Keon,
why not fit a loglinear independence model which as far as I know is the
same.
Gerard
Here's an example from Agresti - Intro to Cat Data analysis
Example: Alcohol, cigarette, marijuana use
|------------------+------------------+------------------------------------|
| Alcohol | Cigarette | Marijuana Use |
| | | |
| use | use | |
|------------------+------------------+------------------------------------|
| | | Yes No |
|------------------+------------------+------------------------------------|
| Yes | Yes | 911 538 |
|------------------+------------------+------------------------------------|
| | No | 44 456 |
|------------------+------------------+------------------------------------|
| No | Yes | 3 43 |
|------------------+------------------+------------------------------------|
| | No | 2 279 |
|------------------+------------------+------------------------------------|
Coding and Models
table8.3 = read.table(textConnection("alc cig mar count
Yes Yes Yes 911
Yes Yes No 538
Yes No Yes 44
Yes No No 456
No Yes Yes 3
No Yes No 43
No No Yes 2
No No No 279"),header=TRUE)
closeAllConnections()
# independence model (A,C,M)
fit1.a.c.m = glm(count ~ mar+cig+alc, family=poisson, data=table8.3)
fit1.glm$fitted.values
# intermediate model
fit2.m.ca = glm(count ~ mar+cig:alc, family=poisson, data=table8.3)
fit2.m.ca$fitted.values
# homogeneous association model
fit3.m.c.a = glm(count ~ mar:cig+mar:alc+cig:alc, family=poisson,
data=table8.3)
fit3.m.c.a$fitted.values
# saturated model
fits = glm(count ~ mar*cig*alc, family=poisson, data=table8.3)
fits$fitted.values
The coding for variables in the above program and the fitted values are
given below ? they show that the homogeneous association model is the only
model that fits these data well.
|---------+------------+------------+--------+------------+---------+-----------|
| Alcohol | Cigarette | Marijuana | Actual | (A,C,M) | (AC,M) | (AC:AM:CM)|
| use | Use | Use | (ACM) | Independenc| | homogeneou|
| | | | | e | | s |
|---------+------------+------------+--------+------------+---------+-----------|
| Yes | Yes | Yes | 911 | 540.0 | 611.2 | 910.4 |
|---------+------------+------------+--------+------------+---------+-----------|
| | | No | 538 | 740.2 | 837.8 | 538.6 |
|---------+------------+------------+--------+------------+---------+-----------|
| | No | Yes | 44 | 282.1 | 210.9 | 44.6 |
|---------+------------+------------+--------+------------+---------+-----------|
| | | No | 456 | 386.7 | 289.1 | 455.4 |
|---------+------------+------------+--------+------------+---------+-----------|
| No | Yes | Yes | 3 | 90.6 | 19.4 | 3.6 |
|---------+------------+------------+--------+------------+---------+-----------|
| | | No | 43 | 124.2 | 26.6 | 42.4 |
|---------+------------+------------+--------+------------+---------+-----------|
| | No | Yes | 2 | 47.3 | 118.5 | 1.4 |
|---------+------------+------------+--------+------------+---------+-----------|
| | | No | 279 | 64.9 | 162.5 | 279.6 |
|---------+------------+------------+--------+------------+---------+-----------|
Koen Hufkens
<koen.hufkens at ua.
ac.be> To
Sent by: r-help <r-help at r-project.org>
r-help-bounces at r- cc
project.org
Subject
[R] Iterative Proportional Fitting,
23/03/2009 12:13 use
Hi list,
I would like to normalize a matrix (two actually for comparison) using
iterative proportional fitting.
Using ipf() would be the easiest way to do this, however I can't get my
head around the use of the function. More specifically, the margins
settings...
for a matrix:
mat <- matrix(c(65,4,22,24,6,81,5,8,0,11,85,19,4,7,3,90),4,4)
using
fit <- ipf(mat,margins=c(1,1,1,1,0,1,1,1,1))
generates a matrix with just 1's.
using
fit <- ipf(mat,margins=c(100,100,100,100,0,100,100,100,100))
gives a segmentation fault and crashes R !
so how do you define the margin values to which to sum the row and
column values in your matrix correctly?
Kind regards,
Koen
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