SOG <- c(4,4,0,0,0,3,4,5,0,0,1,2,0,0,0) #run length encoding: tmp <- rle(SOG) #turn values into logicals tmp$values <- tmp$values == 0 #cumulative sum of TRUE values tmp$values[tmp$values] <- cumsum(tmp$values[tmp$values]) #inverse the run length encoding inverse.rle(tmp) #[1] 0 0 1 1 1 0 0 0 2 2 0 0 3 3 3
One option seems to be the use of {} as in: dat %>% group_by(yy = {yy = rle(x); rep(seq_along(yy$lengths), yy$lengths)}) %>% summarize(mean(y)) #Source: local data frame [4 x 2] # # yy mean(y) # (int) (dbl) #1 1 2.0 #2 2 4.5 #3 3 6.0 #4 4 7.0 It would be nice if future dplyr … Read more
Here is a compact solution without loop, cellfun or arrayfun: chainCode=”11012321170701000700000700766666666666665555555544443344444333221322222322″; numCode = chainCode – ‘0’; % turn to numerical array J=find(diff([numCode(1)-1, numCode])); relMat=[numCode(J); diff([J, numel(numCode)+1])];
Core logic: # Example vector and rle object x = rev(rep(6:10, 1:5)) rle_x = rle(x) # Compute endpoints of run end = cumsum(rle_x$lengths) start = c(1, lag(end)[-1] + 1) # Display results data.frame(start, end) # start end # 1 1 5 # 2 6 9 # 3 10 12 # 4 13 14 # 5 … Read more
Here’s one way I like to accomplish this: >> index = zeros(1,sum(rep)); >> index(cumsum([1 rep(1:end-1)])) = 1; index = 1 0 1 0 0 1 1 0 0 0 0 >> index = cumsum(index) index = 1 1 2 2 2 3 4 4 4 4 4 >> vv = v(index) vv = 3 3 … Read more
To find out which one is the most efficient solution, we provide a test-script that evaluates the performance. The first plot depicts runtimes for growing length of the vector runLengths, where the entries are uniformly distributed with maximum length 200. A modification of gnovice‘s solution is the fastest, with Divakar‘s solution being second place. This … Read more
Using rleid from data.table, library(data.table) rleid(x, prefix = “Group_”) #[1] “Group_1” “Group_1” “Group_1” “Group_2” “Group_2” “Group_2” “Group_3” “Group_4” “Group_4” “Group_5” “Group_5”
You need to use sequence and rle: > sequence(rle(as.character(dataset$input))$lengths) [1] 1 1 2 1 2 1 1 2 3 4 1 1
I’m a fan of the KRON function: >> a = 1:3; >> N = 3; >> b = kron(a,ones(1,N)) b = 1 1 1 2 2 2 3 3 3 You can also look at this related question (which dealt with replicating elements of 2-D matrices) to see some of the other solutions involving matrix … Read more
Problem Statement We have an array of values, vals and runlengths, runlens: vals = [1,3,2,5] runlens = [2,2,1,3] We are needed to repeat each element in vals times each corresponding element in runlens. Thus, the final output would be: output = [1,1,3,3,2,5,5,5] Prospective Approach One of the fastest tools with MATLAB is cumsum and is … Read more