R Different Prediction Result for Formula Containing "%>%" - r

Question

In R, when I use "predict" to get a confidence interval for a certain x (x=42) under the model: y = (centered x) + (centered x)^2. I found two possible ways:
model1 = lm(y ~ scale(x, center=T, scale=F) + I( (scale(x, center=T, scale=F))^2 ), data=data)
model2 = lm(y ~ (x %>% scale(center=T, scale=F)) + I( (x %>% scale(center=T, scale=F))^2 ), data=data)
The summary results for the two models are the same. But when I ran:
predict(model1, data.frame(x=42), interval="confidence", level=0.95)
predict(model2, data.frame(x=42), interval="confidence", level=0.95)
The results are different. I am wondering why. Does R treat the above two formulas differently because of the usage of "%>%"?
The dataset is a practice dataset from Kutner's textbook SENIC.txt, y is the 11th column, x is the 12th column.

Answer 1

The issue here is with scale and not %>%. scale returns a matrix which seems to affect the outcome.
One solution is to write vector --> vector equivalent of scale and use it:
library("magrittr") # for `%>%`
set.seed(1)
dataset = data.frame(x = rnorm(30))
dataset[["y"]] = 1 + (3 * dataset[["x"]]) + rnorm(30, mean = 0, sd = 0.1)
scale_vector = function(x, ...){
stopifnot(inherits(x, "numeric"))
scale(x, ...)[, 1]
}
lm(y ~ scale_vector(x, center=T, scale=F) + I( (scale_vector(x, center=T, scale=F))^2 ), data=dataset)
#>
#> Call:
#> lm(formula = y ~ scale_vector(x, center = T, scale = F) + I((scale_vector(x,
#> center = T, scale = F))^2), data = dataset)
#>
#> Coefficients:
#> (Intercept)
#> 1.27423
#> scale_vector(x, center = T, scale = F)
#> 2.99296
#> I((scale_vector(x, center = T, scale = F))^2)
#> -0.01645
lm(y ~ (x %>% scale_vector(center=T, scale=F)) + I( (x %>% scale_vector(center=T, scale=F))^2 ), data=dataset)
#>
#> Call:
#> lm(formula = y ~ (x %>% scale_vector(center = T, scale = F)) +
#> I((x %>% scale_vector(center = T, scale = F))^2), data = dataset)
#>
#> Coefficients:
#> (Intercept)
#> 1.27423
#> x %>% scale_vector(center = T, scale = F)
#> 2.99296
#> I((x %>% scale_vector(center = T, scale = F))^2)
#> -0.01645
Besides, if you do not mind using tidyverse, this might be cleaner:
library("magrittr") # for `%>%`
set.seed(1)
dataset = tibble::tibble(x = rnorm(30),
y = 1 + (3 * x) + rnorm(30, mean = 0, sd = 0.1)
)
dataset %>%
dplyr::mutate(x_scaled = scale(x, center = TRUE, scale = FALSE)[, 1]) %>%
lm(y ~ x_scaled + I(x_scaled^2), data = .)
#>
#> Call:
#> lm(formula = y ~ x_scaled + I(x_scaled^2), data = .)
#>
#> Coefficients:
#> (Intercept) x_scaled I(x_scaled^2)
#> 1.27423 2.99296 -0.01645