(no subject)
Hi
Hi,
The parentheses are in the wrong places in the two if() statements.
Look here:
(length(table(x)>1))
^ ^
(length(table(x)==1))
^ ^
In both cases you're checking whether
the length of the comparison (table(x) > 1) or (table(x) == 1)
is 1, which it always is regardless of whether the
comparison itself is true or false. If you move those, it
should be fine. Although I think I'd use length(unique(x)) instead.
R's scale function is intended to do such things. zt2 <- function(x) as.numeric(ifelse(is.nan(scale(x)), 0, scale(x))) Regards Petr
Sarah On Thu, Feb 23, 2012 at 9:19 AM, Jonathan Williams <jonathan.williams at dpag.ox.ac.uk> wrote:
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 __
-- Sarah Goslee http://www.functionaldiversity.org
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