I am looking to calculate the percent match for a string in R. For example:
x <- "asdf"
y <- "fdjk"
I would like this to return .5 (e.g. 2 matches, irrespective of order). Any thoughts are greatly appreciated.
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1will the strings always be of the same length? what if they are different? – MrFlick May 15 '14 at 16:30
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they will not always be the same length; in that case, the denominator will the length of the longer vector. – coding_heart May 15 '14 at 16:32
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2And what would be the numerator if there where duplicate letters? ie "ab" and "aaabbb", or what about "aaaab" and "abbbb"? what are the right answers here? – MrFlick May 15 '14 at 16:37
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Great followup; I'd like them to be unique such that for the first example it is 1/3 + 1/3 and for the second 1/4 + 1/4 – coding_heart May 15 '14 at 16:44
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1So what is the rule you have that would result in those answers (2/3 and 1/2)? – MrFlick May 15 '14 at 16:46
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I was doing unique() for each letter over the total number the times that letter appears. – coding_heart May 15 '14 at 17:26
1 Answers
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You can split up a string into its specific characters with strsplit:
char.x <- strsplit(x, "")[[1]]
char.x
# [1] "a" "s" "d" "f"
char.y <- strsplit(y, "")[[1]]
char.y
# [1] "f" "d" "j" "k"
Now, you can use intersect and length to compute your metric (the exact formula is not clear because your post didn't specify, for instance, how to handle duplicate characters):
length(intersect(char.x, char.y)) /
max(length(unique(char.x)), length(unique(char.y)))
# [1] 0.5
josliber
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1BTW, if you do want to count dupes, (blowing my own horn again), the package `vecsets` will do intersections including duplicate elements. – Carl Witthoft May 15 '14 at 16:50
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You can use `strsplit` instead of `substring`: `strings <- sapply(list(x, y), strsplit, ""); intersect(strings[[1]], strings[[2]])` – Jilber Urbina May 15 '14 at 17:27
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1@Jilber thanks -- `microbenchmark` tells me that's much faster than `substring` – josliber May 15 '14 at 17:32
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