问题
I'd like to run 10 regressions against the same regressor, then pull all the standard errors without using a loop.
depVars <- as.matrix(data[,1:10]) # multiple dependent variables
regressor <- as.matrix([,11]) # independent variable
allModels <- lm(depVars ~ regressor) # multiple, single variable regressions
summary(allModels)[1] # Can "view" the standard error for 1st regression, but can't extract...
allModels is stored as an "mlm" object, which is really tough to work with. It'd be great if I could store a list of lm objects or a matrix with statistics of interest.
Again, the objective is to NOT use a loop. Here is a loop equivalent:
regressor <- as.matrix([,11]) # independent variable
for(i in 1:10) {
tempObject <- lm(data[,i] ~ regressor) # single regressions
table1Data[i,1] <- summary(tempObject)$coefficients[2,2] # assign std error
rm(tempObject)
}
回答1:
If you put your data in long format it's very easy to get a bunch of regression results using lmList from the nlme or lme4 packages. The output is a list of regression results and the summary can give you a matrix of coefficients, just like you wanted.
library(lme4)
m <- lmList( y ~ x | group, data = dat)
summary(m)$coefficients
Those coefficients are in a simple 3 dimensional array so the standard errors are at [,2,2].
回答2:
Given an "mlm" model object model, you can use the below function written by me to get standard errors of coefficients. This is very efficient: no loop, and no access to summary.mlm().
std_mlm <- function (model) {
Rinv <- with(model$qr, backsolve(qr, diag(rank)))
## unscaled standard error
std_unscaled <- sqrt(rowSums(Rinv ^ 2)[order(model$qr$pivot)])
## residual standard error
sigma <- sqrt(colSums(model$residuals ^ 2) / model$df.residual)
## return final standard error
## each column corresponds to a model
"dimnames<-"(outer(std_unscaled, sigma), list = dimnames(model$coefficients))
}
A simple, reproducible example
set.seed(0)
Y <- matrix(rnorm(50 * 5), 50) ## assume there are 5 responses
X <- rnorm(50) ## covariate
fit <- lm(Y ~ X)
We all know that it is simple to extract estimated coefficients via:
fit$coefficients ## or `coef(fit)`
# [,1] [,2] [,3] [,4] [,5]
#(Intercept) -0.21013925 0.1162145 0.04470235 0.08785647 0.02146662
#X 0.04110489 -0.1954611 -0.07979964 -0.02325163 -0.17854525
Now let's apply our std_mlm:
std_mlm(fit)
# [,1] [,2] [,3] [,4] [,5]
#(Intercept) 0.1297150 0.1400600 0.1558927 0.1456127 0.1186233
#X 0.1259283 0.1359712 0.1513418 0.1413618 0.1151603
We can of course, call summary.mlm just to check our result is correct:
coef(summary(fit))
#Response Y1 :
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) -0.21013925 0.1297150 -1.6200072 0.1117830
#X 0.04110489 0.1259283 0.3264151 0.7455293
#
#Response Y2 :
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 0.1162145 0.1400600 0.8297485 0.4107887
#X -0.1954611 0.1359712 -1.4375183 0.1570583
#
#Response Y3 :
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 0.04470235 0.1558927 0.2867508 0.7755373
#X -0.07979964 0.1513418 -0.5272811 0.6004272
#
#Response Y4 :
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 0.08785647 0.1456127 0.6033574 0.5491116
#X -0.02325163 0.1413618 -0.1644831 0.8700415
#
#Response Y5 :
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 0.02146662 0.1186233 0.1809646 0.8571573
#X -0.17854525 0.1151603 -1.5504057 0.1276132
Yes, all correct!
回答3:
Here an option:
- put your data in the long format using regressor as an id key.
- do your regression against value by group of variable.
For example , using mtcars data set:
library(reshape2)
dat.m <- melt(mtcars,id.vars='mpg') ## mpg is my regressor
library(plyr)
ddply(dat.m,.(variable),function(x)coef(lm(variable~value,data=x)))
variable (Intercept) value
1 cyl 1 8.336774e-18
2 disp 1 6.529223e-19
3 hp 1 1.106781e-18
4 drat 1 -1.505237e-16
5 wt 1 8.846955e-17
6 qsec 1 6.167713e-17
7 vs 1 2.442366e-16
8 am 1 -3.381738e-16
9 gear 1 -8.141220e-17
10 carb 1 -6.455094e-17
来源：https://stackoverflow.com/questions/19740292/obtain-standard-errors-of-regression-coefficients-for-an-mlm-object-returned-b