问题
Hi guys I asked a similar question about this here previously and I got some really awesome answers. But it turns out I have more data I need to work with sadly. So I have sample data in this format which is generated using head(data). So I have 3 specimens with their own Time and speed data....I'm not exactly using speed for my actual data
Time Speed Time.1 Speed.1 Time.2 Speed.2
Error in unique.default(x) : unique() applies only to vectors
Would it be possible for you guys to tell me what to do? I can do this in excel with a formula but I have so much data that excel crashes so I really need 'R' but have very minimal knowledge about it...thanks guys....
回答1:
Run the same code. Why are you editing by=3
? You want to create idx = 1, 3, 5 ...
. by=3
, will create 1, 4, 7...
. Your data, as long as it is in the format Time, data, Time, data, Time, data, ..., ...
, you can use the same code
require(IRanges)
# by equals 2 because we want to get the `Time` column index every time
idx <- seq(1, ncol(data), by=2)
# idx is now 1, 3, 5. It will be passed one value at a time to `i`.
# that is, `i` will take values 1 first, then 3 and then 5 and each time
# the code within is executed.
o <- lapply(idx, function(i) {
ir1 <- IRanges(start=seq(0, max(data[[i]]), by=401), width=401)
ir2 <- IRanges(start=data[[i]], width=1)
t <- findOverlaps(ir1, ir2)
d <- data.frame(mean=tapply(data[[i+1]], queryHits(t), mean))
cbind(as.data.frame(ir1), d)
})
gives me for this data:
# > o
# [[1]]
# start end width mean
# 1 0 400 401 1.05
#
# [[2]]
# start end width mean
# 1 0 400 401 1.1
#
# [[3]]
# start end width mean
# 1 0 400 401 1.383333
来源:https://stackoverflow.com/questions/15018668/r-programming-fixing-error