提取按因子分组的每个回归的R ^ 2（R平方）值

Question

I'm wondering if there is a way to extract R2 for each regression equation.

d <- data.frame(
  state = rep(c('NY', 'CA'), 10),
  year = rep(1:10, 2),
  response= rnorm(20)
)

library(plyr)
models <- dlply(d, "state", function(df) 
  lm(response ~ year, data = df))

ldply(models, coef)
l_ply(models, summary, .print = TRUE)

I tried

l_ply(models, summary$r.squared, .print = TRUE)

But this throws the following error message

Error in summary$r.squared : object of type 'closure' is not subsettable

Answer 1

You can do this to get the R squared value and the coefficients:

ldply(models, function(x) {r.sq <- summary(x)$r.squared
                           intercept <- summary(x)$coefficients[1]
                           beta <- summary(x)$coefficients[2]
                           data.frame(r.sq, intercept, beta)})
#  state        r.sq intercept        beta
#1    CA 0.230696121 0.4915617 -0.12343947
#2    NY 0.003506936 0.1971734 -0.01227367

Answer 2

Using the broom package for converting statistical analysis objects into data.frames and dplyr for bind_rows :

library(dplyr) ; library(broom)
cbind(
  state = attr(models, "split_labels"),
  bind_rows(lapply(models, function(x) cbind(
    intercept = tidy(x)$estimate[1],
    beta = tidy(x)$estimate[2],
    glance(x))))
)

  state  intercept        beta  r.squared adj.r.squared    sigma statistic   p.value df    logLik      AIC      BIC deviance df.residual
1    CA 0.38653551 -0.05459205 0.01427426   -0.10894146 1.434599 0.1158477 0.7423473  2 -16.68252 39.36505 40.27280 16.46460           8
2    NY 0.09028554 -0.08462742 0.04138985   -0.07843642 1.287909 0.3454155 0.5729312  2 -15.60387 37.20773 38.11549 13.26968           8

Answer 3

you can try this

sapply(models, function(x) summary(x)$r.squared)
     CA      NY 
0.05639 0.23751 

Answer 4

If you try

> typeof( summary )
[1] "closure"

you see that 'summary' is a function. You are trying to access a field of the result, but summary$r.squared tries to access that field on the function / closure.

Using an anonymous function,

> l_ply( models, function( m ) summary( m )$r.squared, .print = TRUE )
[1] 0.2319583
[1] 0.01295825

will work and print the result. However, you say that you want to "extract the result". This probably means that you want to use the result and not just print it.

From the documentation of l_ply (which you'll get by typing ?l_ply at the R prompt):

For each element of a list, apply function and discard results.

(So this function will not work if you want to hang on to the result.)

Using the standard sapply / lapply will result in

> a <- sapply( models, function( t ) summary( t )$r.squared )
> a
        CA         NY 
0.23195825 0.01295825 
> typeof( a )
[1] "double"
> is.vector( a )
[1] TRUE
> # or alternatively
> l <- lapply( models, function( t ) summary( t )$r.squared )
> l
$CA
[1] 0.2319583

$NY
[1] 0.01295825
> typeof( l )
[1] "list"

Either one should work -- pick whichever result (vector or list) is easier to use for what you want to do. (If unsure, just pick sapply .)

(Or, if you want to use functions from the plyr package, laply , ldply , and llply seem to work too. But I've never used that package, so I can't say what's best.)