Às 18:40 de 02/06/2024, Rui Barradas escreveu:
Às 18:34 de 02/06/2024, Leo Mada via R-help escreveu:
Dear Shadee,
If you have a data.frame with the following columns:
n = 100; # population size
x = data.frame(
Sex = sample(c("M","F"), n, T),
Country = sample(c("AA", "BB", "US"), n, T),
Income = as.factor(sample(1:3, n, T))
)
# Dummy variable
ONE = rep(1, nrow(x))
r = aggregate(ONE ~ Sex + Income + Country, length, data = x)
r = r[, c("Country", "Income", "Sex")]
print(r)
It is possible to write more simple code, if you need only the
particular combination of variables (which you specified in your
mail). But this is the more general approach.
Note: you may want to use "sum" instead of "length", e.g. if you have
a column specifying the number of individuals in that category.
Hope this helps,
Leonard
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Hello,
The following is simpler.
r2 <- xtabs(~ ., x) |> as.data.frame()
r2[-4L] # or r2[names(r2) != "Freq"]
Hope this helps,
Rui Barradas
Hello,
This is the same solution but the code to keep only the columns in the
original data set is better. And it's a MRE.
n <- 100; # population size
x <- data.frame(
Sex = sample(c("M","F"), n, T),
Country = sample(c("AA", "BB", "US"), n, T),
Income = as.factor(sample(1:3, n, T))
)
r2 <- xtabs(~ ., x) |> as.data.frame()
# no need for constants, find the columns
# to keep from the data
r2[names(r2) %in% names(x)]
Hope this helps,
Rui Barradas
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
