You need to use interp1 with ‘linear’ interpolation method: >> v = [1 2 3 4 10]; >> newNum = 13; % new number of elements in the “buffed” vector >> iv = interp1( linspace(0,1,numel(v)), v, linspace(0,1,newNum) ) iv = 1.0000 1.3333 1.6667 2.0000 2.3333 2.6667 3.0000 3.3333 3.6667 4.0000 6.0000 8.0000 10.0000
There is an Android class called Visualizer that lets you access audio being played http://developer.android.com/reference/android/media/audiofx/Visualizer.html In addition, since Android is all open source, you can look at how the live wallpaper does it! https://android.googlesource.com/platform/packages/wallpapers/MusicVisualization/
Two main ways. Use Random#nextInt(int): List<Foo> list = createItSomehow(); Random random = new Random(); Foo foo = list.get(random.nextInt(list.size())); It’s however not guaranteed that successive n calls returns unique elements. Use Collections#shuffle(): List<Foo> list = createItSomehow(); Collections.shuffle(list); Foo foo = list.get(0); It enables you to get n unique elements by an incremented index (assuming that the … Read more
I would suggest using either stratified from my “splitstackshape” package, or sample_n from the “dplyr” package: ## Sample data set.seed(1) n <- 1e4 d <- data.table(age = sample(1:5, n, T), lc = rbinom(n, 1 , .5), ants = rbinom(n, 1, .7)) # table(d$age, d$lc) For stratified, you basically specify the dataset, the stratifying columns, and … Read more