Thank you all for your help,
This is what I was able to do
aggregate(items,items[2],mean)
Where items has all my data, and items[2] has my group id number.
Here are my 2 questions
1) items[1] has a unquie number in it, how do I tell mean to take the mean
of items[3] not items[1]
2) Is there an
Hi all,
I have data that looks like
number|grouping
I would like to preform stats for each grouping
so
1|5
6|5
10|5
11|5
3|9
5|9
10|9
Say I would like to take the median for above, I should be returned 2 lines,
one for group #5 and one for group #9
Does this make sense?
I am sorry for the
# I am sure there is a better way than this
a - c(1,6,10,11,3,5,10)
group - c(5,5,5,5,9,9,9)
my.df - cbind(a,group)
a.5 - subset(my.df, group==5)
median(a.5[,1])
a.9 - subset(my.df, group==9)
median(a.9[,1])
# MASS 4 is a good book
On Fri, Aug 1, 2008 at 1:30 PM, Lotta R [EMAIL PROTECTED] wrote:
?aggregate
Something like this should do it.
aggregate(xx[,1], list(xx[,2], median)
--- On Fri, 8/1/08, stephen sefick [EMAIL PROTECTED] wrote:
From: stephen sefick [EMAIL PROTECTED]
Subject: Re: [R] simple help request
To: Lotta R [EMAIL PROTECTED]
Cc: r-help@r-project.org
Received
Dear Lotta,
Try
my.df - data.frame(a - c(1,6,10,11,3,5,10), group - c(5,5,5,5,9,9,9))
tapply(my.df$a,my.df$group,median)
5 9
8 5
See ?tapply and/or ?aggregate for more information.
HTH,
Jorge
On Fri, Aug 1, 2008 at 1:30 PM, Lotta R [EMAIL PROTECTED] wrote:
Hi all,
I have data that
5 matches
Mail list logo
