Find the last index of a character in a string - string

Question

I want to have ability to use a lastIndexOf method for the strings in my Lua (Luvit) project. Unfortunately there's no such method built-in and I'm bit stuck now.
In Javascript it looks like:
'my.string.here.'.lastIndexOf('.') // returns 14

Answer 1

function findLast(haystack, needle)
local i=haystack:match(".*"..needle.."()")
if i==nil then return nil else return i-1 end
end
s='my.string.here.'
print(findLast(s,"%."))
print(findLast(s,"e"))
Note that to find . you need to escape it.

Answer 2

If you have performance concerns, then this might be a bit faster if you're using Luvit which uses LuaJIT.
local find = string.find
local function lastIndexOf(haystack, needle)
local i, j
local k = 0
repeat
i = j
j, k = find(haystack, needle, k + 1, true)
until j == nil
return i
end
local s = 'my.string.here.'
print(lastIndexOf(s, '.')) -- This will be 15.
Keep in mind that Lua strings begin at 1 instead of 0 as in JavaScript.

Answer 3

Here’s a solution using
LPeg’s position capture.
local lpeg = require "lpeg"
local Cp, P = lpeg.Cp, lpeg.P
local lpegmatch = lpeg.match
local cache = { }
local find_last = function (str, substr)
if not (str and substr)
or str == "" or substr == ""
then
return nil
end
local pat = cache [substr]
if not pat then
local p_substr = P (substr)
local last = Cp() * p_substr * Cp() * (1 - p_substr)^0 * -1
pat = (1 - last)^0 * last
cache [substr] = pat
end
return lpegmatch (pat, str)
end
find_last() finds the last occurence of substr in the string
str, where substr can be a string of any length.
The first return value is the position of the first character of
substr in str, the second return value is the position of the
first character following substr (i.e. it equals the length of the
match plus the first return value).
Usage:
local tests = {
A = [[fooA]], --> 4, 5
[""] = [[foo]], --> nil
FOO = [[]], --> nil
K = [[foo]], --> nil
X = [[X foo X bar X baz]], --> 13, 14
XX = [[foo XX X XY bar XX baz X]], --> 17, 19
Y = [[YYYYYYYYYYYYYYYYYY]], --> 18, 19
ZZZ = [[ZZZZZZZZZZZZZZZZZZ]], --> 14, 17
--- Accepts patterns as well!
[P"X" * lpeg.R"09"^1] = [[fooX42barXxbazX]], --> 4, 7
}
for substr, str in next, tests do
print (">>", substr, str, "->", find_last (str, substr))
end

Answer 4

To search for the last instance of string needle in haystack:
function findLast(haystack, needle)
--Set the third arg to false to allow pattern matching
local found = haystack:reverse():find(needle:reverse(), nil, true)
if found then
return haystack:len() - needle:len() - found + 2
else
return found
end
end
print(findLast("my.string.here.", ".")) -- 15, because Lua strings are 1-indexed
print(findLast("my.string.here.", "here")) -- 11
print(findLast("my.string.here.", "there")) -- nil
If you want to search for the last instance of a pattern instead, change the last argument to find to false (or remove it).

Answer 5

Can be optimized but simple and does the work.
function lastIndexOf(haystack, needle)
local last_index = 0
while haystack:sub(last_index+1, haystack:len()):find(needle) ~= nil do
last_index = last_index + haystack:sub(last_index+1, haystack:len()):find(needle)
end
return last_index
end
local s = 'my.string.here.'
print(lastIndexOf(s, '%.')) -- 15