应用于'NULL'类型的非（列表或向量）的is.na（）是什么意思？

Question

I want to select a Cox model with the forward procedure from a data.frame with no NA. Here is some sample data:

test <- data.frame(
  x_1   = runif(100,0,1),
  x_2   = runif(100,0,5),
  x_3   = runif(100,10,20),
  time  = runif(100,50,200),
  event = c(rep(0,70),rep(1,30))
)

This table has no signification but if we try to build a model anyway :

modeltest <- coxph(Surv(time, event) ~1, test)
modeltest.forward <- step(
  modeltest, 
  data      = test, 
  direction = "forward", 
  scope     = list(lower = ~ 1, upper = ~ x_1 + x_2 + x_3)
)

The forward ends at the first step and says:

In is.na(fit$coefficients) : is.na() applied to non-(list or vector) of type 'NULL'

(three times)

I tried to change the upper model, I even tried upper = ~ 1 but the warning stays. I don't understand: I have no NAs and my vectors are all numerics (I checked it). I searched if people had the same issue but all I could find was problems due to the name or class of the vectors.

What's wrong with my code?

Answer 1

The problem in this specific case

The right hand side of your formula is 1 , which makes it a null model . coxph calls coxph.fit , which (perhaps lazily) doesn't bother to return coefficients for null models.

Later coxph calls extractAIC , which erroneously assumes that the model object contains an element named coefficients .

The general case

is.na assumes that its input argument is an atomic vector or a matrix or a list or a data.frame. Other data types cause the warning. It happens with NULL , as you've seen:

is.na(NULL)
## logical(0)
## Warning message:
## In is.na(NULL) : is.na() applied to non-(list or vector) of type 'NULL'

One common cause of this problem is trying to access elements of a list, or columns of a data frame that don't exist.

d <- data.frame(x = c(1, NA, 3))
d$y # "y" doesn't exist is the data frame, but NULL is returned
## NULL
is.na(d$y)
## logical(0)
## Warning message:
## In is.na(d$y) : is.na() applied to non-(list or vector) of type 'NULL'

You can protect against this by checking that the column exists before you manipulate it.

if("y" in colnames(d))
{
  d2 <- d[is.na(d$y), ]
}

The warning with other data types

You get a simliar warning with formulae, functions, expressions, etc.:

is.na(~ NA)
## [1] FALSE FALSE
## Warning message:
## In is.na(~NA) : is.na() applied to non-(list or vector) of type 'language'

is.na(mean)
## [1] FALSE
## Warning message:
## In is.na(mean) : is.na() applied to non-(list or vector) of type 'closure'

is.na(is.na)
## [1] FALSE
## Warning message:
## In is.na(is.na) : is.na() applied to non-(list or vector) of type 'builtin'

is.na(expression(NA))
## [1] FALSE
## Warning message:
## In is.na(expression(NA)) :
##   is.na() applied to non-(list or vector) of type 'expression'