[R] Iterative Proportional Fitting, use

Koen Hufkens koen.hufkens at ua.ac.be
Mon Mar 23 14:32:42 CET 2009


The data I used was just an example to work upon.

My real dataset is a confusion matrix of 24x24 (and 17x17), so coding it
into a model with all different kinds of combinations seems tedious.
That's why I hoped to use the ipf() function as it accepts a matrix as
input.

Thanks for the suggestion but I hope I can get around this without
recoding the original data.

Kind regards,
Koen

-----Original Message-----
From: Gerard M. Keogh <GMKeogh at justice.ie>
To: Koen Hufkens <koen.hufkens at ua.ac.be>
Cc: r-help <r-help at r-project.org>, r-help-bounces at r-project.org
Subject: Re: [R] Iterative Proportional Fitting, use
Date: Mon, 23 Mar 2009 13:11:15 +0000

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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