Ca data <- cut(data$Time, breaks=seq(0, max(data$Time)+400, 400)) by(data$Oxytocin, cuts, mean)
but this would only work for only one person's data....But I have ten people with their own Time and oxytocin data....How would I get their averages simultaneously? Also instead of having this type output :
cuts: (0,400]
[1] 0.7
------------------------------------------------------------
cuts: (400,800]
[1] 0.805
Is there a way I can get a list of those cuts?
Here's a solution using IRanges package.
idx assumes your data format is Time, data, Time, data, ... and so on.. So, it creates indices 1,3,5,...ncol(df)-1.
ir1 is the intervals you would want the mean for. It's width is 400. It goes from 0 to max(Time) for each Time column (here columns 1 and 3).
ir2 is the corresponding Time column of interval width = 1.
Then I get the overlaps of ir1 with ir2, which basically tells me which intervals from ir2 overlap with ir1 (which we want), from which I calculate the mean and output the data.frame.
idx <- seq(1, ncol(df), by=2)
o <- lapply(idx, function(i) {
ir1 <- IRanges(start=seq(0, max(df[[i]]), by=401), width=401)
ir2 <- IRanges(start=df[[i]], width=1)
t <- findOverlaps(ir1, ir2)
d <- data.frame(mean=tapply(df[[i+1]], queryHits(t), mean))
cbind(as.data.frame(ir1), d)
})
> o
# [[1]]
# start end width mean
# 1 0 400 401 0.6750000
# 2 401 801 401 0.8050000
# 3 802 1202 401 0.8750000
# 4 1203 1603 401 0.2285333
# [[2]]
# start end width mean
# 1 0 400 401 0.73508
# 2 401 801 401 0.13408
# 3 802 1202 401 0.26408
# 4 1203 1603 401 1.06408
# 5 1604 2004 401 3.06408
For each Time column, you'll get a list with the intervals and mean for that interval.
