Question

我需要解析多个日期和时间。这是一个例子：

library(dplyr)
library(tidyr)
library(lubridate)

date.called <- c("28-Feb","10/1/16","8/13/15")
call.times <- c("912","1513","922, 1810")
df<- data.frame(call.times,date.called,stringsAsFactors = F )
df <- df %>% 
separate(call.times, c("time.1", "time.2"), sep = "\\,", remove=FALSE) %>%
  mutate(time.1 = trimws(time.1), time.2 = trimws(time.2))
#parse each new date and time
df %>% mutate(time.1 = ifelse( !is.na(time.1) ,
                             parse_date_time(sprintf("%s %04s", date.called, 
                             gsub('^([0-9]+)([0-9]{2})$', '\\1:\\2', time.1)), 
                             c('%m/%d/%y %H:%M','%d-%b %H%M')), time.1),
    time.2 = ifelse( !is.na(time.2) ,
                             parse_date_time(sprintf("%s %04s", date.called, 
                             gsub('^([0-9]+)([0-9]{2})$', '\\1:\\2', time.2)), 
                             c('%m/%d/%y %H:%M','%d-%b %H%M')), time.2) )

我得到df的以下输出：

  call.times       time.1     time.2 date.called
1        912 -62162174880       <NA>      28-Feb
2       1513   1475334780       <NA>     10/1/16
3  922, 1810   1439457720 1439489400     8/13/15

相反，我试图生成此输出：

  call.times              time.1              time.2 date.called
1        912 0000-02-28 09:12:00                <NA>      28-Feb
2       1513 2016-10-01 15:13:00                <NA>     10/1/16
3  922, 1810 2015-08-13 09:22:00 2015-08-13 18:10:00     8/13/15

感谢您的帮助。

Answer 1

我会做

df %>% mutate_at(vars(starts_with("time")), function(x) 
  parse_date_time(
    sprintf("%s %04s", date.called, gsub('^([0-9]+)([0-9]{2})$', '\\1:\\2', x)), 
    c('%m/%d/%y %H:%M','%d-%b %H%M')
))
#   call.times              time.1              time.2 date.called
# 1        912 0000-02-28 09:12:00                <NA>      28-Feb
# 2       1513 2016-10-01 15:13:00                <NA>     10/1/16
# 3  922, 1810 2015-08-13 09:22:00 2015-08-13 18:10:00     8/13/15

Answer 2

正如LukeA在评论中所说，ifelse剥离了属性，因此删除了日期格式。您可以通过在parse_date_time之外移动ifelse函数来规避此行为。修改后的代码版本会提供您想要的结果：

df %>% 
  mutate(time.1 = parse_date_time(ifelse(!is.na(time.2), 
                                         sprintf("%s %04s", date.called, gsub('^([0-9]+)([0-9]{2})$', '\\1:\\2', time.1)),
                                         NA), 
                                  c('%m/%d/%y %H:%M','%d-%b %H%M')),
         time.2 = parse_date_time(ifelse(!is.na(time.2), 
                                         sprintf("%s %04s", date.called, gsub('^([0-9]+)([0-9]{2})$', '\\1:\\2', time.2)),
                                         NA), 
                                  c('%m/%d/%y %H:%M','%d-%b %H%M'))
         )

结果：

  call.times              time.1              time.2 date.called
1        912 0000-02-28 09:12:00                <NA>      28-Feb
2       1513 2016-10-01 15:13:00                <NA>     10/1/16
3  922, 1810 2015-08-13 09:22:00 2015-08-13 18:10:00     8/13/15

解析dplyr中的日期时间

2 个答案: