Às 12:51 de 07/02/2023, Naresh Gurbuxani escreveu:
Consider a data.frame whose different columns have numeric, character,
and factor data. In apply function, R seems to pass all elements of a
row as character. Is it possible to preserve numeric class?
mydf <- data.frame(x = rnorm(10), y = runif(10))
apply(mydf, 1, function(row) {row["x"] + row["y"]})
[1] 0.60150197 -0.74201827 0.80476392 -0.59729280 -0.02980335 0.31351909
[7] -0.63575990 0.22670658 0.55696314 0.39587314
mydf[, "z"] <- sample(letters[1:3], 10, replace = TRUE)
apply(mydf, 1, function(row) {row["x"] + row["y"]})
Error in row["x"] + row["y"] (from #1) : non-numeric argument to binary operator
apply(mydf, 1, function(row) {as.numeric(row["x"]) + as.numeric(row["y"])})
[1] 0.60150194 -0.74201826 0.80476394 -0.59729282 -0.02980338 0.31351912
[7] -0.63575991 0.22670663 0.55696309 0.39587311
apply(mydf[,c("x", "y")], 1, function(row) {row["x"] + row["y"]})
[1] 0.60150197 -0.74201827 0.80476392 -0.59729280 -0.02980335 0.31351909
[7] -0.63575990 0.22670658 0.55696314 0.39587314
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Hello,
The last form,
apply(mydf[,c("x", "y")], 1, function(row) {row["x"] + row["y"]})
is the right one. If your data has columns of a mix of classes, then the
rows which are vectors are coerced to the greatest common denominator
class.
From ?c:
Details
The output type is determined from the highest type of the components in
the hierarchy NULL < raw < logical < integer < double < complex <
character < list < expression.
Also, since you have a data.frame the following is another possible way:
apply(mydf[c("x", "y")], 1, function(row) {row["x"] + row["y"]})
This doesn't work with matrices.
Hope this helps,
Rui Barradas
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https://stat.ethz.ch/mailman/listinfo/r-help
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and provide commented, minimal, self-contained, reproducible code.
