问题:

R beginner here.
I would like to convert 0s into NaNs in these 6 of the 9 columns: "int_t", "ext_t", "int_h", ext_h", "db", and "pa". Thanks.

回答1:

There is an is.na.data.frame function, but no is.na<-.data.frame, so ... only perhaps:

is.na( dfrm[ , c( "int_t", "ext_t", "int_h", ext_h", "db", "pa")] ) <- 
     dfrm[ , c( "int_t", "ext_t", "int_h", ext_h", "db", "pa")]==0

You should provide a test case. (But now that I have tested it I think ti should succeed.)

回答2:

Try

nm <- c("int_t", "ext_t", "int_h", "ext_h", "db", "pa")
df[nm] <- (NaN^!df[nm])*df[nm]
df
#   int_t ext_t int_h ext_h ColN  db pa
#1      5     2     5     4    2   2  4   
#2      5     4   NaN     4    2   2  5
#3      1     5     5     2    0   2  4
#4      4     1     5     4    5   4  3
#5      3     2   NaN   NaN    2 NaN  5
#6      3     5     3     4    5   4  1
#7      4     5     2   NaN    5   4  1
#8    NaN   NaN     5     1    3   1  4
#9      3     2     2     5    5   1  4
#10     4     3     5     3    3   3  1

data

set.seed(42)
df <- as.data.frame(matrix(sample(0:5, 7*10, replace=TRUE), 
    ncol=7, dimnames=list(NULL, c( "int_t", "ext_t", "int_h", 
   "ext_h", "ColN", "db", "pa"))))