Suppose I have two dependent categorical attributes A, B. I have a dataframe X that lists probabilities (or expected counts) for all combinations of all categories of A and B. Let's assume two categories in each attribute, the dataframe could then look like this:

X <- data.frame(A = c(1,1,2,2), B = 1:2, PROB = c(0.1,0.3,0.2,0.4))


1 1 1   .1
2 1 2   .3
3 2 1   .2
4 2 2   .4

Now I would like to add a B column to another dataframe Y that has only an A column. Y exists and has many more rows than X, and also additional columns. A synthetic Y could look like this:

Y <- data.frame(A=sample(2, size=10000, replace=TRUE),
                C=sample(5, size=10000, replace=TRUE),
                D=sample(3, size=10000, replace=TRUE))

The contents of the new B column should be sampled at random using the conditional probabilities for B given the value of the A column.

I am new to R, and I was wondering whether this operation can be handled more elegantly than writing a loop that iterates over all A categories. Perhaps a dedicated model class for which predict could be applied? Also, I lack the knowledge of statistical terminology -- I would appreciate any hint on how this kind of imputation (?) is referred to in the literature.

  • $\begingroup$ PROB is the overall probability, it sums up to 1 in my toy dataframe. In Y$B, I want the value 2 to appear about three times as often as the value 1 in those rows that have A=1. For the rows with A=2, this ratio should be 2:1 (=0.4/0.2). $\endgroup$ – krlmlr Sep 22 '11 at 11:58
  • $\begingroup$ OK, so in the final dataframe only these conditional amounts should count and PROB won't come out the same as here. That had me stumped for a bit because your toy Y has equal probability 1 and 2 for A. $\endgroup$ – John Sep 22 '11 at 14:22

It seems to me that if Y doesn't actually already exist you probably want to just make it from scratch like this.

Y <- X[sample(1:nrow(X), size = 10000, replace = TRUE, prob = X$PROB),]
Y <- Y[,-3]

BTW, you can test to see that you got what you want with...


If Y really does already exist then yes, you have to go through the unique values of A, but not all values of A.

b <- lapply(unique(X$A), function(x){
    n <- nrow(Y[Y$A==x,])
    df <- X[X$A == x,]
    Y$B[Y$A==x] <<- with( df, sample(B, n, replace = TRUE, prob = PROB) )

In that case b is garbage. A for loop version of this wouldn't even make the garbage b

for(x in unique(X$A)){
    n <- nrow(Y[Y$A == x,])
    df <- X[X$A == x,]
    Y$B[Y$A == x] <- with( df, sample(B, n, replace = TRUE, prob = PROB) )

And thus... this might be a case where using a loop is the cleanest way to go. :)

I did some quick time testing of these two methods and the plyr version. The fastest is the for loop, followed by lapply, followed by (far in the back), ddply (50% with given dataset). The cost gets bigger as Y grows in either size or complexity such that the for loop can get 2 to 5x the performance of ddply and it uses much less memory.

  • $\begingroup$ +1 for the table test. Is there really no shortcut for this assignment in R? I was thinking about a dedicated model class for which predict could be applied. $\endgroup$ – krlmlr Sep 22 '11 at 12:07
  • $\begingroup$ predict couldn't work since it doesn't give random outcomes but fixed ones. $\endgroup$ – John Sep 22 '11 at 14:23

Use ddply (from plyr package) first to create the conditional probabilities:

Xcond<-ddply(X, .(A),

Now, given Y with column A, you can get the result you want e.g. by (if the order is not of importance to you):

ddply(Y, .(A), .fun=function(curdfr){
  curB<-sample(curOrg$B, size=nrow(curdfr), replace=TRUE, prob=curOrg$CPROB)

No doubt there are more elegant/efficient solutions. Read its help if you don't understand the ddply call - if that fails, get back to us.

Edit as per @John's suggestion: X can be used immediately instead of Xcond:

ddply(Y, .(A), .fun=function(curdfr){
  curB<-sample(curOrg$B, size=nrow(curdfr), replace=TRUE, prob=curOrg$PROB)
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  • $\begingroup$ +1 Thank you, that's definitely more elegant (and should be faster, too!) than looping through categories. $\endgroup$ – krlmlr Sep 22 '11 at 12:43
  • $\begingroup$ Actually, that's less efficient than looping through categories. $\endgroup$ – John Sep 22 '11 at 13:23
  • $\begingroup$ The first run to modify X is unnecessary. It will work fine with the original X. The values in prob for sample don't have to add to 1.0. The relative amounts will work. $\endgroup$ – John Sep 22 '11 at 13:25

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