Find most repeated sequence of values - r

Question

Suppose I have a data frame that looks like this
ITEM
1 X
2 A
3 B
4 C
5 A
6 F
7 U
8 A
9 B
10 C
11 F
12 U
How can I obtain the most common sequence of values in the 'ITEM' column?. In this case the most frequent sequence would be A, B, C since it appears in row 2 to 4 and 8 to 10.
I have already tried the function rle, as well as some of the solutions found here, and I haven't been lucky. Can I have a suggestion, hint, or package recommendation?

Answer 1

I guess you want the longest non-overlapping sub-string. There's some good explanation about the dynamic programming solution here.
x = c("X", "A", "B", "C", "A", "F", "U", "A", "B", "C", "F", "U")
n = length(x)
m1 = sapply(x, function(i) sapply(x, function(j) as.integer(i == j)))
diag(m1) = 0
m1[lower.tri(m1)] = 0
m1
# X A B C A F U A B C F U
# X 0 0 0 0 0 0 0 0 0 0 0 0
# A 0 0 0 0 1 0 0 1 0 0 0 0
# B 0 0 0 0 0 0 0 0 1 0 0 0
# C 0 0 0 0 0 0 0 0 0 1 0 0
# A 0 0 0 0 0 0 0 1 0 0 0 0
# F 0 0 0 0 0 0 0 0 0 0 1 0
# U 0 0 0 0 0 0 0 0 0 0 0 1
# A 0 0 0 0 0 0 0 0 0 0 0 0
# B 0 0 0 0 0 0 0 0 0 0 0 0
# C 0 0 0 0 0 0 0 0 0 0 0 0
# F 0 0 0 0 0 0 0 0 0 0 0 0
# U 0 0 0 0 0 0 0 0 0 0 0 0
m2 = m1
for (i in 2:nrow(m1)){
for (j in 2:nrow(m1)){
if (m1[i-1, j-1] == 1 & m1[i, j] == 1){
if (j - i > m2[i - 1, j - 1]){
m2[i, j] = m2[i - 1, j - 1] + m2[i, j]
m2[i - 1, j - 1] = 0
} else {
m2[i, j] = 0
}
}
}
}
m2
# X A B C A F U A B C F U
# X 0 0 0 0 0 0 0 0 0 0 0 0
# A 0 0 0 0 1 0 0 0 0 0 0 0
# B 0 0 0 0 0 0 0 0 0 0 0 0
# C 0 0 0 0 0 0 0 0 0 3 0 0
# A 0 0 0 0 0 0 0 1 0 0 0 0
# F 0 0 0 0 0 0 0 0 0 0 0 0
# U 0 0 0 0 0 0 0 0 0 0 0 2
# A 0 0 0 0 0 0 0 0 0 0 0 0
# B 0 0 0 0 0 0 0 0 0 0 0 0
# C 0 0 0 0 0 0 0 0 0 0 0 0
# F 0 0 0 0 0 0 0 0 0 0 0 0
# U 0 0 0 0 0 0 0 0 0 0 0 0
ans_len = max(m2)
inds = c(which(m2 == ans_len, arr.ind = TRUE)[,2])
lapply(inds, function(ind) x[(ind - ans_len + 1):ind])
# [[1]]
# [1] "A" "B" "C"

Answer 2

A tidyverse solution mixed with nested apply functions. The solution is generalized and will report the most frequent non-trivial consecutive sequence that appears at least twice--tie goes to the longer sequence.
library(tidyverse)
# Data
x <- data.frame(ITEM = c("X", "A", "B", "C", "A", "F", "U", "A", "B", "C", "F", "U"), stringsAsFactors = F)
# convert x to vector
y <- x$ITEM
# Create list to check for sequence of each length 2 through n/2
l <- lapply(2:floor(length(y)/2), function(a) sapply(1:a, function(x) y[(0 + x):(length(y) - a + x)])) %>%
lapply(as.data.frame) %>%
setNames(sapply(2:(length(.) + 1), function(a) paste0("Consecutive", a)))
# Show most frequent sequence(s), choosing the longest
lapply(1:length(l), function(x) (as.data.frame(table(do.call(paste, l[[x]])), stringsAsFactors = F) %>%
dplyr::mutate(length = nchar(Var1)) %>%
dplyr::filter(length == max(length) & Freq == max(Freq) & Freq > 1)) ) %>%
.[which(sapply(., nrow) > 0)] %>%
dplyr::bind_rows() %>%
dplyr::filter(Freq == max(Freq)) %>%
dplyr::filter(length == max(length)) %>%
dplyr::rename(Sequence = Var1) %>%
dplyr::select(-length)
# Sequence Freq
#1 A B C 2