I have scraped news data, including time in form of:
time <- c("11 hours ago", "2 days ago", "3 days ago")
How can I translate this into a standard date time format? BTW: I assume that for intra-day differences (e.g. "11 hours ago") the browser recognizes my system time? Since news come from around the globe.
Thank you
If you only have hours and days as units of time, then,
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"
You can use seq when removing the ago and adding in front a -. This will work for times given as sec, min, hour, day, DSTday, week, month, quarter or 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"
To get a vector use c with 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"
a lubridate solution;
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
output;
[1] "2021-08-30 00:29:21 +03"
[1] "2021-08-28 11:29:21 +03"
[1] "2021-08-27 11:29:21 +03"