如何将 "X hours/days ago" 转换为标准日期时间

Question

我抓取了新闻数据，包括以下形式的时间：

time <- c("11 hours ago", "2 days ago", "3 days ago")

如何将其转换为标准日期时间格式？顺便说一句：我假设对于日内差异（例如“11 小时前”），浏览器可以识别我的系统时间？由于新闻来自全球各地。

谢谢

Answer 1

润滑剂溶液；

library(lubridate)
    
time <- c("11 hours ago", "2 days ago", "3 days ago")

time_numeric <- as.numeric(gsub("([0-9]+).*$", "\1", time))
time_logical <- ifelse(grepl('hour',time),'hour',ifelse(grepl('day',time),'day','unnknown'))

time_tidy <- data.frame(diff=time_numeric,type=time_logical)

current_time <- Sys.time()

time_tidy$new <- as_datetime(ifelse(time_tidy$type=='hour',current_time-hours(time_tidy$diff),ifelse(time_tidy$type=='day',current_time-days(time_tidy$diff),current_time)))

time_tidy$new

输出；

[1] "2021-08-30 00:29:21 +03"
[1] "2021-08-28 11:29:21 +03"
[1] "2021-08-27 11:29:21 +03"

Answer 2

如果你只有小时和天作为时间单位，那么，

Sys.time() - ifelse(grepl('hours', time), 
                   as.numeric(gsub('\D+', '', time)) * 3600, 
                               as.numeric(gsub('\D+', '', time)) * 24 * 3600)

#[1] "2021-08-30 00:31:32 +03" "2021-08-28 11:31:32 +03" "2021-08-27 11:31:32 +03"

Answer 3

去掉ago前面加-就可以用seq了。这将适用于 sec、min、hour、day、DSTday、week、month、quarter 或 year.

lapply(sub(" ago", "", time), function(x) seq(Sys.time(), by=paste0("-", x),
 length.out = 2)[2])
#[[1]]
#[1] "2021-08-29 23:41:26 CEST"
#
#[[2]]
#[1] "2021-08-28 10:41:26 CEST"
#
#[[3]]
#[1] "2021-08-27 10:41:26 CEST"

要获得向量，请使用 c 和 do.call:

do.call(c, lapply(sub(" ago", "", time), function(x) seq(Sys.time(),
  by=paste0("-",x), length.out = 2)[2]))
#[1] "2021-08-30 00:11:15 CEST" "2021-08-28 11:11:15 CEST"
#[3] "2021-08-27 11:11:15 CEST"

如何将 "X hours/days ago" 转换为标准日期时间

How to transform "X hours/days ago" into standard date time

r

datetime

as.date