如何匹配R中的简单交替模式列表

Question

In R, I have data a vector of integers.

run <- sample.int(9, 1000, replace=T)
run[sample.int(1000, 100)] <- NA

If at least one of the following patterns, c(1, x, 1, y) or c(x, 1, y, 1) where x and y are either whole numbers or NA, is present, I would like to print out the start index of each pattern and update a count variable for each instance of a pattern. What is the most efficient way of doing this?

I was thinking of using the rle function and testing for every 4 consecutive values for a length of 1, and then testing whether they conform to one of the patterns. However, I am having problems with NAs with this approach since each NA is treated separately. Perhaps there is a better way to do this.

Answer 1

Taking your usage of sample.int as implying your vector only contains values from 1:9 and NA , here's a regular expressions approach:

run <- c(1, NA, 1, 3, 1, 1, NA, NA, NA, 1)
run[is.na(run)] <- 0
pat1 <- "(?=1[0-9]1[0-9])" # using a lookahead assertion around the pattern is a way to allow overlapping matches
pat1.idxs <- unlist(gregexpr(pat1, paste(run, collapse=''), perl=TRUE))
pat1.idxs
# match indexes
# [1] 1 3
length(pat1.idxs)
# counts
# [1] 2

Then you would do second pattern similarly.

Answer 2

This kind of task could be done with the rollapply function from the zoo package.

set.seed(42)
run <- sample.int(9, 1000, replace=T)
run[sample.int(1000, 100)] <- NA

# a list of the patterns
pattern <- list(c(1, NA, 1, NA), c(NA, 1, NA, 1))

library(zoo)

colSums(rollapply(run, length(pattern[[1]]),
                  function(x) sapply(pattern, identical, x)))

The result is a vector including the counts of the patterns in the pattern list:

[1] 0 0

Note . If the lengths of the patterns were different, rollapply had to be executed multiple times.