HI Greg,
Sorry, I misunderstand your question.
I am not sure whether it works with numSummary() from library(Rcmdr).
You could use other ways, such as:
test<-read.table(text="
id year incidents
100 1 0
101 1 1
102 1 21
100 1 27
101 1 3
102 1 12
100 2 5
101 2 5
102 2 19
100 2 10
101 2 2
102 2 12
100 3 0
101 3 0
102 3 22
100 3 14
101 3 16
102 3 13
",sep="",header=TRUE)
test1<-test
test1<-within(test1,{id<-factor(id);year<-factor(year)})
library(plyr)
ddply(test,.(id,year),summarise,mean=mean(incidents),sd=sd(incidents),n=length(incidents))
#or
by(test1$incidents,list(test1$id,test1$year),FUN=function(x)
c(mean=mean(x),sd=sd(x),n=length(x)))
#or
with(test,aggregate(incidents,by=list(id,year),function(x)
c(mean=mean(x),sd=sd(x),n=length(x))))
#or
library(data.table)
test1<- data.table(test)
test1[,list(mean=mean(incidents),sd=sd(incidents),n=length(incidents)),by=list(id,year)]
library(psych)
res<-describeBy(test[,3],group=list(test$id,test$year),mat=FALSE)
res
#: 100
#: 1
# var n mean sd median trimmed mad min max range skew kurtosis se
#1 1 2 13.5 19.09 13.5 13.5 20.02 0 27 27 0 -2.75 13.5
#------------------------------------------------------------
#: 101
#: 1
# var n mean sd median trimmed mad min max range skew kurtosis se
1 1 2 2 1.41 2 2 1.48 1 3 2 0 -2.75 1
------------------------------------------------------------------
A.K.
----- Original Message -----
From: "agr...@gmail.com" <agr...@gmail.com>
To: smartpink...@yahoo.com
Cc:
Sent: Friday, December 28, 2012 6:09 PM
Subject: thanks -- Re: syntax for identifying more than one
Hi Arun,
Thanks for responding to my posted question. Your example showed how to use
numsummary to produce mean incidents, sd incidents, etc. for each level of "id"
and for each level of "year." I'm trying to generate these descriptive stats
for each *combination* of "id" and "year." I could do that in SAS with the
following code
proc means mean std n data=test;
by id year;
var incidents;
output out = testsummary
mean = mnincidents
std = sdincidents
n = nincidnets;
proc print data=testsummary;
Again, thanks for responding to the posted question.
Best wishes,
Greg
<quote author='arun kirshna'>
HI,
You could try that in a list if that is okay.
test<-read.table(text="
id year incidents
100 1 0
101 1 1
102 1 21
103 1 27
104 1 3
105 1 12
100 2 5
101 2 5
102 2 19
103 2 10
104 2 2
105 2 12
100 3 0
101 3 0
102 3 22
103 3 14
104 3 16
105 3 13
",sep="",header=TRUE)
library(e1071)
library(Rcmdr)
res<- lapply(seq_len(ncol(test[,-3])),function(i)
numSummary(test[,3],groups=test[,i]))
names(res)<-names(test)[-3]
res
$id
# mean sd IQR 0% 25% 50% 75% 100% data:n
#100 1.666667 2.8867513 2.5 0 0.0 0 2.5 5 3
#101 2.000000 2.6457513 2.5 0 0.5 1 3.0 5 3
#102 20.666667 1.5275252 1.5 19 20.0 21 21.5 22 3
#103 17.000000 8.8881944 8.5 10 12.0 14 20.5 27 3
#104 7.000000 7.8102497 7.0 2 2.5 3 9.5 16 3
#105 12.333333 0.5773503 0.5 12 12.0 12 12.5 13 3
#
#$year
# mean sd IQR 0% 25% 50% 75% 100% data:n
#1 10.666667 11.325487 17.25 0 1.50 7.5 18.75 27 6
#2 8.833333 6.177918 6.50 2 5.00 7.5 11.50 19 6
#3 10.833333 8.953584 12.25 0 3.25 13.5 15.50 22 6
A.K.
</quote>
Quoted from:
http://r.789695.n4.nabble.com/syntax-for-identifying-more-than-one-group-in-numsummary-tp4654172p4654185.html
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