0
$\begingroup$
nsam=100
X1 = runif(nsam)
X2 = runif(nsam)
data = cbind(X1,X2)

pairmin = matrix(NA,nn*(nn-1)/2,3)
k = 0
for (i in 1:(nn-1)) {
  for (j in (i+1):nn) {
    k = k + 1
    pairmin[k,1:2] = pmin(data[i,],data[j,])
    pairmin[k,3] = sign((data[i,1]-data[j,1])*(data[i,2]-data[j,2]))
  }
}
pairmin = cbind(pairmin,pairmin[,3]*0.5 + 0.5)
pairmin = data.frame(pairmin)

I have two explanatory variables X1, X2 and zero-one variable X4. I could do logistic regression by

lm2 <- glm(X4 ~ X1 + X2, data = pairmin, family = "binomial")

but I tried

library(plyr)
pairmin1 = pairmin[order(pairmin[,1],pairmin[,2],pairmin[,3]),] 
freqdat = ddply(pairmin1,.(X1,X2,X3,X4),nrow)

lm22 <- glm(X4 ~ X1 + X2,weights=V1,data=freqdat, family = "binomial")

That is, counting the redundant observations and use them as weights. The reason I do this is to reduce computation time. (Actual sample size in my study is >100,000 and have to repeat the task several hundred time). I found that using redundancy counts as weights considerably reduces computation while maintaining same estimates. (As I found this in GAM : https://stackoverflow.com/questions/40101775/weights-option-in-gam)

But I got warnings. 1) Algorithms did not converge 2) Fitted probabilities are 0 or 1.

The response variables of this random data are concordances and discordances of all pairs, and it is obviously not 'completely separated.' as seen in the plot. (Blue is 1, and green is 0)

enter image description here

I think this is not complete separation. Then why does this happen?

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  • $\begingroup$ It's not "complete separation": You didn't get convergence, so you have no answers at all. $\endgroup$ – whuber Nov 9 '16 at 16:32
  • $\begingroup$ @whuber Then do you have any idea why it didn't converge? $\endgroup$ – user67275 Nov 14 '16 at 5:23
  • $\begingroup$ None at all, because you haven't shared your data. If you want your code debugged, first you need to reduce the data to a smaller set you can share and then you ought to consider posting the question on Stack Overflow. $\endgroup$ – whuber Nov 14 '16 at 14:44

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