Is there way to replace n occurrences starting from the end of a string?

General Usage
1

Let’s say you have:

foo = "fizz_buzz_fizz_buzz_fizz_buzz_fizz_buzz_fizz_buzz"

and you want to change the last two buzzes to bars:

nu_foo = "fizz_buzz_fizz_buzz_fizz_buzz_fizz_bar_fizz_bar"

What is the easiest way to do this with the least code.

For clarity, this would be the opposite of the current replace function with an limit count given:

julia> replace(foo, "buzz", "baz", 2)

"fizz_baz_fizz_baz_fizz_buzz_fizz_buzz_fizz_buzz"
2

In case nothing better is found, a rookery of reversions could do the trick

 reverse(replace(reverse(foo), reverse("buzz") => reverse("baz"), count=2))
"fizz_buzz_fizz_buzz_fizz_buzz_fizz_baz_fizz_baz"
2 Likes
3

I think this is the way to go, even if there’s a function for OP, probably would just be a wrapper

4

Yup, that’s the way to do it:

function rreplace(varargs...; kwargs...) 
    varargs = map(
        vararg -> isa(vararg, AbstractString) ? reverse(vararg) : vararg,
        varargs
    )
    
    reverse(replace(varargs...;kwargs...))
end
5

be careful:

julia> rreplace(foo, r"b.zz"=>"baz", count=2)
"fizz_buzz_fizz_buzz_fizz_buzz_fizabuzz_fizabuzz"
2 Likes
6

I’ll leave that as an exercise for the reader

7

Another option, more concise and efficient than the previous:

julia> join(rsplit(foo, "buzz", limit=3), "baz")
"fizz_buzz_fizz_buzz_fizz_buzz_fizz_baz_fizz_baz"

According to the documentation, this solution should also support regexes, but the code for that seems to be missing.

1 Like