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: "agreg2 at gmail.com" <agreg2 at gmail.com>
To: smartpink111 at 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
