记录连续几天通过集团中的R

Question

我有一个数据帧如下：

DATE <- as.Date(c('2016-12-01', '2016-12-02', '2016-12-03', '2016-12-04', '2016-12-01', '2016-12-03', '2016-12-04', '2016-12-04')) 
Parent <- c('A','A','A','A','A','A','A','B') 
Child <- c('ab', 'ab', 'ab', 'ab', 'ac','ac', 'ac','bd') 
salary <- c(1000, 100, 4000, 2000,1000,3455,1234,600) 
avg_child_salary <- c(500, 500, 500, 500, 300, 300, 300, 9000) 
Callout <- c('HIGH', 'LOW', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'HIGH', 'LOW') 
employ.data <- data.frame(DATE, Parent, Child, avg_child_salary, salary, Callout) 

employ.data 

     DATE Parent Child avg_child_salary salary Callout 
1 2016-12-01  A ab    500 1000 HIGH 
2 2016-12-02  A ab    500 100  LOW 
3 2016-12-03  A ab    500 4000 HIGH 
4 2016-12-04  A ab    500 2000 HIGH 
5 2016-12-01  A ac    300 1000 HIGH 
6 2016-12-03  A ac    300 3455 HIGH 
7 2016-12-04  A ac    300 1234 HIGH 
8 2016-12-04  B bd    9000 600  LOW 

我过滤掉了就在昨天的数据是2016-12-04如下：

yesterday <- as.Date(Sys.Date()-1) 
df2<-filter(employ.data, DATE == yesterday) 
df2 

      DATE Parent Child avg_child_salary salary Callout 
    4 2016-12-04  A ab    500 2000 HIGH 
    7 2016-12-04  A ac    300 1234 HIGH 
    8 2016-12-04  B bd    9000 600  LOW 

我的目标是包括列Callout显示从2016-12-04连续天的数额，标注已HIGH或LOW由Child根据employ.data数据帧。这是我需要的最终输出：

  DATE Parent Child avg_child_salary salary Callout Consec. Days with Callout 
    4 2016-12-04  A ab    500 2000 HIGH       2 
    7 2016-12-04  A ac    300 1234 HIGH       2 
    8 2016-12-04  B bd    9000 600  LOW       1 

谢谢！

Answer 1

这里是另一种方法那是相当混乱，但我觉得你想要做什么：

library(dplyr) 
yesterday <- as.Date(Sys.Date()-1) 
df2 <- employ.data %>% group_by(Child) %>% 
    mutate(`Consec. Days with Callout`=cumsum(rev(cumprod(rev((yesterday-DATE)==(which(DATE == yesterday)-row_number()) & Callout==Callout[DATE == yesterday]))))) %>% 
    filter(DATE == yesterday) 
##Source: local data frame [3 x 7] 
##Groups: Child [3] 
## 
##  DATE Parent Child avg_child_salary salary Callout Consec. Days with Callout 
##  <date> <fctr> <fctr>   <dbl> <dbl> <fctr>      <dbl> 
##1 2016-12-04  A  ab    500 2000 HIGH       2 
##2 2016-12-04  A  ac    300 1234 HIGH       2 
##3 2016-12-04  B  bd    9000 600  LOW       1 

注：

1. (yesterday-DATE)==(which(DATE == yesterday)-row_number()) & Callout==Callout[DATE == yesterday]计算出如果Callout为t，则该行的条件将为TRUE他与Callout的yesterday相同，并且与行yesterday的行距离与日期的距离相同。这给了Cond列如下图所示：

   Source: local data frame [8 x 7] 
   Groups: Child [3] 
   
        DATE Parent Child avg_child_salary salary Callout Cond 
        <date> <fctr> <fctr>   <dbl> <dbl> <fctr> <lgl> 
   1 2016-12-01  A  ab    500 1000 HIGH TRUE 
   2 2016-12-02  A  ab    500 100  LOW FALSE 
   3 2016-12-03  A  ab    500 4000 HIGH TRUE 
   4 2016-12-04  A  ab    500 2000 HIGH TRUE 
   5 2016-12-01  A  ac    300 1000 HIGH FALSE 
   6 2016-12-03  A  ac    300 3455 HIGH TRUE 
   7 2016-12-04  A  ac    300 1234 HIGH TRUE 
   8 2016-12-04  B  bd    9000 600  LOW TRUE 
   

2. 有鉴于此，我们希望从为yesterday（由Child分组）排倒数的连续TRUE数量。为此，我们可以使用rev来反转矢量，做一个cumprod，一旦它遇到FALSE就会从1切换到0，使用rev再次反转矢量，最后做cumsum累积连续的天数。这样做可以使其中Consec. Days with Callout列被解释为以前的连续天数与同Callout为yesterday如下：

   Source: local data frame [8 x 7] 
   Groups: Child [3] 
   
        DATE Parent Child avg_child_salary salary Callout Consec. Days with Callout 
        <date> <fctr> <fctr>   <dbl> <dbl> <fctr>      <dbl> 
   1 2016-12-01  A  ab    500 1000 HIGH       0 
   2 2016-12-02  A  ab    500 100  LOW       0 
   3 2016-12-03  A  ab    500 4000 HIGH       1 
   4 2016-12-04  A  ab    500 2000 HIGH       2 
   5 2016-12-01  A  ac    300 1000 HIGH       0 
   6 2016-12-03  A  ac    300 3455 HIGH       1 
   7 2016-12-04  A  ac    300 1234 HIGH       2 
   8 2016-12-04  B  bd    9000 600  LOW       1 
   

3. 最后，做filter像你一样，产生最终结果。

Answer 2

试试这个我的男人

library(lubridate) 

df3 <- df2 %>% 
     group_by(child, callout) %>%       
     mutate(DATE = ymd(DATE), 
       consecutive_day_flag = if_else(DATE == (lag(DATE) + days(1)), 1, 0), 
       how_many = sum(consecutive_day_flag))