[Rd] Match .3 in a sequence

[Duncan Murdoch]

On 3/17/2009 11:26 AM, Daniel Murphy wrote:
> Is this a reasonably fast way to do an approximate match of a vector x to
> values in a list?
> 
> match.approx  <- function(x,list,tol=.0001)
>     sapply(apply(abs(outer(list,x,"-"))<tol,2,which),"[",1)

If you are willing to assume that the list values are all multiples of 
2*tol, then it's easy:  just divide both x and list by 2*tol, round to 
nearest integer, and use the regular match function.

If not, it becomes harder; I'd probably use a solution like yours.

Duncan Murdoch

> 
> Thanks.
> -Dan
> 
> On Mon, Mar 16, 2009 at 8:24 AM, Stavros Macrakis <macrakis at alum.mit.edu>wrote:
> 
>> Well, first of all, seq(from=.2,to=.3) gives c(0.2), so I assume you
>> really mean something like seq(from=.2,to=.3,by=.1), which gives
>> c(0.2, 0.3).
>>
>> %in% tests for exact equality, which is almost never a good idea with
>> floating-point numbers.
>>
>> You need to define what exactly you mean by "in" for floating-point
>> numbers.  What sort of tolerance are you willing to allow?
>>
>> Some possibilities would be for example:
>>
>> approxin <- function(x,list,tol) any(abs(list-x)<tol)   # absolute
>> tolerance
>>
>> rapproxin <- function(x,list,tol) (x==0 && 0 %in% list) ||
>> any(abs((list-x)/x)<=tol,na.rm=TRUE)
>>     # relative tolerance; only exact 0 will match 0
>>
>> Hope this helps,
>>
>>          -s
>>
>> On Mon, Mar 16, 2009 at 9:36 AM, Daniel Murphy <chiefmurphy at gmail.com>
>> wrote:
>> > Hello:I am trying to match the value 0.3 in the sequence seq(.2,.3). I
>> get
>> >> 0.3 %in% seq(from=.2,to=.3)
>> > [1] FALSE
>> > Yet
>> >> 0.3 %in% c(.2,.3)
>> > [1] TRUE
>> > For arbitrary sequences, this "invisible .3" has been problematic. What
>> is
>> > the best way to work around this?
>>
> 
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