Question

在每个家庭中，我想定义一个指标，该指标确定驾驶员是否适合乘客。如果他/她的旅程在乘客旅程后最多1个小时开始，那么驾驶员将可用。

示例：

      household    person     mode       start
           1         1         car        7:20
           1         1         car        8:00
           1         1         car        8:30
           1         2       non-car      7:30
           1         3       non-car      7:15
           1         4         car        7:00
           2         1          car       7:00
           2         2        non-car      9:00

第一家庭驾驶员可以乘车，因为他的旅行比第二人称旅行晚了30分钟，他也可以对第三人乘车。在第二个家庭。

输出

      household    person     mode       start      indicator
           1         1         car        8:00          1 
           1         2       non-car      7:30          1
           1         3       non-car      7:15          1
           2         1          car       7:00          0
           2         2        non-car      9:00         0

然后我要将这些匹配的行（指示符为1）彼此相邻

输出

      household    person     mode       start      indicator   household    person     mode       start      indicator
           1         1         car        8:00          1       2         2       non-car      7:30          1
           1         1         car        8:00          1       3         2       non-car      7:15          1

Answer 1

我们用as.POSIXct将'start'转换为datetime类，并按'household'分组，检查{start的diff ernece是否小于或等于1，将逻辑强制为带有as.integer

的二进制文件

library(dplyr)
df1 %>% 
 mutate(start = as.POSIXct(start, format = '%H:%M')) %>% 
 group_by(household) %>% 
 mutate(indicator = as.integer(any(diff(start) <= 1)))
# A tibble: 4 x 5
# Groups:   household [2]
#  household person mode    start               indicator
#      <int>  <int> <chr>   <dttm>                  <int>
#1         1      1 car     2019-09-03 08:00:00         1
#2         1      2 non-car 2019-09-03 07:30:00         1
#3         2      1 car     2019-09-03 07:00:00         0
#4         2      2 non-car 2019-09-03 09:00:00         0

要获取第二个输出，我们可以使用pivot_wider开发版本中的tidyr

df1 %>% 
  mutate(startn = as.POSIXct(start, format = '%H:%M')) %>% 
  group_by(household) %>% 
  mutate(indicator = as.integer(any(diff(startn) <= 1))) %>%  
  filter(indicator == 1) %>% 
  select(-startn) %>% 
  group_by(household) %>% 
  mutate(n = row_number()) %>%
  pivot_wider(names_from = n, values_from = c(household, person, mode, start, indicator))
# A tibble: 1 x 10
#  household_1 household_2 person_1 person_2 mode_1 mode_2  start_1 start_2 indicator_1 indicator_2
#        <int>       <int>    <int>    <int> <chr>  <chr>   <chr>   <chr>         <int>       <int>
#1           1           1        1        2 car    non-car 8:00    7:30              1           1

数据

df1 <- structure(list(household = c(1L, 1L, 2L, 2L), person = c(1L, 
2L, 1L, 2L), mode = c("car", "non-car", "car", "non-car"), start = c("8:00", 
"7:30", "7:00", "9:00")), class = "data.frame", row.names = c(NA, 
-4L))

在小组中寻找重叠的时间

1 个答案:

数据