# Iterative proportional fitting in R

The mission

I am trying to find a way to do Iterative Proportional Fitting in R. The logic of the procedure is like this: one has a table with e.g. sample distribution of some variables. Let us say it is this one:

sample1 <- structure(c(6L, 14L, 46L, 16L, 6L, 21L, 62L, 169L, 327L, 174L,
44L, 72L, 43L, 100L, 186L, 72L, 23L, 42L), .Dim = c(6L, 3L), .Dimnames = list(
c("Primary", "Lowersec", "Highersec", "Highershort", "Higherlong",
"University"), c("B", "F", "W")))


Another table is from some other source, say another sample:

sample2 <- structure(c(171796L, 168191L, 240671L, 69168L, 60079L, 168169L,
954045L, 1040981L, 1872732L, 726410L, 207366L, 425786L, 596239L,
604826L, 991640L, 323215L, 134066L, 221696L), .Dim = c(6L, 3L
), .Dimnames = list(c("Primary", "Lowerse", "Highersec", "Highershort",
"Higherlong", "University"), c("B", "F", "W")))


Now, we want to preserve the relationships between variables found in sample1, but we want to apply these relationships to the marginal distribution we find in sample2. Iterative Proportional Fitting does this as is described here (I could not possibly offer a better explanation). I have tried doing it in LEM, with the following result:

                    B         F        W
Primary     124204.64  960173.6 637701.7
Lowerse     119749.12 1081459.0 612789.9
Highersec   336934.21 1792001.6 976107.2
Highershort  90512.27  736464.1 291816.6
Higherlong   43486.91  238593.0 119431.0
University  163186.85  418628.6 233835.5


I am not 100% sure that this result is, but chances are (say 99%) that it is. The odds ratios from the first table are preserved in the resulting table, while the marginal distributions (row and column sums) are identical to the second input table.

The problem

Strangely, this quite useful algorithm is not readily available in R, at least not in a user-friendly form. One function that is likely to be relevant is cat::ipf(). However, I cannot figure out how to use the margins= argument. I am certainly not alone in this question. The help example uses a 3-dimensional table, which makes things even more confusing.

In addition, there exist some user-written functions, one is here and the other is to be found here. The first one gives an erroneous result, unfortunately. The second one is also very non-transparent, requiring specifically pre-formatted CSV files as input, instead of R matrix objects.

The question

1. Can anyone please explain how to actually use the cat::ipf() function?
2. Are there any alternative functions to achieve the IPF adjustment task, using matrices as input?
3. (SOLVED) Can this function be fixed to deliver a proper result?

Thank you.

ADDENDUM: I was able to get a proper output from the function in (3). After some consideration in turned out that the function does not accept matrix as input for marginal distribution, but simply a list of these marginal distributions. So practically the question is solved. However, a proper answer for 1 and 2 would be of use to the larger community, since IPF is quite essential in loglinear models.

Also reproducing a working IPF function here seems like a good idea, for those who search in the future:

ipf <- function(Margins_, seedAry, maxiter=100, closure=0.001) {
#Check to see if the sum of each margin is equal
MarginSums. <- unlist(lapply(Margins_, sum))
if(any(MarginSums. != MarginSums.[1])) warning("sum of each margin
not equal")

#Replace margin values of zero with 0.001
Margins_ <- lapply(Margins_, function(x) {
if(any(x == 0)) warning("zeros in marginsMtx replaced with
0.001")
x[x == 0] <- 0.001
x
})

#Check to see if number of dimensions in seed array equals the number of
#margins specified in the marginsMtx
numMargins <- length(dim(seedAry))
if(length(Margins_) != numMargins) {
stop("number of margins in marginsMtx not equal to number of
margins in seedAry")
}

#Set initial values
resultAry <- seedAry
iter <- 0
marginChecks <- rep(1, numMargins)
margins <- seq(1, numMargins)

#Iteratively proportion margins until closure or iteration criteria are met
while((any(marginChecks > closure)) & (iter < maxiter)) {
for(margin in margins) {
marginTotal <- apply(resultAry, margin, sum)
marginCoeff <- Margins_[[margin]]/marginTotal
marginCoeff[is.infinite(marginCoeff)] <- 0
resultAry <- sweep(resultAry, margin, marginCoeff, "*")
marginChecks[margin] <- sum(abs(1 - marginCoeff))
}
iter <- iter + 1
}

#If IPF stopped due to number of iterations then output info
if(iter == maxiter) cat("IPF stopped due to number of iterations\n")

#Return balanced array
resultAry
}


Note that this function will not accept a matrix for marginal distribution. The following will help:

m1 <- rowSums(sample2)
m2 <- colSums(sample2)
m <- list(m1,m2)


And then supply m as the first argument.

-

## migrated from stackoverflow.comMay 15 '13 at 13:41

This question came from our site for professional and enthusiast programmers.

This look more like a statistics question (or at least 1. and maybe 2.). You could flag the question for transfer to CrossValidated... – Paul Hiemstra May 15 '13 at 8:43
@PaulHiemstra I have considered that. It is the matter of proper input for 1, general R knowledge for 2 and pure programming for 3. – Maxim.K May 15 '13 at 8:52
Maybe even ask there instead of here? They are more into R and Matlab than the general SO user and be in a position to be able to offer you direction. – hd1 May 15 '13 at 8:57
Another function for IPF is rake from the survey package. – Henrico May 15 '13 at 9:26
@Henrico I have seen it, but it is for survey weights, requires surveydesign object as input, and thus works in an entirely different fashion. – Maxim.K May 15 '13 at 9:36

Make a glm fit to the marginals with Poisson errors (yielding a log-linear model) and then use predict on expand.grid data.frame from the the row and column values based of the second sample. (There's no particular advantage that I can see in using IPF to estimate a log-linear model of this sort.)

require(reshape2)
> melt(sample1)
Var1 Var2 value
1      Primary    B     6
2     Lowersec    B    14
3    Highersec    B    46
4  Highershort    B    16
5   Higherlong    B     6
6   University    B    21
7      Primary    F    62
8     Lowersec    F   169
9    Highersec    F   327
10 Highershort    F   174
11  Higherlong    F    44
12  University    F    72
13     Primary    W    43
14    Lowersec    W   100
15   Highersec    W   186
16 Highershort    W    72
17  Higherlong    W    23
18  University    W    42
> m_sample1<- melt(sample1)
> glm( value ~ Var1+Var2, data=m_sample1)

Call:  glm(formula = value ~ Var1 + Var2, data = msample)

Coefficients:
(Intercept)    Var1Highersec  Var1Highershort     Var1Lowersec      Var1Primary
-36.56           162.00            63.00            70.00            12.67
Var1University            Var2F            Var2W
20.67           123.17            59.50

Degrees of Freedom: 17 Total (i.e. Null);  10 Residual
Null Deviance:      121200
Residual Deviance: 22510    AIC: 197.4


That was the linear model. This is the multiplicative (log-linear) model:

> glm( value ~ Var1+Var2, data= m_sample1, family="poisson")

Call:  glm(formula = value ~ Var1 + Var2, family = "poisson", data = m_sample1)

Coefficients:
(Intercept)    Var1Highersec  Var1Highershort     Var1Lowersec      Var1Primary
1.7213           2.0357           1.2779           1.3550           0.4191
Var1University            Var2F            Var2W
0.6148           2.0515           1.4528

Degrees of Freedom: 17 Total (i.e. Null);  10 Residual
Null Deviance:      1287
Residual Deviance: 21.05    AIC: 139.1

> predict(glm( value ~ Var1+Var2, data=msample,family="poisson"), data.frame(Var1="Lowersec", Var2="B") )
1
3.076272


Edit; More detail requested:

Multiply the grand sum by combinations of the appropriate entries from exp(coef(fit)). The non-Intercept entries in coef(fit) let you compute the estimated ratios of proportions in "non-corner" cells to the "corner cell". The Var1:University with Var2:F cell would have an estimate of exp( 1.7213 + 0.6148+ 2.0515) in the original model (which is what predict(fit) or predict(fit, expand.grid(data.frame( rows=rowMeans(m_sample1), cols=colMeans(m_sample1)))) should give you). You then need to multiply by the ratio of the grand sums of the new data to the grand sum of the fit data.

-
Would you please care to expand your answer to use expand.grid() as you mention? I don't see how prediction in its current form leads to the derivation of the resulting table above. Thanks. – Maxim.K May 16 '13 at 10:29
Please don't mind me pinging (one and only time) to see if you have missed the follow up question. Thanks! (And sorry for bothering if you chose not to elaborate). – Maxim.K May 22 '13 at 14:47