desired output:

library(dplyr)
df <- data.frame(customerId = c(1,2,2,3,3,3), date = c('2015-01-01','2015-02-01','2015-02-02', '2015-03-01','2015-03-02','2015-03-03'),isFailure = c(0,0,1,0,1,1))
> df
  customerId       date isFailure
1          1 2015-01-01         0
2          2 2015-02-01         0
3          2 2015-02-02         1
4          3 2015-03-01         0
5          3 2015-03-02         1
6          3 2015-03-03         1

所需结果：

> desired.df
  customerId maxConsecutiveFailures
1          1                      0
2          2                      1
3          3                      2

我正在fl fl bit through through through through through b b b b b b b b b b b b b：：：：：：：：：：：：：：：：：：：：：：：：

I'm flailing quite a bit and searching through other rle questions isn't helping me yet - this is what I was "expecting" a solution to resemble:

df %>% 
  group_by(customerId) %>%
  summarise(maxConsecutiveFailures = 
    max(rle(isFailure[isFailure == 1])$lengths))

推荐答案

我们通过'customerId'分组，并使用 do 在'isFailure'列执行 rle 。对值（ length [values] 长度 c $ c>），并创建一个如果/ else 条件的'Max'列返回0，那些没有任何1值。

We group by the 'customerId' and use do to perform the rle on 'isFailure' column. Extract the lengths that are 'TRUE' for values (lengths[values]), and create the 'Max' column with an if/else condition to return 0 for those that didn't have any 1 value.

 df %>%
    group_by(customerId) %>%
    do({tmp <- with(rle(.$isFailure==1), lengths[values])
     data.frame(customerId= .$customerId, Max=if(length(tmp)==0) 0 
                    else max(tmp)) }) %>% 
     slice(1L)
#   customerId Max
#1          1   0
#2          2   1
#3          3   2

这篇关于用dplyr和rle总结连续的故障的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！