weight cases?
I missed your second question. See ?cov.wt
On 10/14/06, Gabor Grothendieck <ggrothendieck at gmail.com> wrote:
Try this (and round the result to make to it comparable to your calculation): xtabs(weight ~ var1 + var2, my.data) On 10/14/06, Adrian Dusa <dusa.adrian at gmail.com> wrote:
Thanks for this Gabor, Sometimes weights can take various values, like 0.9
rep(letters[1:3], c(1, 0.9, 1.6))
[1] "a" "c" What if the weight variable would be: my.data$weight <- c(0.4, 2, 1.3, 0.9, 1) The way I found the solution was to compute the unweighted table, then find the weight for each unique combination and multiply that with the corresponding row-column entry in the table. The solution though is not very satisfactory: my.data$var1 <- as.factor(my.data$var1) my.data$var2 <- as.factor(my.data$var2) total <- expand.grid(levels(my.data$var1), levels(my.data$var2)) rowsmy.data <- apply(unique(my.data[,1:2]), 1, paste, collapse="") rowstotal <- apply(total, 1, paste, collapse="") total$weight <- 0 total$weight[sapply(rowsmy.data, function(x) which(rowstotal == x))] <- unique(my.data)[,3] (unweighted <- table(my.data$var1, my.data$var2)) round(unweighted*total$weight, 0) Yet another question: how would the weight variable be applied to correlate two numerical variables? Best, Adrian On Saturday 14 October 2006 16:00, Gabor Grothendieck wrote:
Try this:
table(lapply(my.data, rep, my.data$weight)[1:2])
On 10/14/06, Adrian Dusa <dusa.adrian at gmail.com> wrote:
Dear all, This is probably a stupid question for which I have a solution, which unfortunately is not as straighforward as I'd like. I wonder if there's a simple way to apply a weighting variable for the cases of a dataframe (well I'm sure there is, I just cannot find it). My toy example:
my.data <- data.frame(var1=c("c", "e", "a", "d", "b"),
var2=c("E", "B", "A", "C", "D"),
weight=c(1, 2, 1, 1, 1))
table(my.data$var1, my.data$var2)
A B C D E a 1 0 0 0 0 b 0 0 0 1 0 c 0 0 0 0 1 d 0 0 1 0 0 e 0 1 0 0 0 Applying the weight variable, the table should yield a value of 2 for the "eB" combination:
table(my.data$var1, my.data$var2)
A B C D E
a 1 0 0 0 0
b 0 0 0 1 0
c 0 0 0 0 1
d 0 0 1 0 0
e 0 2 0 0 0
Thanks in advance,
Adrian
--
Adrian Dusa
Romanian Social Data Archive
1, Schitu Magureanu Bd
050025 Bucharest sector 5
Romania
Tel./Fax: +40 21 3126618 \
+40 21 3120210 / int.101
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--
Adrian Dusa
Romanian Social Data Archive
1, Schitu Magureanu Bd
050025 Bucharest sector 5
Romania
Tel./Fax: +40 21 3126618 \
+40 21 3120210 / int.101