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我已经从以下代码的其他 SO 答案中对信息进行了三角测量，但遇到了错误消息。搜索了类似的错误和解决方案，但无法弄清楚，因此感谢您的帮助。

对于每个组（“id”），我想获得连续行的开始时间之间的差异。

可重现的数据：

require(dplyr)
df <-data.frame(id=as.numeric(c("1","1","1","2","2","2")), 
            start= c("1/31/17 10:00","1/31/17 10:02","1/31/17 10:45", 
                             "2/10/17 12:00", "2/10/17 12:20","2/11/17 09:40"))
time <- strptime(df$start, format = "%m/%d/%y %H:%M")
df %>%
group_by(id)%>%
mutate(diff = time - lag(time),
     diff_mins = as.numeric(diff, units = 'mins'))

得到我的错误：

mutate_impl(.data, dots) 中的错误：列diff的长度必须为 3（组大小）或 1，而不是 6 另外：警告消息：在 unclass(time1) - unclass(time2) 中：更长的对象长度不是的倍数更短的物体长度

4

2 回答 2

1

Do you mean something like this?

There is no need for lag here, a simple diff on the grouped times is sufficient.

df %>%
    mutate(start = as.POSIXct(start, format = "%m/%d/%y %H:%M")) %>%
    group_by(id) %>%
    mutate(diff = c(0, diff(start)))
## A tibble: 6 x 3
## Groups:   id [2]
#     id start                diff
#  <dbl> <dttm>              <dbl>
#1    1. 2017-01-31 10:00:00    0.
#2    1. 2017-01-31 10:02:00    2.
#3    1. 2017-01-31 10:45:00   43.
#4    2. 2017-02-10 12:00:00    0.
#5    2. 2017-02-10 12:20:00   20.
#6    2. 2017-02-11 09:40:00 1280.
于 2018-05-12T05:41:10.343 回答
0

您可以使用lagand difftime（根据Hadley）：

df %>%
  mutate(time = as.POSIXct(start, format = "%m/%d/%y %H:%M")) %>%
  group_by(id) %>%
  mutate(diff = difftime(time, lag(time)))

# A tibble: 6 x 4
# Groups:   id [2]
     id start         time                diff  
  <dbl> <fct>         <dttm>              <time>
1    1. 1/31/17 10:00 2017-01-31 10:00:00 <NA>  
2    1. 1/31/17 10:02 2017-01-31 10:02:00 2     
3    1. 1/31/17 10:45 2017-01-31 10:45:00 43    
4    2. 2/10/17 12:00 2017-02-10 12:00:00 <NA>  
5    2. 2/10/17 12:20 2017-02-10 12:20:00 20    
6    2. 2/11/17 09:40 2017-02-11 09:40:00 1280
于 2018-05-12T06:04:08.423 回答