median time period
Typical - you post to the list and then work it out for yourself! Anyway here's my solution Toy code as before then: intervalsMonths <- 12 * intervals$year + intervals$month #convert whole years to months then add the remaining months for that entry in intervals medianMonths <- median(as.numeric(intervalsMonths)) Best iain
--- On Wed, 29/6/11, Iain Gallagher <iaingallagher at btopenworld.com> wrote:
From: Iain Gallagher <iaingallagher at btopenworld.com>
Subject: [R] median time period
To: r-help at r-project.org
Date: Wednesday, 29 June, 2011, 16:24
Hello List
I'm trying to calculate the median period (in months) of a
set of time intervals (between two interventions).
I have been playing with the lubridate package to create
the intervals but I can't think of the right approach to get
the median timeperiod.
Toy code:
library(lubridate)
test <- c('08-04-22', '08-07-28', '09-03-02',
'09-03-03', '09-01-30', '09-03-09', '10-02-24', '10-03-05')
test <- ymd(test)
intervals <- as.period(test[5:8] - test[1:4])
intervals
[1] 9 months and 8 days? ? 7 months and 9
days? ? 11 months and 22 days
[4] 1 year and 2 days
How can I convert this 'period' object to months? From
there I think I should just convert to 'numeric' and
calculate the median.
Garrett if you're out there - great package but could you
help please!?
Best
iain
sessionInfo()
R version 2.13.0 (2011-04-13) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_GB.utf8? ? ???LC_NUMERIC=C? ? ? ? ? ??? [3] LC_TIME=en_GB.utf8? ? ? ? LC_COLLATE=en_GB.utf8? ? [5] LC_MONETARY=C? ? ? ? ? ???LC_MESSAGES=en_GB.utf8??? [7] LC_PAPER=en_GB.utf8? ? ???LC_NAME=C? ? ? ? ? ? ? ? [9] LC_ADDRESS=C? ? ? ? ? ? ? LC_TELEPHONE=C? ? ? ? ??? [11] LC_MEASUREMENT=en_GB.utf8 LC_IDENTIFICATION=C? ? ? attached base packages: [1] stats? ???graphics? grDevices utils? ???datasets? methods???base? ??? other attached packages: [1] lubridate_0.2.4 loaded via a namespace (and not attached): [1] plyr_1.5.2? stringr_0.4
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