[R] help with column substaction with a twist

[jim holtman]

try this:


> # generate test data
> set.seed(1)
> n <- 100
> test <- data.frame(p = sample(10, n, TRUE)
+                 , b = sample(10, n, TRUE)
+                 )
> test$e <- sample(5, n, TRUE) + test$b  # make sure e > b
> # add distance
> test$dist <- ifelse(test$p < test$b
+                 , test$p - test$b
+                 , ifelse(test$p > test$e
+                     , test$p - test$e
+                     , 0L  # within
+                     )
+                 )
> head(test, 10)
    p  b  e dist
1   3  7  9   -4
2   4  4  6    0
3   6  3  6    0
4  10 10 12    0
5   3  7  8   -4
6   9  3  6    3
7  10  2  5    5
8   7  5  6    1
9   7 10 12   -3
10  1  6 10   -5

Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.


On Sun, Jul 20, 2014 at 7:33 AM, Ubernerdy <ubernerdy at gmail.com> wrote:
> Hello guys!
>
> I have been messing around with R for a while now, but this situation has
> me a bit stumped. I was unable to solve it by reading documentation.
>
> So I have this table (currently in Excel - could export it as csv) with
> values in 3 columns. Let's call them value P (for position), value B
> (beginning) and E (end). P represents the position of a mutation in the
> genome, B and E are the beginnings and ends of a nearby gene that either
> contains the mutation or not.
>
> I am trying to compute the distance between the mutation and the gene.
>
> If the mutation is contained in the gene, that is value P is greater than B
> and lesser than E, the result is 0.
>
> If the mutation is "left" of the gene, the distance is negative and is
> equal to P-B.
>
> If the mutation is "right" of the gene, the distance is positive and is
> equal to P-E.
>
> How would i achieve this in R?
>
> Regards and thanks, S.
>
>         [[alternative HTML version deleted]]
>
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