extracting standardized coefficients from lm in R

Question

My apologies for the dumb question...but I can\'t seem to find a simple solution

I want to extract the standardized coefficients from a fitted linear model (in R) there

Answer 1

There is a convenience function in the QuantPsyc package for that, called lm.beta. However, I think the easiest way is to just standardize your variables. The coefficients will then automatically be the standardized "beta"-coefficients (i.e. coefficients in terms of standard deviations).

For instance,

 lm(scale(your.y) ~ scale(your.x), data=your.Data)

will give you the standardized coefficient.

Are they really the same? The following illustrates that both are identical:

library("QuantPsyc")
mod <- lm(weight ~ height, data=women)
coef_lmbeta <- lm.beta(mod)

coef_lmbeta
> height 
  0.9955 

mod2 <- lm(scale(weight) ~ scale(height), data=women)
coef_scale <- coef(mod2)[2]

coef_scale
> scale(height) 
  0.9955 

all.equal(coef_lmbeta, coef_scale, check.attributes=F)
[1] TRUE

which shows that both are identical, as they should be.

How to avoid clumsy variable names? In case you don't want to deal with these clumsy variable names such as scale(height), one option is to standardize the variables outside the lm call in the dataset itself. For instance,

women2 <- lapply(women, scale) # standardizes all variables

mod3 <- lm(weight ~ height, data=women2)
coef_alt <- coef(mod3)[2]
coef_alt
> height 
  0.9955 

all.equal(coef_lmbeta, coef_alt)
[1] TRUE

How do I standardize multiple variables conveniently? In the likely event that you don't want to standardize all variables in your dataset, you could pick out all that occur in your formula. For instance, referring to the mtcars-dataset now (since women only contains height and weight):

Say the following is the regression model I want to estimate:

 modelformula <- mpg ~ cyl + disp + hp + drat + qsec

We can use the fact that all.vars gives me a vector of the variable names.

 all.vars(modelformula)
 [1] "mpg"  "cyl"  "disp" "hp"   "drat" "qsec"

We can use this to subset the dataset accordingly. For instance,

mycars <- lapply(mtcars[, all.vars(modelformula)], scale) 

will give me a dataset in which all variables have been standardized. Linear regressions using mycars will now give standardized betas. Please make sure that standardizing all these variables makes sense, though!

Potential issue with only one variable: In case you model formula only contains one explanatory variable and you are working with the built-in dataframes (and not with tibbles), the following adjustment is advisable (credits go to @JerryT in the comments):

mycars <- lapply(mtcars[, all.vars(modelformula), drop=F], scale) 

This is because when you extract only one column from a standard data frame, R retuns a vector instead of a dataframe. drop=F will prevent this from happening. This also won't be a problem if e.g. tibbles are used. See e.g.

class(mtcars[, "mpg"])
[1] "numeric"
class(mtcars[, "mpg", drop=F])
[1] "data.frame"
library(tidyverse)
class(as.tibble(mtcars)[, "mpg"])
[1] "tbl_df"     "tbl"        "data.frame"

Another issue with missing values in the dataframe (credits go again to @JerryT in the comments): By default, R's lm removes all rows where at least one column is missing. scale, on the other hand, would take all values that are non-missing, even if an observation has a missing value in a different column. If you want to mimick the action of lm, you may want to first drop all rows with missing values, like so:

all_complete <- complete.cases(df)
df[all_complete,]

Answer 2

Package lm.beta has several functions to work with standardised coefficients, including lm.beta() which requires an lm object:

res <- lm(y~x)
lm.beta(res) 

Answer 3

Just use colnames(data) with lapply or sapply.
For example:

lapply(data[, colnames(data)], scale)