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我有 2 个数据框，一个 ( df1) 记录每天发生的不同活动，另一个 ( df2) 记录白天发生的活动的属性。

从中df1可以识别活动的重复发生以及持续时间。一天的开始时间由Date变量指定。

例如：

  • id12 事件从第 1 天开始，到第 7 天结束。在这种情况下，发生次数为 7，持续时间为 11。
  • 对于id123，一周从第 5 天开始，到第 7 天结束；重复顺序发生，因为第 6 天有间隔天，持续时间为 6，id 123（从第 6 天开始，到第 7 天结束）连续发生 2 次，持续时间为 6。

在df1变量 Date 中定义记录开始的日期。例如 id 12 记录从第 1 天开始，依此类推。

我想确定在连续发生期间是否有关于活动属性的记录df2。

例如 id 12，发生 7 次，持续时间为 12 有周三（day3 in df1）的记录，该记录对应于连续发生的第 3 天。对于 id 123，没有数据（例如，没有连续出现），但对于 id 10 的 6 天出现和持续时间 18，在第 6 天有记录。

DF1：

id   day1 day2 day3 day4 day5 day6  day7   Date
 12    2    1    2    1    1    3    1     Mon
123    0    3    0    3    3    0    3     Fri
 10    0    3    3    3    3    3    3     Sat

DF2：

    id   c1 c2  Date
    12   3   3   Wednesday
   123   3   2   Fri
     10  3   1   Sat

结果：

 id c1 c2  Occurrence Position
 12 3   3     7          3
123 0   0     0          0
 10 3   1     2          1

样本数据：df1

structure(list(id = c(12L, 123L, 10L), day1 = c(2L, 0L, 3L), 
    day2 = c(1L, 3L, 3L), day3 = c(2L, 0L, 3L), day4 = c(1L, 
    3L, 3L), day5 = c(1L, 3L, 3L), day6 = c(3L, 0L, 3L), day7 = c(1L, 
    3L, 3L), Date = c("Monday", "Friday", "Saturday")), row.names = c(NA, 
-3L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x000002a81a571ef0>)

df2:

structure(list(id = c(12, 123, 10), c1 = c(3, 3, 3), c2 = c(3, 
2, 1), Date = structure(c(3L, 1L, 2L), .Label = c("Friday", "Saturday", 
"Wednesday"), class = "factor")), row.names = c(NA, -3L), class = "data.frame")
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1 回答 1

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一个解决方案dplyr（也许不是最短的）：

# library
library(tidyverse)

# get data
df1 <- structure(list(id = c(12L, 123L, 10L), 
               day1 = c(2L, 0L, 3L), 
               day2 = c(1L, 3L, 3L), 
               day3 = c(2L, 0L, 3L), 
               day4 = c(1L,3L, 3L), 
               day5 = c(1L, 3L, 3L), 
               day6 = c(3L, 0L, 3L), 
               day7 = c(1L,3L, 3L), 
               Date = c("Monday", "Friday", "Saturday")),
               row.names = c(NA,-3L), class = c("data.table", "data.frame"))


df2 <- structure(list(id = c(12, 123, 10),
                      c1 = c(3, 3, 3), 
                      c2 = c(3, 2, 1),
                      Date = structure(c(3L, 1L, 2L), .Label = c("Friday", "Saturday","Wednesday"),
                                       class = "factor")), row.names = c(NA, -3L), class = "data.frame")


# change days to nummeric (will help you later)
df1 %>% mutate(
  Date_nr_df1=case_when(
    Date=="Monday" ~ 1,
    Date=="Tuesday" ~2,
    Date=="Wednesday" ~3,
    Date=="Thursday" ~4,
    Date=="Friday" ~5,
    Date=="Saturday" ~6,
    Date=="Sunday" ~7)) -> df1

df2 %>% mutate(
  Date_nr_df2=case_when(
    Date=="Monday" ~ 1,
    Date=="Tuesday" ~2,
    Date=="Wednesday" ~3,
    Date=="Thursday" ~4,
    Date=="Friday" ~5,
    Date=="Saturday" ~6,
    Date=="Sunday" ~7)) -> df2

# combine data by the id column
left_join(df1,df2, by=c("id")) -> df     

# adjust data   
df %>%
  group_by(id) %>% # to make changes per row
  mutate(days=paste0(day1,day2,day3,day4,day5,day6,day7)) %>% #pastes the values together
  mutate(days_correct=substring(days,Date_nr_df1)) %>% # applies the start day
  mutate(Occurrence_seq=str_split(days_correct, fixed("0"))[[1]][1]) %>% # extracts all days before 0
  mutate(Occurrence=nchar(Occurrence_seq)) %>%  ## counts these days
  mutate(Occurrence=case_when(Occurrence==1 ~ 0, TRUE ~ as.numeric(Occurrence))) %>% # sets Occurrence to 0 if there is no consecutive occurrence
  mutate(Position=Date_nr_df2-Date_nr_df1+1) %>% ## calculates the position you wanted
  mutate(c1=case_when(Occurrence==0 ~0, TRUE ~ c1),
         c2=case_when(Occurrence==0 ~0, TRUE ~c1),
         Position=case_when(Occurrence==0 ~ 0, TRUE ~ as.numeric(Position))) %>% 
  ungroup() %>% ungroups the df
  select(id,c1,c2,Occurrence,Position) # selects the wanted variables
#> # A tibble: 3 x 5
#>      id    c1    c2 Occurrence Position
#>   <dbl> <dbl> <dbl>      <dbl>    <dbl>
#> 1    12     3     3          7        3
#> 2   123     0     0          0        0
#> 3    10     3     3          2        1

由reprex 包于 2020-04-10 创建(v0.2.1)

于 2020-04-10T16:44:09.830 回答