It is not possible, apply() converts its argument to an array. You might be
able to use split() and lapply() to solve your problem.
On Tue, Feb 7, 2023, 07:52 Naresh Gurbuxani <naresh_gurbux...@hotmail.com>
wrote:
>
> > 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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