In the rigr
package, we have set out to make regression and analysis easier by
This capability is implemented in the function regress()
. The basic arguments to this function are
fnctl
: the functionalformula
: the formula for the linear modeldata
: the data to use for the modelfnctl
: the functionalWe use the concept of a functional to handle our first goal: allowing you to specify different types of regression models using a single function. A functional takes a function as its argument and returns a number. The most common example of a functional in regression is the mean. The allowed functionals to regress()
are
Functional | Type of Regression | base R command |
---|---|---|
"mean" |
Linear Regression | lm() |
"geometric mean" |
Linear Regression on logarithmically transformed Y | lm() , with Y log-transformed |
"odds" |
Logistic Regression | glm(family = binomial) |
"rate" |
Poisson Regression | glm(family = poisson) |
"hazard" |
Proportional Hazards Regression | coxph() |
formula
and data
The formula to regress()
is the same as a formula given to lm()
or glm()
, but with additional optional functionality to enable sophisticated analyses (multiple partial F-tests) with fewer headaches.
The data argument is exactly the same as that in lm()
or any of the other regression commands.
As a first example, we run a linear regression of atrophy (a measure of global brain activity) on age, sex and race, from the mri
data. This dataset is included in the rigr
package; see its documentation by running ?mri
.
## Preparing our R session
library(rigr)
data(mri)
regress("mean", atrophy ~ age + sex + race, data = mri)
##
## Call:
## regress(fnctl = "mean", formula = atrophy ~ age + sex + race,
## data = mri)
##
## Residuals:
## Min 1Q Median 3Q Max
## -34.120 -8.331 -0.434 7.325 53.915
##
## Coefficients:
## Estimate Naive SE
## [1] Intercept -17.60 6.341
## [2] age 0.6866 0.08134
## [3] sexMale 5.988 0.8867
## race
## [4] Black -2.375 2.109
## [5] Subject did not identify White, Black or Asian -3.078 3.885
## [6] White -0.2664 1.822
## Robust SE 95%L
## [1] Intercept 6.893 -31.14
## [2] age 0.08836 0.5132
## [3] sexMale 0.8895 4.242
## race
## [4] Black 2.049 -6.397
## [5] Subject did not identify White, Black or Asian 4.157 -11.24
## [6] White 1.780 -3.761
## 95%H F stat
## [1] Intercept -4.072 6.52
## [2] age 0.8601 60.38
## [3] sexMale 7.734 45.32
## race 1.14
## [4] Black 1.647 1.34
## [5] Subject did not identify White, Black or Asian 5.082 0.55
## [6] White 3.228 0.02
## df Pr(>F)
## [1] Intercept 1 0.0109
## [2] age 1 < 0.00005
## [3] sexMale 1 < 0.00005
## race 3 0.3315
## [4] Black 1 0.2467
## [5] Subject did not identify White, Black or Asian 1 0.4592
## [6] White 1 0.8810
##
## Residual standard error: 12 on 729 degrees of freedom
## Multiple R-squared: 0.1439, Adjusted R-squared: 0.138
## F-statistic: 21.32 on 5 and 729 DF, p-value: < 2.2e-16
Notice that by default robust standard error estimates are returned in addition to the naive estimates. The robust estimates are also used to perform inference. Thus, the confidence intervals, statistics, and p-values use these estimates of the standard error.
F-statistics are also displayed by default, including the multiple partial F-tests for the levels of a multi-level category (such as race
) as well as the overall F-test for the variable.
We can also run generalized linear regression using regress()
. For example, to model the odds of having diabetes for males compared to females, we could run a logistic regression as follows:
##
## Call:
## regress(fnctl = "odds", formula = diabetes ~ sex, data = mri)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -0.5593 -0.5593 -0.3823 -0.3823 2.3034
##
## Coefficients:
##
## Raw Model:
## Estimate Naive SE Robust SE F stat df Pr(>F)
## [1] Intercept -2.580 0.2034 0.2037 160.39 1 < 0.00005
## [2] sexMale 0.8037 0.2519 0.2522 10.15 1 0.0015
##
## Transformed Model:
## e(Est) e(95%L) e(95%H) F stat df Pr(>F)
## [1] Intercept 0.07580 0.05082 0.1131 160.39 1 < 0.00005
## [2] sexMale 2.234 1.361 3.665 10.15 1 0.0015
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 501.59 on 734 degrees of freedom
## Residual deviance: 490.82 on 733 degrees of freedom
## AIC: 494.82
##
## Number of Fisher Scoring iterations: 5
In all of the generalized linear regression output we see two tables. The Raw Model
table displays estimated coefficients (and their standard errors) on the log-odds scale. The Transformed Model
table exponentiates the estimated coefficients and their confidence intervals so that the estimated parameters can be interpreted on the odds scale.
Note that the only possible link function in regress
with fnctl = odds"
is the logit link. Similarly, the only possible link function in regress
with fnctl = "rate"
is the log link.
The next functional that regress
supports is "rate"
, for use in Poisson regression. To regress yrsquit
on age
, we would run:
##
## Call:
## regress(fnctl = "rate", formula = yrsquit ~ age, data = mri)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -5.385 -4.365 -4.186 2.395 9.920
##
## Coefficients:
##
## Raw Model:
## Estimate Naive SE Robust SE F stat df Pr(>F)
## [1] Intercept 1.011 0.1571 0.7239 1.95 1 0.1629
## [2] age 0.01680 2.087e-03 9.688e-03 3.01 1 0.0834
##
## Transformed Model:
## e(Est) e(95%L) e(95%H) F stat df Pr(>F)
## [1] Intercept 2.749 0.6637 11.39 1.95 1 0.1629
## [2] age 1.017 0.9978 1.036 3.01 1 0.0834
##
## (Dispersion parameter for poisson family taken to be 1)
##
## Null deviance: 14574 on 734 degrees of freedom
## Residual deviance: 14511 on 733 degrees of freedom
## AIC: 16008
##
## Number of Fisher Scoring iterations: 6
Note that again we have two tables of output, denoted by Raw Model
and Transformed Model
, with Transformed Model
displaying exponentiated estimated coefficients.
The final functional that regress
supports is "hazard"
, for use in proportional hazards regression. To regress age
on the death status (note that we need to create a Surv
object first), we would run:
##
## Call:
## regress(fnctl = "hazard", formula = Surv(obstime, death) ~ age,
## data = mri)
##
##
## Coefficients:
##
## Raw Model:
## Estimate Naive SE Robust SE F stat df Pr(>F)
## [1] age 0.06795 0.01354 0.01412 23.17 1 < 0.00005
##
## Transformed Model:
## e(Est) e(95%L) e(95%H) F stat df Pr(>F)
## [1] age 1.070 1.041 1.100 23.17 1 < 0.00005
## n = 735, number of events= 133
## Overall significance test:
## Likelihood ratio test= 22.33 on 1 df, p=2.292e-06
## Wald test = 25.17 on 1 df, p=5.24e-07
## Score (logrank) test = 25.44 on 1 df, p=4.554e-07
Similar to the Poisson regression case, we have two tables of output, denoted by Raw Model
and Transformed Model
, with Transformed Model
displaying exponentiated estimated coefficients (i.e., on the hazard scale).
Most often in linear regression we are interested in modeling the mean of the response variable. However, we are sometimes interested in modeling the mean of the log-transformed response variable, which allows us to make statements about the geometric mean of the response. In regress()
, we can use the "geometric mean"
functional to fit this model. Regression on the geometric mean of the packyrs
variable in the mri
dataset can be performed as follows:
## ( 1 cases deleted due to missing values)
##
##
## Call:
## regress(fnctl = "geometric mean", formula = packyrs ~ age, data = mri)
##
## Residuals:
## Min 1Q Median 3Q Max
## -4.136 -3.417 1.283 2.970 5.219
##
## Coefficients:
##
## Raw Model:
## Estimate Naive SE Robust SE F stat df Pr(>F)
## [1] Intercept 6.410 1.586 1.537 17.39 1 < 0.00005
## [2] age -0.07986 0.02121 0.02050 15.17 1 1e-04
##
## Transformed Model:
## e(Est) e(95%L) e(95%H) F stat df Pr(>F)
## [1] Intercept 608.1 29.74 12431 17.39 1 < 0.00005
## [2] age 0.9232 0.8868 0.9612 15.17 1 1e-04
##
## Residual standard error: 3.13 on 732 degrees of freedom
## (1 observation deleted due to missingness)
## Multiple R-squared: 0.01899, Adjusted R-squared: 0.01765
## F-statistic: 15.17 on 1 and 732 DF, p-value: 0.0001074
It should be noted that many of the packyrs
observations are zero, but the geometric mean of data including an observation of zero is zero… regardless of how many non-zeros were also observed. Therefore, by default, zeroes in the outcome variable are replaced by a value equal to one-half the lowest nonzero value in the outcome variable. This is based on the idea that the lowest observed value could be a proxy for the lower limit of detection. If you wish to specify a different value with which to replace zeroes, you may do say using the replaceZeroes
argument.
## ( 1 cases deleted due to missing values)
##
##
## Call:
## regress(fnctl = "geometric mean", formula = packyrs ~ age, data = mri,
## replaceZeroes = 1)
##
## Residuals:
## Min 1Q Median 3Q Max
## -3.6437 -1.7460 0.0625 1.7259 3.8246
##
## Coefficients:
##
## Raw Model:
## Estimate Naive SE Robust SE F stat df Pr(>F)
## [1] Intercept 5.119 0.8887 0.8462 36.60 1 < 0.00005
## [2] age -0.04498 0.01189 0.01127 15.93 1 1e-04
##
## Transformed Model:
## e(Est) e(95%L) e(95%H) F stat df Pr(>F)
## [1] Intercept 167.2 31.76 880.7 36.60 1 < 0.00005
## [2] age 0.9560 0.9351 0.9774 15.93 1 1e-04
##
## Residual standard error: 1.754 on 732 degrees of freedom
## (1 observation deleted due to missingness)
## Multiple R-squared: 0.01919, Adjusted R-squared: 0.01785
## F-statistic: 15.93 on 1 and 732 DF, p-value: 7.218e-05
In the output from regress using the geometric mean functional, we see a table for the Raw Model
and the Transformed Model
. The e(Est)
, e(95%L)
, and e(95%H)
columns in the Transformed Model
table correspond to exponentiated values from the Raw Model
- you’ll notice that \(e^{5.119} \approx 167.2.\)
There are two special functions in rigr
which allow us to re-parameterize variables:
dummy
: create dummy variablespolynomial
: create a polynomialBoth of these functions may be used in a regress()
call, and will additionally give a multiple partial F-test of the entire variable automatically.
dummy
The dummy
function is useful for specifying the reference group that you wish to use with categorical variables. Below we show an example of using the reference group “Female” vs. the reference group “Male” in a regression on sex
.
##
## Call:
## regress(fnctl = "mean", formula = atrophy ~ dummy(sex, reference = "Male"),
## data = mri)
##
## Residuals:
## Min 1Q Median 3Q Max
## -29.087 -9.087 -0.905 8.095 49.095
##
## Coefficients:
## Estimate Naive SE Robust SE
## [1] Intercept 39.09 0.6563 0.6733
## [2] dummy(sex, reference = "Male") -6.182 0.9263 0.9265
## 95%L 95%H F stat df
## [1] Intercept 37.77 40.41 3370.22 1
## [2] dummy(sex, reference = "Male") -8.001 -4.363 44.53 1
## Pr(>F)
## [1] Intercept < 0.00005
## [2] dummy(sex, reference = "Male") < 0.00005
##
## Dummy terms calculated from sex, reference = Male
##
## Residual standard error: 12.56 on 733 degrees of freedom
## Multiple R-squared: 0.05729, Adjusted R-squared: 0.05601
## F-statistic: 44.53 on 1 and 733 DF, p-value: 4.944e-11
##
## Call:
## regress(fnctl = "mean", formula = atrophy ~ dummy(sex, reference = "Female"),
## data = mri)
##
## Residuals:
## Min 1Q Median 3Q Max
## -29.087 -9.087 -0.905 8.095 49.095
##
## Coefficients:
## Estimate Naive SE Robust SE
## [1] Intercept 32.91 0.6536 0.6364
## [2] dummy(sex, reference = "Female") 6.182 0.9263 0.9265
## 95%L 95%H F stat df
## [1] Intercept 31.66 34.15 2673.43 1
## [2] dummy(sex, reference = "Female") 4.363 8.001 44.53 1
## Pr(>F)
## [1] Intercept < 0.00005
## [2] dummy(sex, reference = "Female") < 0.00005
##
## Dummy terms calculated from sex, reference = Female
##
## Residual standard error: 12.56 on 733 degrees of freedom
## Multiple R-squared: 0.05729, Adjusted R-squared: 0.05601
## F-statistic: 44.53 on 1 and 733 DF, p-value: 4.944e-11
Notice that below the coefficients table in the output, the reference category is reported.
You can fit higher-order polynomials using polynomial
:
##
## Call:
## regress(fnctl = "mean", formula = atrophy ~ polynomial(age, degree = 2),
## data = mri)
##
## Residuals:
## Min 1Q Median 3Q Max
## -36.828 -9.063 -1.018 7.816 50.887
##
## Coefficients:
## Estimate Naive SE Robust SE 95%L
## [1] Intercept 35.34 0.5598 0.5622 34.24
## polynomial(age, degree = 2)
## [2] age^1 0.5869 0.1009 0.09808 0.3943
## [3] age^2 0.02159 0.01099 0.01150 -9.897e-04
## 95%H F stat df Pr(>F)
## [1] Intercept 36.45 3952.13 1 < 0.00005
## polynomial(age, degree = 2) 32.91 2 < 0.00005
## [2] age^1 0.7794 35.80 1 < 0.00005
## [3] age^2 0.04417 3.52 1 0.0609
##
## Polynomial terms calculated from age, centered at 74.566
##
## Residual standard error: 12.33 on 732 degrees of freedom
## Multiple R-squared: 0.09148, Adjusted R-squared: 0.089
## F-statistic: 32.91 on 2 and 732 DF, p-value: 2.06e-14
Note that all polynomials less than or equal to the degree specified are included in the model, and that the variables in the polynomial specification are mean-centered by default. You can change the centering using the center
parameter in the polynomial function, an example of which is as follows.
##
## Call:
## regress(fnctl = "mean", formula = atrophy ~ polynomial(age, degree = 2,
## center = 65), data = mri)
##
## Residuals:
## Min 1Q Median 3Q Max
## -36.828 -9.063 -1.018 7.816 50.887
##
## Coefficients:
## Estimate Naive SE
## [1] Intercept 31.70 1.528
## polynomial(age, degree = 2, center = 65)
## [2] age^1 0.1738 0.2797
## [3] age^2 0.02159 0.01099
## Robust SE 95%L
## [1] Intercept 1.497 28.77
## polynomial(age, degree = 2, center = 65)
## [2] age^1 0.2811 -0.3781
## [3] age^2 0.01150 -9.897e-04
## 95%H F stat df
## [1] Intercept 34.64 448.45 1
## polynomial(age, degree = 2, center = 65) 32.91 2
## [2] age^1 0.7257 0.38 1
## [3] age^2 0.04417 3.52 1
## Pr(>F)
## [1] Intercept < 0.00005
## polynomial(age, degree = 2, center = 65) < 0.00005
## [2] age^1 0.5367
## [3] age^2 0.0609
##
## Polynomial terms calculated from age, centered at 65
##
## Residual standard error: 12.33 on 732 degrees of freedom
## Multiple R-squared: 0.09148, Adjusted R-squared: 0.089
## F-statistic: 32.91 on 2 and 732 DF, p-value: 2.06e-14
You can also perform multiple partial F-tests using formulas and the U
function. This is useful when want to test a subset of variables all at once in your regression. For example, to test whether both the variables packyrs
and yrsquit
are associated with atrophy
in a model with age as a predictor, we can run
## ( 1 cases deleted due to missing values)
##
##
## Call:
## regress(fnctl = "mean", formula = atrophy ~ age + sex + U(`Smoking variables` = ~packyrs +
## yrsquit), data = mri)
##
## Residuals:
## Min 1Q Median 3Q Max
## -33.593 -8.613 -0.306 7.279 52.399
##
## Coefficients:
## Estimate Naive SE Robust SE 95%L 95%H
## [1] Intercept -18.82 6.167 6.690 -31.95 -5.683
## [2] age 0.6917 0.08212 0.08900 0.5170 0.8664
## [3] sexMale 5.628 0.9386 0.9596 3.744 7.512
## Smoking variables
## [4] packyrs 9.782e-03 0.01684 0.01693 -0.02346 0.04302
## [5] yrsquit 0.02071 0.03282 0.03276 -0.04361 0.08503
## F stat df Pr(>F)
## [1] Intercept 7.91 1 0.0050
## [2] age 60.40 1 < 0.00005
## [3] sexMale 34.39 1 < 0.00005
## Smoking variables 0.37 2 0.6905
## [4] packyrs 0.33 1 0.5636
## [5] yrsquit 0.40 1 0.5275
##
## Residual standard error: 11.99 on 729 degrees of freedom
## (1 observation deleted due to missingness)
## Multiple R-squared: 0.1419, Adjusted R-squared: 0.1372
## F-statistic: 26.44 on 4 and 729 DF, p-value: < 2.2e-16
"Smoking variables"
is what we name the group of variables packyrs
and yrsquit
. The overall F statistic and p-value associated with the inclusion of these two smoking variables variables in the model are 4.37 and 0.0130, respectively.
You may be interested in testing a null hypothesis about a linear combination of coefficients in our regression model. For example, to investigate “Is the mean atrophy for a female subject equal to the mean atrophy for a male subject who is 10 years younger?”, our hypothesis test involves both the coefficients on age and sex. We can test this hypothesis using the lincom
function. First, we need to fit a linear model of age and sex on atrophy:
##
## Call:
## regress(fnctl = "mean", formula = atrophy ~ age + sex, data = mri)
##
## Residuals:
## Min 1Q Median 3Q Max
## -33.765 -8.582 -0.356 7.344 52.100
##
## Coefficients:
## Estimate Naive SE Robust SE 95%L 95%H
## [1] Intercept -17.83 6.081 6.557 -30.70 -4.959
## [2] age 0.6819 0.08129 0.08769 0.5097 0.8540
## [3] sexMale 5.964 0.8857 0.8845 4.227 7.700
## F stat df Pr(>F)
## [1] Intercept 7.40 1 0.0067
## [2] age 60.47 1 < 0.00005
## [3] sexMale 45.46 1 < 0.00005
##
## Residual standard error: 12 on 732 degrees of freedom
## Multiple R-squared: 0.14, Adjusted R-squared: 0.1376
## F-statistic: 52.18 on 2 and 732 DF, p-value: < 2.2e-16
We then create a vector giving the linear combination of the coefficient that we hypothesized to be zero, and perform the test using lincom
. The elements in mod_combo
correspond to Intercept
, age
, and sexMale
, because this was their order in the coefficient table shown above.
##
## H0: -10*age+1*sexMale = 0
## Ha: -10*age+1*sexMale != 0
## Estimate Std. Err. 95%L 95%H T Pr(T > |t|)
## [1,] -0.855 1.236 -3.282 1.571 -0.692 0.489
Note that the standard errors returned by default are robust, as are the associated confidence intervals and p-values.
We could also test the null hypothesis that the mean difference in atrophy between these two groups (females, and males 10 years younger) is equal to -1 as follows:
##
## H0: -10*age+1*sexMale = -1
## Ha: -10*age+1*sexMale != -1
## Estimate Std. Err. 95%L 95%H T Pr(T > |t|)
## [1,] -0.855 1.236 -3.282 1.571 0.117 0.907