What are you trying to do? Why use apply when there is already a vector
addition operation?
df$x+df$y or as.numeric(df$x)+as.numeric(df$y) or
rowSums(as.numeric(df[c('x','y')])).
As noted in other answers, apply will coerce your data frame to a matrix, and
all entries of a matrix must have the same type.
Regards,
Jorgen Harmse.
Message: 1
Date: Tue, 7 Feb 2023 07:51:50 -0500
From: Naresh Gurbuxani <naresh_gurbux...@hotmail.com>
To: "r-help@r-project.org" <r-help@R-project.org>
Subject: [R] preserve class in apply function
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<ia1p223mb049955f19238028d7a818cb2fa...@ia1p223mb0499.namp223.prod.outlook.com>
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> 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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