NaN from function

Dear Helpers,
I wrote a simple function to standardise variables if they
contain more than one value. If the elements of the variable
are all identical, then I want the function to return zero.

When I submit variables whose elements are all identical to
the function, it returns not zero, but NaNs.

zt=function(x){if (length(table(x)>1)) y=(x-mean(x))/sd(x) else if
(length(table(x)==1)) y=0; return(y)}

zt(c(1:10))
#[1] -1.4863011 -1.1560120 -0.8257228 -0.4954337 -0.1651446  0.1651446 
#0.4954337  0.8257228  1.1560120  1.4863011

zt(rep(1,10))
#[1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN

Would you be so kind as to point out what I am doing wrong, here?
How can I obtain zeros from my function, instead of NaNs?
(I obtain NaNs also if I set the function to zt=function(x){
 if (length(table(x)>1)) y=(x-mean(x))/sd(x) else if
(length(table(x)==1)) y=rep(0, length(x)); return(y)} ).

Thanks, in advance, for your help,
Jonathan Williams

The issue here, Jonathan, is that when you evaluate
(x-mean(x))/sd(x) for a vector x whose elements are all equal,
not only is (x-mean(x)) = 0, but also sd(x) = 0, so you are
asking the function to return the result of 0/0. Since this
is undefined, the result is NaN.

A basic solution for this special case would be

  zt=function(x){
    if (sd(x) == 0) return(0*x) else return( (x-mean(x))/sd(x) )
  }

This should cover the case where length(table(x))==1 (see also below).

I'm not happy about your conditions

  if (length(table(x)>1))
  if (length(table(x)==1))

since they ask for "length(table(x)>1)", which doesn't seem
to represent any natural criterion. E.g.:

  length(table(1:10)>1)
  # [1] 10
  length(table(rep(1,10))>1)
  # [1] 1

  if(length(table(1:10)>1)) y <- "Yes" else y <- "No" ; y
  # [1] "Yes"
  if(length(table(rep(1,10))>1)) y <- "Yes" else y <- "No" ; y
  # [1] "Yes"

  length(table(1:10)==1)
  # [1] 10
  length(table(rep(1,10))==1)
  # [1] 1
  
  if(length(table(1:10)==1)) y <- "Yes" else y <- "No" ; y
  # [1] "Yes"
  if(length(table(rep(1,10))==1)) y <- "Yes" else y <- "No" ; y
  # [1] "Yes"

I suspect you meant to write

  if (length(table(x))>1)
and
  if (length(table(x)))==1)

since this distinguishes between two more more different values
(length(table(x)) > 1) and all equal values (length(table(x)) == 1).

Ted.



-------------------------------------------------
E-Mail: (Ted Harding) <Ted.Harding at wlandres.net>
Date: 23-Feb-2012  Time: 16:40:03
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