I guess I don't know which.min as well as I thought.
I'm trying to find the occurrence in a vector of a minimum value that is positive.
TIME <- c(0.00000, 4.47104, 6.10598, 6.73993, 8.17467, 8.80862, 10.00980, 11.01080, 14.78110, 15.51520, 16.51620, 17.11680)
I want to know for the values z of 1 to 19, the index of the above vector TIME containing the value that is closest to but above z. I tried the following code:
vec <- sapply(seq(1,19,1), function(z) which.min((z-TIME > 0)))
vec
#[1] 2 2 2 2 3 3 5 5 7 7 8 9 9 9 10 11 12 1 1
To my mind, the last two values of vec should be '12, 12'. The reason it's doing this is because it thinks that '0.0000' is closest to 0.
So, I thought that maybe it was because I exported the data from external software and that 0.0000 wasn't really 0. But,
TIME[1]==0 #TRUE
Then I got further confused. Why do these give the answer of index 1, when really they should be an ERROR?
which.min(0 > 0 ) #1
which.min(-1 > 0 ) #1
I'll be glad to be put right.
EDIT:
I guess in a nutshell, what is the better way to get this result:
#[1] 2 2 2 2 3 3 5 5 7 7 8 9 9 9 10 11 12 12 12
which shows the index of TIME that gives the smallest possible positive value, when subtracting each element of TIME from the values of 1 to 19.
The natural function to use here (both to limit typing and for efficiency) is actually not which.min + sapply but the cut function, which will determine which range of times each of the values 1:19 falls into:
cut(1:19, breaks=TIME, right=FALSE)
# [1] [0,4.47) [0,4.47) [0,4.47) [0,4.47) [4.47,6.11) [4.47,6.11) [6.74,8.17)
# [8] [6.74,8.17) [8.81,10) [8.81,10) [10,11) [11,14.8) [11,14.8) [11,14.8)
# [15] [14.8,15.5) [15.5,16.5) [16.5,17.1) <NA> <NA>
# 11 Levels: [0,4.47) [4.47,6.11) [6.11,6.74) [6.74,8.17) [8.17,8.81) ... [16.5,17.1)
From this, you can easily determine what you're looking for, which is the index of the smallest element in TIME greater than the cutoff:
(x <- as.numeric(cut(1:19, breaks=TIME, right=FALSE))+1)
# [1] 2 2 2 2 3 3 5 5 7 7 8 9 9 9 10 11 12 NA NA
The last two entries appear as NA because there is no element in TIME that exceeds 18 or 19. If you wanted to replace these with the largest element in TIME, you could do so with replace:
replace(x, is.na(x), length(TIME))
# [1] 2 2 2 2 3 3 5 5 7 7 8 9 9 9 10 11 12 12 12
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