Binning average in R?

Here is another way to do it: set. Seed(5) x = matrix(runif(1000), ncol = 10) nr = nrow(x) gr = rep(1:floor(nr/10), each = 10) aggregate(x ~ gr, FUN=mean),-1 which results in NA NA.1 NA.2 NA.3 NA.4 NA.5 NA.6 NA.7 1 0.5295264 0.5957229 0.4502069 0.5168083 0.3398190 0.4075922 0.6059122 0.5127865 2 0.4778341 0.3967321 0.4069635 0.4514742 0.6172677 0.2486085 0.6340686 0.4052600 3 0.5168132 0.5117207 0.5202261 0.5068593 0.5218041 0.4925462 0.5169584 0.4919296 4 0.3299557 0.3314723 0.4503393 0.3965103 0.6166598 0.5525628 0.4943880 0.6048207 5 0.6145423 0.5853235 0.4822182 0.3377771 0.3540784 0.5974846 0.5202577 0.5769518 6 0.5009249 0.5203701 0.3940540 0.4237508 0.3199265 0.4817713 0.4655320 0.6124400 7 0.7335082 0.5856578 0.3929621 0.6403662 0.5347719 0.5658542 0.4226456 0.7196593 8 0.4976663 0.5205538 0.4529273 0.4757352 0.6980300 0.5694570 0.4384924 0.5481236 9 0.5275932 0.5014861 0.5363340 0.5664576 0.5006055 0.5611069 0.3803889 0.4680865 10 0.4560031 0.5527328 0.4419076 0.6893043 0.5161281 0.5895931 0.3965911 0.3842419 NA.8 NA.9 1 0.3711607 0.5541607 2 0.4379255 0.4159131 3 0.5048523 0.5884052 4 0.4642687 0.4572388 5 0.6054209 0.5174784 6 0.4659952 0.5332438 7 0.4568273 0.3943798 8 0.6978356 0.5087778 9 0.4897584 0.4710949 10 0.6310546 0.4775762.

T( sapply(1:(NROW(A)/10), function(x) colMeans(A x:(x+9), ) ) ) You need the transpose operation to re-orient the result. One often needs to do so after an 'apply' operation.

I cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.