Identify a common pattern [duplicate]

问题:

This question already has answers here:

Find common substrings between two character variables (3 answers)

Closed 4 years ago.

Is there a (easy) possibility to identify a common pattern which two strings share? Here is a little example to make clear what I mean:

I have two variables containing a string. Both include the same pattern ("ABC") and also some "noise".

a <- "xxxxxxxxxxxABCxxxxxxxxxxxx"
b <- "yyyyyyyyyyyyyyyyyyyyyyyABC"

Lets say I don't know the common pattern and I want R to find out that both strings contain "ABC". How can I do this?

*edit

The first example was maybe a bit to simplistic. Here is a example from my real data.

a <- "DUISBURG-HAMBORNS"
b <- "DUISBURG (-31.7.29)S"

Both strings contain "DUISBURG" which I want the function to identify.

*edit

I took the solution proposed in the link posted in the comments. But I still have not exactly what I want.

library(qualV)
LCS(strsplit(a[1], '')[[1]],strsplit(b[1], '')[[1]])$LCS

[1] "D" "U" "I" "S" "B" "U" "R" "G" "-" " " " " "S"

If the function is looking for the longest common subsequence of the two vectors, why does it not stop after "D" "U" "I" "S" "B" "U" "R" "G"? .

回答1:

Function LCS from qualV package (in Find common substrings between two character variables; not a possible duplicate) does something else than what you need. It solves the longest common subsequence problem, where subsequences are not required to occupy consecutive positions within the original sequences.

What you have is the longest common substring problem, for which you could use this algorithm, and here is the code assuming that there is a unique (in terms of length) longest common substring:

a <- "WWDUISBURG-HAMBORNS"
b <- "QQQQQQDUISBURG (-31.7.29)S"

A <- strsplit(a, "")[[1]]
B <- strsplit(b, "")[[1]]

L <- matrix(0, length(A), length(B))
ones <- which(outer(A, B, "=="), arr.ind = TRUE)
ones <- ones[order(ones[, 1]), ]
for(i in 1:nrow(ones)) {
  v <- ones[i, , drop = FALSE]
  L[v] <- ifelse(any(v == 1), 1, L[v - 1] + 1)
}
paste0(A[(-max(L) + 1):0 + which(L == max(L), arr.ind = TRUE)[1]], collapse = "")
# [1] "DUISBURG"