我的数据框有带和不带秒的时间戳,并且在月份和小时前随机使用 0,即 01 或 1
library(tidyverse)
df <- data_frame(cust=c('A','A','B','B'), timestamp=c('5/31/2016 1:03:12', '05/25/2016 01:06',
'6/16/2016 01:03', '12/30/2015 23:04:25'))
cust timestamp
A 5/31/2016 1:03:12
A 05/25/2016 01:06
B 6/16/2016 01:03
B 12/30/2015 23:04:25
如何将小时数提取到单独的列中?所需的输出:
cust timestamp hours
A 5/31/2016 1:03:12 1
A 05/25/2016 01:06 1
B 6/16/2016 9:03 9
B 12/30/2015 23:04:25 23
我更喜欢 tidyverse 和 mutate 的答案,但我的尝试未能正确提取小时数:
df %>% mutate(hours=strptime(timestamp, '%H') %>% as.character() )
# A tibble: 4 × 3
cust timestamp hours
<chr> <chr> <chr>
1 A 5/31/2016 1:03:12 2016-10-31 05:00:00
2 A 05/25/2016 01:06 2016-10-31 05:00:00
3 B 6/16/2016 01:03 2016-10-31 06:00:00
4 B 12/30/2015 23:04:25 2016-10-31 12:00:00