CODEWITHSUNDEEP
rlm
NelsonGon
NelsonGon

Reputation: 13319

Supplying several names to a function call

This is a function from an earlier question of mine: How to let R predict user input I would like to make it easier to supply several names to the xname argument but I still can't fugure out just how to do that.

lmfun<-function(df,yname,xname){
  y<-deparse(substitute(yname))
  x<-deparse(substitute(xname))
  f<-as.formula(paste0(y,"~",x))
  lm.fit<-do.call("lm",list(data=quote(df),f))
  coef(lm.fit)
}

Here's what I've tried

vals<-names(mtcars)[-1]
lmfun(mtcars,mpg,disp)#This works

How can I best make this work? I've tried several other ways but showing only this:

for(name in 1:seq_along(vals)){
  name<-eval(substitute(name))
  lmfun(mtcars,mpg,name)
}

This fails:

Error in deparse(substitute(xname)) : 'arg' should be one of “mpg”, “cyl”, “disp”, “hp”, “drat”, “wt”, “qsec”, “vs”, “am”, “gear”, “carb”

Also tried:

for(name in 1:length(vals)){
  vals<-noquote(vals)
 lmfun(mtcars,mpg,vals[name])
}

I would also appreciate if I could be pointed at a way to incorporate multilinear regression. That is xname+xname1+xname2 Thanks!

Upvotes: 0

Views: 89

Answers (2)

Ewen
Ewen

Reputation: 1381

Multiple univariable lm() can be easily done in finalfit. It likes factors specified correctly:

library(finalfit)
dependent = "mpg"
explanatory = names(mtcars)[-1]
mtcars %>% 
  dplyr::mutate(
    cyl = factor(cyl),
    vs = factor(vs),
    am = factor(am),
    gear = factor(gear)
    ) %>% 
  finalfit(dependent, explanatory)

 Dependent: mpg              Mean (sd)         Coefficient (univariable)    Coefficient (multivariable)
            cyl           4 26.7 (4.5)                                 -                              -
                          6 19.7 (1.5)  -6.92 (-10.11 to -3.73, p<0.001) -1.20 (-6.20 to 3.80, p=0.621)
                          8 15.1 (2.6) -11.56 (-14.22 to -8.91, p<0.001) 3.05 (-7.05 to 13.16, p=0.535)
           disp  [71.1,472] 20.1 (6.0)   -0.04 (-0.05 to -0.03, p<0.001)  0.01 (-0.02 to 0.05, p=0.487)
             hp    [52,335] 20.1 (6.0)   -0.07 (-0.09 to -0.05, p<0.001) -0.06 (-0.12 to 0.01, p=0.088)
           drat [2.76,4.93] 20.1 (6.0)     7.68 (4.60 to 10.76, p<0.001)  0.74 (-3.42 to 4.89, p=0.715)
             wt [1.51,5.42] 20.1 (6.0)   -5.34 (-6.49 to -4.20, p<0.001) -3.55 (-7.54 to 0.45, p=0.079)
           qsec [14.5,22.9] 20.1 (6.0)      1.41 (0.27 to 2.55, p=0.017)  0.77 (-0.81 to 2.34, p=0.320)
             vs           0 16.6 (3.9)                                 -                              -
                          1 24.6 (5.4)     7.94 (4.61 to 11.27, p<0.001)  2.49 (-2.83 to 7.81, p=0.340)
             am           0 17.1 (3.8)                                 -                              -
                          1 24.4 (6.2)     7.24 (3.64 to 10.85, p<0.001)  3.35 (-1.44 to 8.14, p=0.160)
           gear           3 16.1 (3.4)                                 -                              -
                          4 24.5 (5.3)     8.43 (4.70 to 12.16, p<0.001) -1.00 (-7.17 to 5.17, p=0.738)
                          5 21.4 (6.7)     5.27 (0.30 to 10.25, p=0.038)  1.06 (-5.27 to 7.40, p=0.729)
           carb       [1,8] 20.1 (6.0)   -2.06 (-3.22 to -0.89, p=0.001)  0.79 (-1.38 to 2.96, p=0.457)

Lots of options in the documentation here finalfit.org.

Upvotes: 1

G. Grothendieck
G. Grothendieck

Reputation: 269501

Invoke lmfun using do.call like this:

lapply(vals, function(val) do.call("lmfun", list(mtcars, quote(mpg), as.name(val))))

This also works although generally code which does not use eval is preferred when possible:

lapply(vals, 
  function(val) eval(substitute(lmfun(mtcars, mpg, val), list(val = as.name(val)))))

Upvotes: 1

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