# One-to-one best string match [closed]

I have two vectors of strings, and each element in one vector matches one or zero elements in the other vector. No one-to-many or many-to-one pairs.

The closest matches are found with a stringdist calculation, see MWE

library(stringdist)
names.variant.1 <- c("Apple", "Banana", "Citrus", "Orange", "Pear", "Apple Pear")
names.variant.2 <- c("Apples (Fruit)", "Banan", "Citrusfruit", "Some really good oranges", "Pear-fruits", "Pear Apple")
dist.mat <- stringdistmatrix(a=names.variant.1, b=names.variant.2, useNames="strings",method="cosine")
dist.mat[, "Apples (Fruit)"]
# Apple      Banana     Citrus     Orange     Pear       Apple Pear
# 0.3385622  1.0000000  0.4896896  0.7958759  0.7500000  0.3318469
dist.mat[, "Pear Apple"]
# Apple      Banana     Citrus     Orange     Pear       Apple Pear
# 0.1918780  0.7857143  0.8908911  0.5635642  0.3318469  0.0000000

Here the closest match of "Apples (Fruit)" is "Apple Pear" (should be "Apple"). But the match "Pear Apple" to "Apple Pear" is stronger (and correct).

Is there any method that uses the knowledge that there can only be one-to-one pairs?

In the MWE, the match "Pear Apple" to "Apple Pear" should have disqualified "Apples (Fruit)" to "Apple Pear".

Note: another method than cosine might solve the MWE, but please disregard that. It is how to use the constraint one-to-one or one-to-zero I'm after.

Update: After realising that the problem was how to minimize pairs in the distance matrix, I found this solution:

# Update with solution
library(clue)
solve_LSAP(dist.mat, maximum=FALSE)
# Optimal assignment:
#   1 => 1, 2 => 2, 3 => 3, 4 => 4, 5 => 5, 6 => 6
• I thought questions about solving linear sum assignment problems using the Hungarian method belonged to this site. – Chris Feb 15 '19 at 16:09