library(survey)
<- 100
N <- data.frame(x1 = rnorm(N), x2 = rnorm(N))
df $y <- 1 + 3 * df$x1 - 2 * df$x2 + rnorm(N, 0, 0.5)
df
<- lm(y ~ x1 + x2, data = df) fit
I still use a version of this occasionally, but the various marginal effects package – specifically `marignaleffects’ seem to replicate most of the functionality I need these days.
One of the few features I miss from Stata is the very-intuitive lincom
command. Most of the time I recreate that functionality with survey::svycontrast
. But, I always forget the syntax. This code demonstrates a minimal working example.
Having fit the model, we can pass unnamed vector with the right number of coefficients to get our desired linear combination:
svycontrast(fit, c(0, 1, 1))
contrast SE
contrast 0.88835 0.0639
Or a named vector with any number of coefficients (as long as the names match). Or a list of named vectors. One “gotcha” to keep in mind: I have found that the latter syntax fails for some versions of survey
and may be OS-dependent.
svycontrast(fit, list(c("x2" = 1, "x1" = 1), c("x2" = 4, "x1" = 2)))
contrast SE
[1,] 0.88835 0.0639
[2,] -2.32800 0.2075
If you want an easy way to pass svycontrast
a data.frame
of where each row is a different linear combination (and each column a different variable), below is how you do that. This is usually if you want to produce similar behavior to, say, running predict
on a fitted model but A) the predict call for that model doesn’t return standard errors (as with felm
or fixest
) and/or B) you only want to linearly combine some of the variables (equivalent to setting all other coefficients = 0). In either case, this will work:
<- function(fit, newdata) {
svycontrast_df # Call surveyconstrast with a data frame
<- newdata
df
# Transform data.frame into a list of its row vectors
<- as.list(as.data.frame(t(newdata)))
df_list <- lapply(df_list, setNames, colnames(df)) # Set all character vector names inside list
df_list
# Return a named list
setNames(as.list(as.data.frame(svycontrast(fit, df_list))), c("est", "se"))
}
<- data.frame(x1 = seq(-3, 3, 1))
newdata $x2 <- newdata$x^2
newdatasvycontrast_df(fit, newdata)
$est
[1] -27.2932457 -14.0907987 -4.9930501 0.0000000 0.8883517 -2.3279951
[7] -9.6490403
$se
[1] 0.44004435 0.20524188 0.06296086 0.00000000 0.06389291 0.20753339 0.44365681