Lapply的稳健线性回归

Question

I'm having problems to run a robust linear regression model (using rlm from the MASS library) over a list of dataframes.

Reproducible example:

var1 <- c(1:100)
var2 <- var1*var1
df1  <- data.frame(var1, var2)
var1 <- var1 + 50
var2 <- var2*2
df2  <- data.frame(var1, var2)
lst1 <- list(df1, df2)

Linear model (works):

lin_mod <- lapply(lst1, lm, formula = var1 ~ var2)
summary(lin_mod[[1]])

My code for the robust model:

rob_mod <- lapply(lst1, MASS::rlm, formula = var1 ~ var2)

gives the following error:

Error in rlm.default(X[[i]], ...) : 
argument "y" is missing, with no default

How could I solve this?

The error in my actual data is:

Error in qr.default(x) : NA/NaN/Inf in foreign function call (arg 1)
In addition: Warning message:
In storage.mode(x) <- "double" : NAs introduced by coercion    

Answer 1

You can also try a purrr:map solution:

library(tidyverse)
map(lst1, ~rlm(var1 ~ var2, data=.))

or as joran commented

map(lst1, MASS:::rlm.formula, formula = var1 ~ var2)

As you can see here ?lm provides only a formula method. In contrast ?rlm provides both ( formula and x, y ). Thus, you have to specify data= to say rlm to explicitly use the formula method. Otherwise rlm wants x and y as input.

Answer 2

Your call is missing the data argument. lapply will call FUN with each member of the list as the first argument of FUN but data is the second argument to rlm .

The solution is to define an anonymous function.

lin_mod <- lapply(lst1, function(DF) MASS::rlm(formula = var1 ~ var2, data = DF))
summary(lin_mod[[1]])
#
#Call: rlm(formula = var1 ~ var2, data = DF)
#Residuals:
#    Min      1Q  Median      3Q     Max 
#-18.707  -5.381   1.768   6.067   7.511 
#
#Coefficients:
#              Value   Std. Error t value
#(Intercept) 19.6977  1.0872    18.1179
#var2         0.0092  0.0002    38.2665
#
#Residual standard error: 8.827 on 98 degrees of freedom