Hi,
I see no need to construct the vector, try this instead,
belong = function(x=4, y=c(1,10)) x = y[2] x = y[1]
see also ?findInterval
HTH,
baptiste
On 13 August 2010 01:10, fishkbob fishk...@gmail.com wrote:
So basically I want to do this -
4 %in% 1:10
should return true
Would
Awesome, that works great, and it cuts my runtime down by a lot.
thanks baptiste
I totally forgot that I could just check to see if the number was inbetween
via less than and greater than rather than having to construct the vector.
findInterval also helps with another problem I was having
So basically I want to do this -
4 %in% 1:10
should return true
Would there be another way of doing this without having to do the 1:10 part?
I am using a very large data set and trying to do
459124 %in% 103000:983000
multiple times for many values, and it is taking quite a long time
Also, I
Hi,
Here is one way:
is.there - function(mynumber, a, b) mynumber %in% a:b
is.there(4, 1, 10)
[1] TRUE
is.there(459124, 103000, 983000)
[1] TRUE
is.there(4, 103000, 983000)
[1] FALSE
HTH,
Jorge
On Thu, Aug 12, 2010 at 7:10 PM, fishkbob wrote:
So basically I want to do this -
4 %in%
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