Problem: Given an atomic vector, find the start and end indices of runs in the vector.
Example vector with runs:
x = rev(rep(6:10, 1:5))
# [1] 10 10 10 10 10 9 9 9 9 8 8 8 7 7 6
Output from rle()
:
rle(x)
# Run Length Encoding
# lengths: int [1:5] 5 4 3 2 1
# values : int [1:5] 10 9 8 7 6
Desired output:
# start end
# 1 1 5
# 2 6 9
# 3 10 12
# 4 13 14
# 5 15 15
The base rle
class doesn't appear to provide this functionality, but the class Rle
and function rle2
do. However, given how minor the functionality is, sticking to base R seems more sensible than installing and loading additional packages.
There are examples of code snippets (here, here and on SO) which solve the slightly different problem of finding start and end indices for runs which satisfy some condition. I wanted something that would be more general, could be performed in one line, and didn't involve the assignment of temporary variables or values.
Answering my own question because I was frustrated by the lack of search results. I hope this helps somebody!
runs.lengths.cumsum = cumsum(runs$lengths) ends = runs.lengths.cumsum[myruns] Next, we find the start positions of these runs. newindex = ifelse(myruns>1, myruns-1, 0) starts = runs.lengths.cumsum[newindex] + 1 if (0 %in% newindex) starts = c(1,starts)
Given an array of integers, task is to find the starting and ending position of a given key. Input : arr [] = {1, 2, 3, 4, 5, 5} Key = 5 Output : Start index: 4 Last index: 5 Explanation: Starting index where 5 is present is 4 and ending address is 5.
Next, we can do a cumulative sum of the run lengths and extract the end positions of the runs with length of at least 5 using the above found indices. runs.lengths.cumsum = cumsum(runs$lengths) ends = runs.lengths.cumsum[myruns]
Q: Given a sequence of random numbers, find the start and end positions of runs of five or more consecutive numbers that are greater than zero. A: Use the rle () function. For example, let’s apply rle () to the following sequence of numbers. seq3 = c(2,2,2,2,2,5,3,7,7,7,2,2,5,5,5,3,3,3) ( rle.seq3 = rle(seq3) )
A data.table
possibility, where .I
and .N
are used to pick relevant indices, per group defined by rleid
runs.
library(data.table)
data.table(x)[ , .(start = .I[1], end = .I[.N]), by = rleid(x)][, rleid := NULL][]
# start end
# 1: 1 5
# 2: 6 9
# 3: 10 12
# 4: 13 14
# 5: 15 15
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 15 15
Tidyverse/dplyr
way (data frame-centric):
library(dplyr)
rle(x) %>%
unclass() %>%
as.data.frame() %>%
mutate(end = cumsum(lengths),
start = c(1, dplyr::lag(end)[-1] + 1)) %>%
magrittr::extract(c(1,2,4,3)) # To re-order start before end for display
Because the start
and end
vectors are the same length as the values
component of the rle
object, solving the related problem of identifying endpoints for runs meeting some condition is straightforward: filter
or subset the start
and end
vectors using the condition on the run values.
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