如何在 r 中将日期和时间列拆分为单独的日、月、小时、分钟、秒、星期几列？

Question

我正在尝试将日期和时间列拆分为单独的日、月、小时、分钟、秒、星期几列。 我正在使用 lubridate 和 mutuate function 但是当我尝试使用以下代码时，我收到此错误：警告消息：所有格式都无法解析。 未找到格式。

我的新列已创建，但它们都包含 NA - 想知道是否有人可以提供帮助？

我的专栏是这样的：

tpep_pickup_datetime

01/07/2019 00:51:15
01/07/2019 00:46:30
01/07/2019 00:25:35

我的代码是这样的：

taxidata3 <- taxidata2 %>%
  mutate(tpep_pickup_datetime = mdy_hms(tpep_pickup_datetime), 
         day = day(tpep_pickup_datetime),
         month = month(tpep_pickup_datetime), 
         year = year(tpep_pickup_datetime),
         dayofweek = wday(tpep_pickup_datetime),
         hour = hour(tpep_pickup_datetime),
         minute = minute(tpep_pickup_datetime),
         second = second(tpep_pickup_datetime))

Answer 1

根据 OP 的评论，日期格式是 day/month/... 而不是 month/day/... 这里，我们需要dmy_hms 。 因此，每个字母表示出现的顺序

library(lubridate)
library(dplyr)
taxidata3 <- taxidata2 %>%
  mutate(tpep_pickup_datetime = dmy_hms(tpep_pickup_datetime), 
         day = day(tpep_pickup_datetime),
         month = month(tpep_pickup_datetime), 
         year = year(tpep_pickup_datetime),
         dayofweek = wday(tpep_pickup_datetime),
         hour = hour(tpep_pickup_datetime),
         minute = minute(tpep_pickup_datetime),
         second = second(tpep_pickup_datetime))

Answer 2

这是一个使用正则表达式匹配日期组件的stringr解决方案：

数据：

df <- data.frame(
  tpep_pickup_datetime = c("01/07/2019 00:51:15", "01/07/2019 00:46:30", "01/07/2019 00:25:35")
)

解决方案：

library(stringr)
df$day <- str_extract(df$tpep_pickup_datetime, "^\\d{2}") 
df$month <- str_extract(df$tpep_pickup_datetime, "(?<=/)\\d{2}")
df$year <- str_extract(df$tpep_pickup_datetime, "\\d{4}")
df$hour <- str_extract(df$tpep_pickup_datetime, "(?<= )\\d{2}(?=:)")
df$minute <- str_extract(df$tpep_pickup_datetime, "(?<=:)\\d{2}(?=:)")
df$second <- str_extract(df$tpep_pickup_datetime, "(?<=:)\\d{2}$")

结果：

df
  tpep_pickup_datetime day month year hour minute second
1  01/07/2019 00:51:15  01    07 2019   00     51     15
2  01/07/2019 00:46:30  01    07 2019   00     46     30
3  01/07/2019 00:25:35  01    07 2019   00     25     35

Answer 3

这是使用separate的 function 的另一种选择。 代码和output如下：-

library(tidyverse)
df <- data.frame(
    tpep_pickup_datetime = c("01/07/2019 00:51:15", "01/07/2019 00:46:30", 
                             "01/07/2019 00:25:35"))

df %>% 
    separate(tpep_pickup_datetime, c("Day", "Month", "Year_time"),
             sep = "/", remove = FALSE) %>% 
    separate(Year_time, c("Year", "Time"),
             sep = " ", remove = TRUE) %>% 
    separate(Time, c("Hour", "Minute", "Second"),
             sep = ":", remove = TRUE) 

# tpep_pickup_datetime Day Month Year Hour Minute Second
#1  01/07/2019 00:51:15  01    07 2019   00     51     15
#2  01/07/2019 00:46:30  01    07 2019   00     46     30
#3  01/07/2019 00:25:35  01    07 2019   00     25     35