This question is a "why", not a how. In the following code I'm trying to understand why dplyr::mutate
evaluates one custom function (f()
) with the entire vector but not with the other custom function (g()
). What exactly is mutate
doing?
set.seed(1);sum(rnorm(100, c(0, 10, 100)))
f=function(m) {
set.seed(1)
sum(rnorm(100, mean=m))
}
g <- function(m) sin(m)
df <- data.frame(a=c(0, 10, 100))
y1 <- mutate(df, asq=a^2, fout=f(a), gout=g(a))
y2 <- rowwise(df) %>%
mutate(asq=a^2, fout=f(a), gout=g(a))
y3 <- group_by(df, a) %>%
summarize(asq=a^2, fout=f(a), gout=g(a))
For all three columns, asq
, fout
, and gout
, evaluation is rowwise in y2
and y3
and the results are identical. However, y1$fout
is 3640.889 for all three rows, which is the result of evaluating sum(rnorm(100, c(0, 10, 100)))
. So the function f()
is evaluating the entire vector for each row.
A closely related question has been asked elsewhere mutate/transform in R dplyr (Pass custom function), but the "why" was not explained.
sin
and ^
are vectorized, so they natively operate on each individual value, rather than on the whole vector of values. f
is not vectorized. But you can do f = Vectorize(f)
and it will operate on each individual value as well.
y1 <- mutate(df, asq=a^2, fout=f(a), gout=g(a))
y1
a asq fout gout 1 0 0 3640.889 0.0000000 2 10 100 3640.889 -0.5440211 3 100 10000 3640.889 -0.5063656
f = Vectorize(f)
y1a <- mutate(df, asq=a^2, fout=f(a), gout=g(a))
y1a
a asq fout gout 1 0 0 10.88874 0.0000000 2 10 100 1010.88874 -0.5440211 3 100 10000 10010.88874 -0.5063656
Some additional info on vectorization here, here, and here.
We can loop through each element of 'a' using map
and apply the function f
library(tidyverse)
df %>%
mutate(asq = a^2, fout = map_dbl(a, f), gout = g(a))
# a asq fout gout
#1 0 0 10.88874 0.0000000
#2 10 100 1010.88874 -0.5440211
#3 100 10000 10010.88874 -0.5063656
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