linear model when all occurrences of independent variables are NA - r

Question

I'm looking for suggestions on how to deal with NA's in linear regressions when all occurrences of an independent/explanatory variable are NA (i.e. x3 below).
I know the obvious solution would be to exclude the independent/explanatory variable in question from the model but I am looping through multiple regions and would prefer not to have a different functional forms for each region.
Below is some sample data:
set.seed(23409)
n <- 100
time <- seq(1,n, 1)
x1 <- cumsum(runif(n))
y <- .8*x1 + rnorm(n, mean=0, sd=2)
x2 <- seq(1,n, 1)
x3 <- rep(NA, n)
df <- data.frame(y=y, time=time, x1=x1, x2=x2, x3=x3)
# Quick plot of data
library(ggplot2)
library(reshape2)
df.melt <-melt(df, id=c("time"))
p <- ggplot(df.melt, aes(x=time, y=value)) +
geom_line() + facet_grid(variable ~ .)
p
I have read the documentation for lm and tried various na.action settings without success:
lm(y~x1+x2+x3, data=df, singular.ok=TRUE)
lm(y~x1+x2+x3, data=df, na.action=na.omit)
lm(y~x1+x2+x3, data=df, na.action=na.exclude)
lm(y~x1+x2+x3, data=df, singular.ok=TRUE, na.exclude=na.omit)
lm(y~x1+x2+x3, data=df, singular.ok=TRUE, na.exclude=na.exclude)
Is there a way to get lm to run without error and simply return a coefficient for the explanatory reflective of the lack of explanatory power (i.e. either zero or NA) from the variable in question?

Answer 1

Here's one idea:
set.seed(23409)
n <- 100
time <- seq(1,n, 1)
x1 <- cumsum(runif(n))
y <- .8*x1 + rnorm(n, mean=0, sd=2)
x2 <- seq(1,n, 1)
x3 <- rep(NA, n)
df <- data.frame(y=y, time=time, x1=x1, x2=x2, x3=x3)
replaceNA<-function(x){
if(all(is.na(x))){
rep(0,length(x))
} else x
}
lm(y~x1+x2+x3, data= data.frame(lapply(df,replaceNA)))
Call:
lm(formula = y ~ x1 + x2 + x3, data = data.frame(lapply(df, replaceNA)))
Coefficients:
(Intercept) x1 x2 x3
0.05467 1.01133 -0.10613 NA
lm(y~x1+x2, data=df)
Call:
lm(formula = y ~ x1 + x2, data = df)
Coefficients:
(Intercept) x1 x2
0.05467 1.01133 -0.10613
So you replace the variables which contain only NA's with variable which contains only 0's. you get the coefficient value NA, but all the relevant parts of the model fits are same (expect qr decomposition, but if information about that is needed, it can be easily modified). Note that component summary(fit)$alias (see ?alias) might be useful.
This seems to relate your other question: Replace lm coefficients in [r]

Answer 2

You won't be able to include a column with all NA values. It does strange things to model.matrix
x1 <- 1:5
x2 <- rep(NA,5)
model.matrix(~x1+x2)
(Intercept) x1 x2TRUE
attr(,"assign")
[1] 0 1 2
attr(,"contrasts")
attr(,"contrasts")$x2
[1] "contr.treatment"
So your alternative is to programatically create the model formula based on the data.
Something like...
make_formula <- function(variables, data, response = 'y'){
if(missing(data)){stop('data not specified')}
using <- Filter(variables,f= function(i) !all(is.na(data[[i]])))
deparse(reformulate(using, response))
}
variables <- c('x1','x2','x3')
make_formula(variables, data =df)
[1] "y ~ x1 + x2"
I've used deparse to return a character string so that there is no environment issues from creating the formula within the function. lm can happily take a character string which is a valid formula.