Question

我的date范围如下。其中两个日期为payDay。对于发薪日前后3天的每个日期，我想返回payDay前后的天数。

下面，whatIHave说明了我的数据，whatIWant显示了结果。我想在dplyr中进行此操作。任何帮助将不胜感激。谢谢。

whatIHave <- data.frame(
  date = seq(as.Date("2019/11/01"), as.Date("2019/12/01"), "days"),
  payDay = c(0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0))

whatIWant <- data.frame(
  date = seq(as.Date("2019/11/01"), as.Date("2019/12/01"), "days"),
  payDay = c(0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0),
  payDayLag = c(0,0,0,0,0,0,0,0,0,-3,-2,-1,0,1,2,3,0,0,0,0,0,-3,-2,-1,0,1,2,3,0,0,0))

Answer 1

一种选择是识别“ payDay”为1的行

library(data.table)
library(dplyr)
ind <- which(whatIHave$payDay == 1)

基于'ind'创建行索引序列

v1 <- unlist(lapply(ind, function(i) (i-3):(i+3)))

然后，使用基于通过检查'v1'%in%行序列（row_number()）创建的逻辑向量的分组变量，通过减去row_number()来创建'payDayLag'从“ payDay”为1的行索引开始

whatIHave %>% 
  group_by(group =  rleid(row_number() %in% v1)) %>% 
  mutate(payDayLag = if(all(payDay == 0)) 0 
         else row_number() - row_number()[payDay==1])  %>%
  ungroup %>%
  select(-group)
# A tibble: 31 x 3
#   date       payDay payDayLag
#   <date>      <dbl>     <dbl>
# 1 2019-11-01      0         0
# 2 2019-11-02      0         0
# 3 2019-11-03      0         0
# 4 2019-11-04      0         0
# 5 2019-11-05      0         0
# 6 2019-11-06      0         0
# 7 2019-11-07      0         0
# 8 2019-11-08      0         0
# 9 2019-11-09      0         0
#10 2019-11-10      0        -3
# … with 21 more rows

如果我们希望在单个链条上

library(tidyverse)
whatIHave %>%
    mutate(ind = row_number() * payDay) %>% 
    filter(payDay == 1) %>% 
    mutate(ind = map(ind, ~ (.x-3):(.x+3))) %>% 
    group_by(grp = row_number()) %>%
    unnest %>% 
    mutate(payDayLag =  row_number() - row_number()[4]) %>% 
    ungroup %>% 
    select(-payDay, -grp, -date) %>% 
    right_join(whatIHave %>%
                   mutate(ind = row_number()))  %>%
    mutate(payDayLag = replace_na(payDayLag, 0))

或者没有加入

whatIHave %>%
   mutate(ind =  list(map(which(payDay == 1), ~  (.x -3):(.x + 3)))) %>%
   group_by(grp = rleid(row_number() %in% unlist(ind[[1]]) )) %>%
   select(-ind) %>% 
   mutate(payDayLag = if(all(payDay == 0)) 0 
         else row_number() - row_number()[payDay == 1]) %>%
   ungroup %>%
   select(-grp)

对于每个日期，如何计算直到/自最近的付款期以来（或称为提前期/滞后）的天数？

1 个答案: