Parameter and Model Standardization
Source:vignettes/standardize_parameters_effsize.Rmd
standardize_parameters_effsize.RmdIntroduction
Standardizing parameters (i.e., coefficients) can allow for their comparison within and between models, variables and studies. Moreover, as it returns coefficients expressed in terms of change of variance (for instance, coefficients expressed in terms of SD of the response variable), it can allow for the usage of effect size interpretation guidelines, such as Cohen’s (1988) famous rules of thumb.
However, standardizing a model’s parameters should not be automatically and mindlessly done: for some research fields, particular variables or types of studies (e.g., replications), it sometimes makes more sense to keep, use and interpret the original parameters, especially if they are well known or easily understood.
Critically, parameters standardization is not a trivial process. Different techniques exist, that can lead to drastically different results. Thus, it is critical that the standardization method is explicitly documented and detailed.
Standardizing Parameters of Simple Models
Standardized Associations
library(parameters)
library(effectsize)
m <- lm(rating ~ complaints, data = attitude)
standardize_parameters(m)> # Standardization method: refit
>
> Parameter | Std. Coef. | 95% CI
> ----------------------------------------
> (Intercept) | -9.80e-16 | [-0.21, 0.21]
> complaints | 0.83 | [ 0.61, 1.04]Standardizing the coefficient of this simple linear regression gives a value of
round(standardize_parameters(m)[2, 2], 2)> [1] 0.83But did you know that for a simple regression this is actually the same as a correlation? Thus, you can eventually apply some (in)famous interpretation guidelines (e.g., Cohen’s rules of thumb).
library(correlation)
correlation(attitude, select = c("rating", "complaints"))> # Correlation Matrix (pearson-method)
>
> Parameter1 | Parameter2 | r | 95% CI | t(28) | p
> -----------------------------------------------------------------
> rating | complaints | 0.83 | [0.66, 0.91] | 7.74 | < .001***
>
> p-value adjustment method: Holm (1979)
> Observations: 30Standardized Differences
How does it work in the case of differences, when factors are entered and differences between a given level and a reference level? You might have heard that it is similar to a Cohen’s d. Well, let’s see.
# Select portion of data containing the two levels of interest
mtcars$am <- factor(mtcars$am, labels = c("Manual", "Automatic"))
m <- lm(mpg ~ am, data = mtcars)
standardize_parameters(m)> # Standardization method: refit
>
> Parameter | Std. Coef. | 95% CI
> --------------------------------------------
> (Intercept) | -0.49 | [-0.87, -0.11]
> am [Automatic] | 1.20 | [ 0.60, 1.80]This linear model suggests that the standardized difference
between Manual (the reference level - the model’s intercept)
and Automatic is of 1.20 standard deviation of mpg
(because the response variable was standardized, right?). Let’s compute
the Cohen’s d between these two levels:
library(effectsize)
cohens_d(mpg ~ am, data = mtcars)> Cohen's d | 95% CI
> --------------------------
> -1.48 | [-2.27, -0.67]
>
> - Estimated using pooled SD.It is larger! Why? How? Both differences should be expressed in units of SD! But which SDs? Different SDs!
When looking at the difference between groups as a
slope, the standardized parameter is the difference
between the means in SD_{mpg}. That is,
the slope between Manual and
Automatic is a change of 1.20 SD_{mpg}s.
However, when looking a the difference as a distance between
two populations, Cohen’s d is the distance between the means in
units of pooled
SDs. That is, the distance between
Manual and Automatic is of 1.48 SDs of
each of the groups (here assumed to be equal).
In this simple model, the pooled SD is the residual SD, so we can also estimate Cohen’s d as:
> amAutomatic
> 1.5And we can also get an approximation of Cohen’s d by
converting the t-statistic from the
regression model via t_to_d():
> Parameter | Coefficient | SE | 95% CI | t(30) | p
> ---------------------------------------------------------------------
> (Intercept) | 17.15 | 1.12 | [14.85, 19.44] | 15.25 | < .001
> am [Automatic] | 7.24 | 1.76 | [ 3.64, 10.85] | 4.11 | < .001
t_to_d(4.11, df_error = 30)> d | 95% CI
> -------------------
> 1.50 | [0.68, 2.30]It is also interesting to note that using the smart
method (explained in detail below) when standardizing parameters will
give you indices equivalent to Glass’ delta,
which is a standardized difference expressed in terms of SD of the
reference group.
m <- lm(mpg ~ am, data = mtcars)
standardize_parameters(m, method = "smart")> # Standardization method: smart
>
> Parameter | Std. Coef. | 95% CI
> ------------------------------------------
> (Intercept) | 0.00 | [0.00, 0.00]
> am [Automatic] | 1.17 | [0.59, 1.76]
glass_delta(mpg ~ am, data = mtcars)> Glass' delta (adj.) | 95% CI
> ------------------------------------
> -1.10 | [-1.80, -0.37]… So note that some standardized differences are different than others! :)
Standardizing Parameters of Linear Models
As mentioned above, standardization of parameters can also be used to compare among parameters within the same model. Essentially, what prevents us from normally being able to compare among different parameters is that their underlying variables are on different scales.[^But also as noted above, this is not always an issue. For example, when the variables scale is important for the interpretation of results, standardization might in fact hinder interpretation!]
For example, in the following example, we use a liner regression
model to predict a worker’s salary (in Shmekels) from their age (years),
seniority (years), overtime (xtra_hours) and how many
compliments they give their boss (n_comps).
Let us explore the different parameter standardization methods provided by parameters.
Standardized Slopes are Not (Always) Correlations
We saw that in simple linear models, the standardized slope is equal to the correlation between the outcome and predictor - does this hold for multiple regression as well? As in each effect in a regression model is “adjusted” for the other ones, we might expect coefficients to be somewhat alike to partial correlations. Let’s first start by computing the partial correlation between numeric predictors and the outcome.
> salary xtra_hours n_comps age seniority is_senior
> 1 19745 4.2 1 32 3 FALSE
> 2 11302 1.6 0 34 3 FALSE
> 3 20636 1.2 3 33 5 TRUE
> 4 23047 7.2 1 35 3 FALSE
> 5 27342 11.3 0 33 4 FALSE
> 6 25657 3.6 2 30 5 TRUE
correlation(
hardlyworking,
select = "salary",
select2 = c("xtra_hours", "n_comps", "age", "seniority"),
partial = TRUE # get partial correlations
)> # Correlation Matrix (pearson-method)
>
> Parameter1 | Parameter2 | r | 95% CI | t(498) | p
> ------------------------------------------------------------------
> salary | xtra_hours | 0.87 | [0.85, 0.89] | 39.80 | < .001***
> salary | n_comps | 0.71 | [0.66, 0.75] | 22.40 | < .001***
> salary | age | 0.09 | [0.01, 0.18] | 2.10 | 0.037*
> salary | seniority | 0.19 | [0.10, 0.27] | 4.30 | < .001***
>
> p-value adjustment method: Holm (1979)
> Observations: 500Let’s compare these to the standardized slopes:
mod <- lm(salary ~ xtra_hours + n_comps + age + seniority,
data = hardlyworking
)
standardize_parameters(mod)> # Standardization method: refit
>
> Parameter | Std. Coef. | 95% CI
> ----------------------------------------
> (Intercept) | 2.77e-16 | [-0.03, 0.03]
> xtra hours | 0.77 | [ 0.73, 0.81]
> n comps | 0.39 | [ 0.36, 0.42]
> age | 0.04 | [ 0.00, 0.07]
> seniority | 0.08 | [ 0.04, 0.12]They are quite different! It seems then that standardized slopes in multiple linear regressions are not the same a correlations or partial correlations :(
However, not all hope is lost yet - we can still try and recover the partial correlations from our model, in another way: by converting the t-statistics (and their degrees of freedom, df) into a partial correlation coefficient r.
params <- model_parameters(mod)
t_to_r(params$t[-1], df_error = params$df_error[-1])> r | 95% CI
> -------------------
> 0.87 | [0.85, 0.89]
> 0.71 | [0.67, 0.74]
> 0.09 | [0.01, 0.18]
> 0.19 | [0.10, 0.27]Wow, the retrieved correlations coefficients from the regression model are exactly the same as the partial correlations we estimated above! So these “r” effect sizes can also be used.
Methods of Standardizing Parameters
Let’s convert age into a 3-level factor:
hardlyworking$age_g <- cut(hardlyworking$age,
breaks = c(25, 30, 35, 45)
)
mod <- lm(salary ~ xtra_hours + n_comps + age_g + seniority,
data = hardlyworking
)
model_parameters(mod)> Parameter | Coefficient | SE | 95% CI | t(494) | p
> -----------------------------------------------------------------------------
> (Intercept) | 9806.10 | 446.71 | [8928.41, 10683.79] | 21.95 | < .001
> xtra hours | 1221.35 | 30.72 | [1161.00, 1281.71] | 39.76 | < .001
> n comps | 2944.87 | 131.12 | [2687.25, 3202.48] | 22.46 | < .001
> age g [>30-35] | 393.36 | 241.01 | [ -80.18, 866.90] | 1.63 | 0.103
> age g [>35-45] | 597.61 | 427.72 | [-242.77, 1437.99] | 1.40 | 0.163
> seniority | 443.76 | 102.38 | [ 242.62, 644.91] | 4.33 | < .001It seems like the best or most important predictor is
n_comps as it has the coefficient. However, it is hard to
compare among predictors, as they are on different scales. To address
this issue, we must have all the predictors on the same scale - usually
in the arbitrary unit of standard deviations.
"refit": Re-fitting the model with
standardized data
This method is based on a complete model re-fit with a standardized version of data. Hence, this method is equal to standardizing the variables before fitting the model. It is the “purest” and the most accurate (Neter, Wasserman, and Kutner 1989), but it is also the most computationally costly and long (especially for heavy models such as Bayesian models, or complex mixed models). This method is particularly recommended for models that include interactions or transformations (e.g., exponentiation, log, polynomial or spline terms).
standardize_parameters(mod, method = "refit")> # Standardization method: refit
>
> Parameter | Std. Coef. | 95% CI
> -------------------------------------------
> (Intercept) | -0.05 | [-0.11, 0.02]
> xtra hours | 0.77 | [ 0.73, 0.81]
> n comps | 0.39 | [ 0.36, 0.43]
> age g [>30-35] | 0.06 | [-0.01, 0.14]
> age g [>35-45] | 0.10 | [-0.04, 0.23]
> seniority | 0.08 | [ 0.04, 0.12]standardize_parameters also has a robust
argument (default to FALSE), which enables a robust
standardization of the data, i.e., based on the
median and MAD instead of the
mean and SD:
standardize_parameters(mod, method = "refit", robust = TRUE)> # Standardization method: refit
>
> Parameter | Std. Coef. | 95% CI
> --------------------------------------------
> (Intercept) | -0.20 | [-0.27, -0.13]
> xtra hours | 0.65 | [ 0.62, 0.68]
> n comps | 0.82 | [ 0.74, 0.89]
> age g [>30-35] | 0.07 | [-0.01, 0.16]
> age g [>35-45] | 0.11 | [-0.05, 0.27]
> seniority | 0.12 | [ 0.07, 0.18]
>
> - Scaled by one MAD from the median.Note that since age_g is a factor, it is not numerically
standardized, and so it standardized parameter is still not directly
comparable to those of numeric variables. To address this, we can set
two_sd = TRUE, thereby scaling parameters on 2 SDs (or
MADs) of the predictors (Gelman 2008).
standardize_parameters(mod, method = "refit", two_sd = TRUE)> # Standardization method: refit
>
> Parameter | Std. Coef. | 95% CI
> -------------------------------------------
> (Intercept) | -0.05 | [-0.11, 0.02]
> xtra hours | 1.54 | [ 1.46, 1.61]
> n comps | 0.78 | [ 0.72, 0.85]
> age g [>30-35] | 0.06 | [-0.01, 0.14]
> age g [>35-45] | 0.10 | [-0.04, 0.23]
> seniority | 0.16 | [ 0.09, 0.24]
>
> - Scaled by two SDs from the mean.parameters also comes with a helper function that returns the re-fit model, without summarizing it, which can then be used as the original model would:
mod_z <- standardize(mod, two_sd = FALSE, robust = FALSE)
mod_z>
> Call:
> lm(formula = salary ~ xtra_hours + n_comps + age_g + seniority,
> data = data_std)
>
> Coefficients:
> (Intercept) xtra_hours n_comps age_g(30,35] age_g(35,45]
> -0.0458 0.7692 0.3921 0.0635 0.0964
> seniority
> 0.0821
model_parameters(mod_z)> Parameter | Coefficient | SE | 95% CI | t(494) | p
> ---------------------------------------------------------------------
> (Intercept) | -0.05 | 0.03 | [-0.11, 0.02] | -1.47 | 0.142
> xtra hours | 0.77 | 0.02 | [ 0.73, 0.81] | 39.76 | < .001
> n comps | 0.39 | 0.02 | [ 0.36, 0.43] | 22.46 | < .001
> age g [>30-35] | 0.06 | 0.04 | [-0.01, 0.14] | 1.63 | 0.103
> age g [>35-45] | 0.10 | 0.07 | [-0.04, 0.23] | 1.40 | 0.163
> seniority | 0.08 | 0.02 | [ 0.04, 0.12] | 4.33 | < .001
"posthoc": Refit without
refitting
Post-hoc standardization of the parameters aims at emulating the
results obtained by "refit" without refitting the model.
The coefficients are divided by the standard deviation (or MAD if
robust) of the outcome (which becomes their expression
‘unit’). Then, the coefficients related to numeric variables are
additionally multiplied by the standard deviation (or MAD if
robust) of the related terms, so that they correspond to
changes of 1 SD of the predictor (e.g., “A change in 1 SD of x
is related to a change of 0.24 of the SD of y). This does not
apply to binary variables or factors, so the coefficients are still
related to changes in levels. This method is not accurate and tend to
give aberrant results when interactions are specified.
But why is interaction for posthoc-standardization problematic? A regression model estimates coefficient between two variables when the other predictors are at 0 (are fixed at 0, that people interpret as “adjusted for”). When a standardized data is passed (in the refit method), the effects and interactions are estimated at the means of the other predictors (because 0 is the mean for a standardized variable). Whereas in posthoc-standardization, this coefficient corresponds to something different (because the 0 has different meanings in standardized and non-standardized data).
standardize_parameters(mod, method = "posthoc")> # Standardization method: posthoc
>
> Parameter | Std. Coef. | 95% CI
> -------------------------------------------
> (Intercept) | 0.00 | [ 0.00, 0.00]
> xtra hours | 0.77 | [ 0.73, 0.81]
> n comps | 0.39 | [ 0.36, 0.43]
> age g [>30-35] | 0.06 | [-0.01, 0.14]
> age g [>35-45] | 0.10 | [-0.04, 0.23]
> seniority | 0.08 | [ 0.04, 0.12]
"smart": Standardization of Model’s
parameters with Adjustment, Reconnaissance and Transformation
Experimental
Similar to method = "posthoc" in that it does not
involve model refitting. The difference is that the SD of the response
is computed on the relevant section of the data. For instance, if a
factor with 3 levels A (the intercept), B and C is entered as a
predictor, the effect corresponding to B vs. A will be scaled by the
variance of the response at the intercept only. As a results, the
coefficients for effects of factors are similar to a Glass’
delta.
standardize_parameters(mod, method = "smart")> # Standardization method: smart
>
> Parameter | Std. Coef. | 95% CI
> -------------------------------------------
> (Intercept) | 0.00 | [ 0.00, 0.00]
> xtra hours | 0.77 | [ 0.73, 0.81]
> n comps | 0.39 | [ 0.36, 0.43]
> age g [>30-35] | 0.06 | [-0.01, 0.14]
> age g [>35-45] | 0.10 | [-0.04, 0.23]
> seniority | 0.08 | [ 0.04, 0.12]
"basic": Raw scaling of the model
frame
This method is similar to method = "posthoc", but treats
all variables as continuous: it scales the coefficient by the standard
deviation of model’s matrix’ parameter of factors levels (transformed to
integers) or binary predictors. Although it can be argued that this
might be inappropriate for these cases, this method allows for easier
importance judgment across all predictor type (numeric, factor,
interactions…). It is also the type of standardization implemented by
default in other software packages (also
lm.beta::lm.beta()), and, such as can be used for
reproducibility and replication purposes.
standardize_parameters(mod, method = "basic")> # Standardization method: basic
>
> Parameter | Std. Coef. | 95% CI
> -------------------------------------------
> (Intercept) | 0.00 | [ 0.00, 0.00]
> xtra hours | 0.77 | [ 0.73, 0.81]
> n comps | 0.39 | [ 0.36, 0.43]
> age g [>30-35] | 0.03 | [-0.01, 0.07]
> age g [>35-45] | 0.03 | [-0.01, 0.06]
> seniority | 0.08 | [ 0.04, 0.12]Standardizing Parameters In Mixed Models
Linear mixed models (LMM/HLM/MLM) offer an additional conundrum to standardization - how does one even calculate the SDs of the various predictors? Or of the response - is it the deviations within each group? Or perhaps between them?
The solution: standardize according to level of the predictor (Hoffman 2015, 342)! Level 1 parameters are
standardized according to variance within groups, while level 2
parameters are standardized according to variance between
groups. The resulting standardized coefficient are also called
pseudo-standardized coefficients.[^Note that like method
"basic", these are based on the model matrix.]
m <- lme4::lmer(Reaction ~ Days + (Days | Subject), data = lme4::sleepstudy)
standardize_parameters(m, method = "pseudo", ci_method = "satterthwaite")> # Standardization method: pseudo
>
> Parameter | Std. Coef. | 95% CI
> ---------------------------------------
> (Intercept) | 0.00 | [0.00, 0.00]
> Days | 0.68 | [0.47, 0.89]
# compare to:
standardize_parameters(m, method = "basic", ci_method = "satterthwaite")> # Standardization method: basic
>
> Parameter | Std. Coef. | 95% CI
> ---------------------------------------
> (Intercept) | 0.00 | [0.00, 0.00]
> Days | 0.54 | [0.37, 0.70]Standardizing Parameters In Generalized Linear Models
Unlike linear (/mixed) models, in generalized linear (/mixed) models (GLMs) there is less of a need for standardization. Why? Because in many GLMs the estimated coefficients are themselves measures of effect size, such as odds-ratios (OR) in logistic regression, or incidence rate ratios (IRR) in Poisson regressions. This is because in such model the outcome is not on an arbitrary scale - that is, the meaning of rates and probabilities are changed by arbitrary linear transformations.
But still, some standardization is sometimes needed, for the
predictors. Luckily, standardize_parameters() (and
standardize()) are smart enough to know when GLMs are
passed so as to only standardize according to the predictors:
mod_b <- glm(am ~ mpg + factor(cyl),
data = mtcars,
family = binomial()
)
standardize_parameters(mod_b, method = "refit", two_sd = TRUE)> # Standardization method: refit
>
> Parameter | Std. Coef. | 95% CI
> -------------------------------------------------------
> (Intercept) | -0.91 | [-3.32, 1.33]
> mpg | 4.46 | [ 0.30, 10.54]
> cyl [-0.0524939042876197] | 0.73 | [-2.04, 3.66]
> cyl [0.507441074780324] | 0.70 | [-3.13, 4.78]
>
> - Scaled by two SDs from the mean.
> - Response is unstandardized.
# standardize_parameters(mod_b, method = "posthoc", two_sd = TRUE)
# standardize_parameters(mod_b, method = "basic")These can then be converted to OR (with exp()) and
discussed as the “change in Odds as a function of a change in one SD
of x”.
std <- standardize_parameters(mod_b, method = "refit", two_sd = TRUE)
exp(std$Std_Coefficient)> [1] 0.4 86.4 2.1 2.0Or we can directly ask for the coefficients to be exponentiated:
standardize_parameters(mod_b, method = "refit", two_sd = TRUE, exponentiate = TRUE)> # Standardization method: refit
>
> Parameter | Std_Odds_Ratio | 95% CI
> -------------------------------------------------------------
> (Intercept) | 0.40 | [0.04, 3.76]
> mpg | 86.40 | [1.36, 37955.41]
> cyl [-0.0524939042876197] | 2.08 | [0.13, 39.04]
> cyl [0.507441074780324] | 2.02 | [0.04, 119.12]
>
> - Scaled by two SDs from the mean.
> - Response is unstandardized.