问题
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$coefficients) : is.na() applied to non-(list or vector) of type 'NULL'
(three times)
I tried to change the upper model, I even tried upper = ~ 1
but the warning stays. I don't understand: I have no NAs and my vectors are all numerics (I checked it).
I searched if people had the same issue but all I could find was problems due to the name or class of the vectors.
What's wrong with my code?
回答1:
The problem in this specific case
The right hand side of your formula is 1
, which makes it a null model. coxph
calls coxph.fit
, which (perhaps lazily) doesn't bother to return coefficients for null models.
Later coxph
calls extractAIC
, which erroneously assumes that the model object contains an element named coefficients
.
The general case
is.na
assumes that its input argument is an atomic vector or a matrix or a list or a data.frame. Other data types cause the warning. It happens with NULL
, as you've seen:
is.na(NULL)
## logical(0)
## Warning message:
## In is.na(NULL) : is.na() applied to non-(list or vector) of type 'NULL'
One common cause of this problem is trying to access elements of a list, or columns of a data frame that don't exist.
d <- data.frame(x = c(1, NA, 3))
d$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'
来源:https://stackoverflow.com/questions/17209643/what-does-is-na-applied-to-non-list-or-vector-of-type-null-mean