Calculating string similarity as a percentage

问题:

The given function uses "stringdist" package in R and tells the minimum changes needed to change one string to another. I wish to find out how much similar is one string to another in "%" format. Please help me and thanks.

stringdist("abc","abcd", method = "lv")

回答1:

You can use RecordLinkage package and use the function levenshteinSim, i.e.

#This gives the similarity
RecordLinkage::levenshteinSim('abc', 'abcd')
#[1] 0.75

#so to get the distance just subtract from 1, 
1 - RecordLinkage::levenshteinSim('abc', 'abcd')
#[1] 0.25

回答2:

Something like this might work:

d <- data.frame(original = c("abcd", "defg", "hij"), new = c("abce", "zxyv", "hijk"))
d$dist <- stringdist(d$original, d$new, method = "lv")
d$similarity <- 1 - d$dist / nchar(as.character(d$original))

#### Returns:
####   original  new dist similarity
#### 1     abcd abce    1  0.7500000
#### 2     defg zxyv    4  0.0000000
#### 3      hij hijk    1  0.6666667

回答3:

Here is a function in base R. I added a check for vectors of equal length as inputs. You could change this logic if desired.

strSim <- function(v1, v2) {
            if(length(v1) == length(v2)) 1 - (adist(v1, v2) / pmax(nchar(v1), nchar(v2)))
            else stop("vector lengths not equal")}

this returns

strSim("abc", "abcd")
     [,1]
[1,] 0.75