R version 3.1.1 (2014-07-10) -- "Sock it to Me"
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> 
>  foo <- read.table("squirrel.txt", header = TRUE)
>  foo
   count active passive
1      1      R       S
2      5      R       T
3      8      R       U
4      9      R       V
5      0      R       W
6     29      S       R
7     14      S       T
8     46      S       U
9      4      S       V
10     0      S       W
11     0      T       R
12     0      T       S
13     0      T       U
14     0      T       V
15     0      T       W
16     2      U       R
17     3      U       S
18     1      U       T
19    38      U       V
20     2      U       W
21     0      V       R
22     0      V       S
23     0      V       T
24     0      V       U
25     1      V       W
26     9      W       R
27    25      W       S
28     4      W       T
29     6      W       U
30    13      W       V
>  bar <- xtabs(count ~ active + passive, data = foo)
>  bar
      passive
active  R  S  T  U  V  W
     R  0  1  5  8  9  0
     S 29  0 14 46  4  0
     T  0  0  0  0  0  0
     U  2  3  1  0 38  2
     V  0  0  0  0  0  1
     W  9 25  4  6 13  0
> 
>  # Distribution of genital display in a colony of six squirrel monkeys,
>  # as reported by Ploog (1967), taken from Fienberg (1980, The Analysis
>  # of Cross-Classified Categorical Data, MIT Press), Table 8-2.
> 
>  # diagonal elements are "structural zeros" or "fixed zeros"
>  # (monkeys do not display to themselves -- the don't have mirrors)
>  # xtabs apparently does not have this concept
> 
>  # we can fit this with loglin by specifying starting values that have
>  # zero on the diagonal
>  sss <- (bar + 1)^0
>  diag(sss) <- 0
>  sss
      passive
active R S T U V W
     R 0 1 1 1 1 1
     S 1 0 1 1 1 1
     T 1 1 0 1 1 1
     U 1 1 1 0 1 1
     V 1 1 1 1 0 1
     W 1 1 1 1 1 0
>  lout <- loglin(bar, list(1, 2), start = sss, fit = TRUE)
4 iterations: deviation 0.05112509 
>  lout
$lrt
[1] 135.1691

$pearson
[1] NaN

$df
[1] 25

$margin
$margin[[1]]
[1] "active"

$margin[[2]]
[1] "passive"


$fit
      passive
active           R           S           T           U           V           W
     R  0.00000000  5.25991683  2.48071962  8.21546265  6.64803183  0.39576634
     S 19.18802539  0.00000000 10.32281246 34.18632223 27.66390257  1.64686959
     T  0.00000000  0.00000000  0.00000000  0.00000000  0.00000000  0.00000000
     U 10.93448629 12.47289692  5.88255691  0.00000000 15.76454884  0.93848495
     V  0.21996467  0.25091226  0.11833704  0.39189979  0.00000000  0.01887913
     W  9.65752365 11.01627399  5.19557398 17.20631533 13.92351676  0.00000000

>  # loglin gets the fit right, I think, but does not count the structural
>  # zeros correctly
>  # OTOH, glm has no problem (because the structural zeros just are not
>  # there in the data to confuse it.
>  gout <- glm(count ~ active + passive, family = poisson, data = foo)
>  sout <- summary(gout)
>  sout

Call:
glm(formula = count ~ active + passive, family = poisson, data = foo)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-5.6438  -0.8886  -0.1449   0.9290   4.7316  

Coefficients:
             Estimate Std. Error z value Pr(>|z|)    
(Intercept)    1.5285     0.2745   5.568 2.58e-08 ***
activeS        1.4257     0.2395   5.952 2.65e-09 ***
activeT      -18.6578  1362.2115  -0.014  0.98907    
activeU        0.8636     0.2631   3.282  0.00103 ** 
activeV       -3.0428     1.0228  -2.975  0.00293 ** 
activeW        0.7393     0.2490   2.969  0.00299 ** 
passiveS       0.1316     0.2548   0.516  0.60557    
passiveT      -0.6199     0.2592  -2.391  0.01679 *  
passiveU       0.5776     0.2110   2.738  0.00618 ** 
passiveV       0.3658     0.2030   1.803  0.07146 .  
passiveW      -2.4554     0.6002  -4.091 4.29e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for poisson family taken to be 1)

    Null deviance: 433.14  on 29  degrees of freedom
Residual deviance: 135.17  on 19  degrees of freedom
AIC: 227.43

Number of Fisher Scoring iterations: 15

>  names(sout)
 [1] "call"           "terms"          "family"         "deviance"      
 [5] "aic"            "contrasts"      "df.residual"    "null.deviance" 
 [9] "df.null"        "iter"           "deviance.resid" "coefficients"  
[13] "aliased"        "dispersion"     "df"             "cov.unscaled"  
[17] "cov.scaled"    
>  sout$deviance
[1] 135.1691
>  sout$df.residual
[1] 19
>  pchisq(sout$deviance, sout$df.residual, lower.tail = FALSE)
[1] 1.52349e-19
> 
>  # This does not get the right degrees of freedom either because of
>  # "solutions at infinity".  More on this later.
>  
>  
> 
> proc.time()
   user  system elapsed 
  0.205   0.024   0.348