weighting means

?weighted.mean
?tapply
Q8 <- rnorm(100)
Faculty <- factor(c(rep(1,50), rep(2,50)))
w <- 1:100
tapply(Q8, Faculty, weighted.mean, w)

Hi everyone,

I've got a dataframe called 'faculty'. I want to do a weighted mean on
the column called 'Q8' weighted by the contents of column 'CETP'. In
addition, I need to operate on the result of splitting 'faculty'
according to the contents of a column 'FACULTY'. For example

lapply(split(faculty$Q8, faculty$FACULTY), mean)

$"1"
[1] 2.2

$"2"
[1] 1.888889

gives the mean of 'Q8' after splitting by 'FACULTY'. When I try this
with weighted.mean, it doesn't work because I don't know how to specify
the result of the split as the argument for weighted.mean. For example,

lapply(split(faculty$Q8, faculty$FACULTY), weighted.mean)

$"1"
[1] 2.2

$"2"
[1] 1.888889

gives the same results as above because I haven't specified the vector
to use for weighting. The following doesn't work at all for the obvious
reason that weighted.mean is not operating on the result of the split.

lapply(split(faculty$Q8, faculty$FACULTY), weighted.mean(faculty$Q8,

faculty$CETP))
Error in match.fun(FUN) : not function, character, or symbol:
"weighted.mean(faculty$Q8, faculty$CETP)"

Any ideas?

-Tim

--
Tim Wilson      |   Visit Sibley online:   | Check out:
Henry Sibley HS |  http://www.isd197.org   | http://www.zope.com
W. St. Paul, MN |                          | http://slashdot.org
wilson at visi.com |  <dtml-var pithy_quote>  | http://linux.com
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What about:

?weighted.mean
?tapply
Q8 <- rnorm(100)
Faculty <- factor(c(rep(1,50), rep(2,50)))
w <- 1:100
tapply(Q8, Faculty, weighted.mean, w)

         1          2 
-0.0976302 -0.1457895 


Kjetil Halvorsen

Tim Wilson wrote:

Hi everyone,

I've got a dataframe called 'faculty'. I want to do a weighted mean on
the column called 'Q8' weighted by the contents of column 'CETP'. In
addition, I need to operate on the result of splitting 'faculty'
according to the contents of a column 'FACULTY'. For example

lapply(split(faculty$Q8, faculty$FACULTY), mean)

$"1"
[1] 2.2

$"2"
[1] 1.888889

gives the mean of 'Q8' after splitting by 'FACULTY'. When I try this
with weighted.mean, it doesn't work because I don't know how to specify
the result of the split as the argument for weighted.mean. For example,

lapply(split(faculty$Q8, faculty$FACULTY), weighted.mean)

$"1"
[1] 2.2

$"2"
[1] 1.888889

gives the same results as above because I haven't specified the vector
to use for weighting. The following doesn't work at all for the obvious
reason that weighted.mean is not operating on the result of the split.

lapply(split(faculty$Q8, faculty$FACULTY), weighted.mean(faculty$Q8,

faculty$CETP))
Error in match.fun(FUN) : not function, character, or symbol:
"weighted.mean(faculty$Q8, faculty$CETP)"

Any ideas?

-Tim

--
Tim Wilson      |   Visit Sibley online:   | Check out:
Henry Sibley HS |  http://www.isd197.org   | http://www.zope.com
W. St. Paul, MN |                          | http://slashdot.org
wilson at visi.com |  <dtml-var pithy_quote>  | http://linux.com

a<-c(rep("A",3),rep("B",7))
b<-c(1:10)
c<-c(rep("X",2),rep("Y",8))
w<-c(10:1)
z<-data.frame(a,b,c,w)
library(Hmisc)
summary(b~a+c,data=z,method="cross",fun=function(y)wtd.mean(y,weights=z$w))

On Wed, 21 Aug 2002 12:10:44 -0400
kjetil halvorsen <kjetilh at umsanet.edu.bo> wrote:

What about:

?weighted.mean
?tapply
Q8 <- rnorm(100)
Faculty <- factor(c(rep(1,50), rep(2,50)))
w <- 1:100
tapply(Q8, Faculty, weighted.mean, w)

         1          2 
-0.0976302 -0.1457895 


Kjetil Halvorsen

Tim Wilson wrote:

Hi everyone,

I've got a dataframe called 'faculty'. I want to do a weighted mean on
the column called 'Q8' weighted by the contents of column 'CETP'. In
addition, I need to operate on the result of splitting 'faculty'
according to the contents of a column 'FACULTY'. For example

lapply(split(faculty$Q8, faculty$FACULTY), mean)

$"1"
[1] 2.2

$"2"
[1] 1.888889

gives the mean of 'Q8' after splitting by 'FACULTY'. When I try this
with weighted.mean, it doesn't work because I don't know how to specify
the result of the split as the argument for weighted.mean. For example,

lapply(split(faculty$Q8, faculty$FACULTY), weighted.mean)

$"1"
[1] 2.2

$"2"
[1] 1.888889

gives the same results as above because I haven't specified the vector
to use for weighting. The following doesn't work at all for the obvious
reason that weighted.mean is not operating on the result of the split.

lapply(split(faculty$Q8, faculty$FACULTY), weighted.mean(faculty$Q8,

faculty$CETP))
Error in match.fun(FUN) : not function, character, or symbol:
"weighted.mean(faculty$Q8, faculty$CETP)"

Any ideas?

-Tim

--
Tim Wilson      |   Visit Sibley online:   | Check out:
Henry Sibley HS |  http://www.isd197.org   | http://www.zope.com
W. St. Paul, MN |                          | http://slashdot.org
wilson at visi.com |  <dtml-var pithy_quote>  | http://linux.com

Also take a look at the many weighted stats functions in the Hmisc library.  Do 
library(Hmisc)
?wtd.mean

-Frank

-- 
Frank E Harrell Jr              Prof. of Biostatistics & Statistics
Div. of Biostatistics & Epidem. Dept. of Health Evaluation Sciences
U. Virginia School of Medicine  http://hesweb1.med.virginia.edu/biostat
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html
Send "info", "help", or "[un]subscribe"
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Hello to all

Is there a way to get weighted means using summary.formula() from Hmisc?

For example, the following doesn't work:

a<-c(rep("A",3),rep("B",7))
b<-c(1:10)
c<-c(rep("X",2),rep("Y",8))
w<-c(10:1)
z<-data.frame(a,b,c,w)
library(Hmisc)
summary(b~a+c,data=z,method="cross",fun=function(y)wtd.mean(y,weights=z$w))

Error in wtd.mean(y, weights = z$w) : dim<- length of dims do not match the length of object

Thanks for the replies,

Oscar Rueda

22/08/02 1:16:15, Frank E Harrell Jr <fharrell at virginia.edu> escribi:

On Wed, 21 Aug 2002 12:10:44 -0400
kjetil halvorsen <kjetilh at umsanet.edu.bo> wrote:

What about:

?weighted.mean
?tapply
Q8 <- rnorm(100)
Faculty <- factor(c(rep(1,50), rep(2,50)))
w <- 1:100
tapply(Q8, Faculty, weighted.mean, w)

         1          2 
-0.0976302 -0.1457895 


Kjetil Halvorsen

Tim Wilson wrote:

Hi everyone,

I've got a dataframe called 'faculty'. I want to do a weighted mean on
the column called 'Q8' weighted by the contents of column 'CETP'. In
addition, I need to operate on the result of splitting 'faculty'
according to the contents of a column 'FACULTY'. For example

lapply(split(faculty$Q8, faculty$FACULTY), mean)

$"1"
[1] 2.2

$"2"
[1] 1.888889

gives the mean of 'Q8' after splitting by 'FACULTY'. When I try this
with weighted.mean, it doesn't work because I don't know how to specify
the result of the split as the argument for weighted.mean. For example,

lapply(split(faculty$Q8, faculty$FACULTY), weighted.mean)

$"1"
[1] 2.2

$"2"
[1] 1.888889

gives the same results as above because I haven't specified the vector
to use for weighting. The following doesn't work at all for the obvious
reason that weighted.mean is not operating on the result of the split.

lapply(split(faculty$Q8, faculty$FACULTY), weighted.mean(faculty$Q8,

faculty$CETP))
Error in match.fun(FUN) : not function, character, or symbol:
"weighted.mean(faculty$Q8, faculty$CETP)"

Any ideas?

-Tim

--
Tim Wilson      |   Visit Sibley online:   | Check out:
Henry Sibley HS |  http://www.isd197.org   | http://www.zope.com
W. St. Paul, MN |                          | http://slashdot.org
wilson at visi.com |  <dtml-var pithy_quote>  | http://linux.com

Also take a look at the many weighted stats functions in the Hmisc library.  Do 
library(Hmisc)
?wtd.mean

-Frank

-- 
Frank E Harrell Jr              Prof. of Biostatistics & Statistics
Div. of Biostatistics & Epidem. Dept. of Health Evaluation Sciences
U. Virginia School of Medicine  http://hesweb1.med.virginia.edu/biostat
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html
Send "info", "help", or "[un]subscribe"
(in the "body", not the subject !)  To: r-help-request at stat.math.ethz.ch
_._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._