On Aug 8, 2011, at 7:51 PM, Jeffrey Joh wrote:
Here is a sample of what I'm trying to do:
structure(list(C_lo = c(0.00392581816943354, 0.00901222644518829,
0.00484396253385175, 0.00822377400482716, 0.00780070460187192,
0.00952688235337435), C_hi = c(0.00697755827622381,
0.0123301031600017,
0.0113207627868435, 0.0112887993422598, 0.018567245397701,
0.0195253894885054
), house = c(1, 1, 1, 1, 1, 1), date = c(719, 1027, 1027,
1027, 1030, 1030), hour = c(18, 8, 8, 8, 11, 11), .Names =
c("1000", "10000",
"10001", "10002", "10003", "10004"), press = structure(c(1L,
1L, 1L, 1L, 1L, 1L), .Names = c("1000", "10000",
"10001", "10002", "10003", "10004"), .Label = c("DEPR",
"PRESS"), class = "factor")), .Names = c("C_lo", "C_hi",
"house", "date", "hour", "number", "press"
), class = "data.frame", row.names = c("1000", "10000",
"10001", "10002", "10003", "10004"))
I'd like to aggregate the data by the date. I'd like to have a
table with the median C_lo and C_hi values grouped by date.
I'd also like to plot these points with date on the x-axis, C on y-
axis, and lines going through these medians.
> aggregate(results[, c("C_lo", "C_hi")], results["date"], median)
date C_lo C_hi
1 719 0.003925818 0.006977558
2 1027 0.008223774 0.011320763
3 1030 0.008663793 0.019046317
dat <-.Last.value
matplot(dat[1], dat[-1], type="b", ylim=with(results, range(c(C_lo,
C_hi)) ) )
with(results, points(date, C_lo))
with(results, points(date, C_hi))
--
David.
For plyr, would it be something like: ddply(results, .(date),median,
na.rm=T)
I tried making a for loop to get the medians, but that doesn't work
either.
splitresults = split (results, results$date, drop=T)
mediann <- matrix (,seq_along(splitresults),2)
for (i in seq_along(splitresults)) {
piece <- splitresults[[i]]
mediann [i,1] <- unique(piece$date)
mediann [i,2] <- median (piece$n, na.rm=T)
}
Jeff
----------------------------------------
Date: Fri, 5 Aug 2011 11:59:37 -0700
Subject: Re: [R] Aggregating data
From: djmu...@gmail.com
To: johjeff...@hotmail.com
CC: r-help@r-project.org
Hi:
This is the type of problem at which the plyr package excels. Write a
utility function that produces the plot you want using a data frame
as
its input argument, and then do something like
library('plyr')
d_ply(results, .(a, b, c), plotfun)
where plotfun is a placeholder for the name of the name of your plot
function. The d in d_ply means to take a data frame as input and _
means return nothing. This is used in particular when a side effect,
such as a plot, is the desired 'output'. See
http://www.jstatsoft.org/v40/i01, which contains an example
(baseball)
where groupwise plots are produced. (Don't actually run the example
unless you're willing to wait for 1100+ ggplots to be rendered :)
If memory serves, you should also be able to produce graphics for
each
data subset using the data.table package as well.
If you want a more concrete solution, provide a more concrete
example.
HTH,
Dennis
On Fri, Aug 5, 2011 at 9:55 AM, Jeffrey Joh
<johjeff...@hotmail.com> wrote:
I aggregated my data: aggresults <-aggregate(results,
by=list(results$a, results$b, results$c), FUN=mean, na.rm=TRUE)
results has about 8000 lines of data, and aggresults has about 80
lines. I would like to create a separate variable for each of the
80 aggregates, each containing the 100 lines that were aggregated.
I would also like to create plots for each of those 80 datasets.
Is there a way of automating this, so that I don't have to do each
of the 80 aggregates individually?
Jeff
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______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
