R agrep：如何匹配超过1个替换

Question

I'm trying to match a string to a vector of strings:

a <- c('abcde', 'abcdf', 'abcdg')

agrep('abcdh', a, max.distance=list(substitutions=1))
# [1] 1 2 3

agrep('abchh', a, max.distance=list(substitutions=2))
# character(0)

I didn't expect the latter result as substituting two characters from the pattern makes the pattern identical to the vector elements. This does, however, work with all instead of substitutions :

agrep('abchh', a, max.distance=list(all=2))
# [1] 1 2 3

What do I need to change to match with more than 1 substitution allowed? Is substitution just a broken option? Thanks.

Note: this question is essentially the same as this one: https://stat.ethz.ch/pipermail/r-help/2011-June/281731.html , but that was never answered.

Answer 1

I did not realize that the questions were that old, anyway:

The function needs cost to be appropiate. As ping said, you must set the maximum number of match cost; in your example:

a <- c('abcde', 'abcdf', 'abcdg')
agrep('abcdh', a, max.distance = list(cost = 1))
[1] 1 2 3
agrep('abchh', a, max.distance = 2)
[1] 1 2 3

Now, if you set cost the program can do insertions, deletions and substitutions. If you want only evaluate substitutions, then:

agrep('abhhh', a, 
        max.distance=list(cost=3, substitutions=3, 
                          deletions=0, insertions=0))
[1] 1 2 3