Lua string.gsub with Multiple Patterns

问题

I am working on renaming the Movie titles that has unwanted letters. The string.gsub can replace a string with "" nil value but I have around 200 string patterns that need to be replaces with "".

Right now I have to string.gsub for every pattern. I was thinking is there is a way to put all the string patterns in to single string.gsub line. I have searched around the web for the solution but still didn't got anything.

The movie title is like this B.A.Pass 2013 Hindi 720p DvDRip CROPPED AAC x264 RickyKT and I want to remove the extra characters like 2013, Hindi, 720p, DvDRip, CROPPED, AAC, x264, RickyKT.

回答1:

You can pass to string.gsub a table as the third argument like this:

local movie = "B.A.Pass 2013 Hindi 720p DvDRip CROPPED AAC x264 RickyKT"
movie = movie:gsub("%S+", {["2013"] = "", ["Hindi"] = "", ["720p"] = "", 
                       ["DvDRip"] = "", ["CROPPED"] = "", ["AAC"] = "", 
                       ["x264"] = "", ["RickyKT"] = ""})

print(movie)

回答2:

Put all of the patterns in a table and then enumerate the table, calling string.gsub() for each pattern:

str = "B.A.Pass 2013 Hindi 720p DvDRip CROPPED AAC x264 RickyKT"

patterns = {"pattern1", "pattern2", "pattern3"}
for i,v in ipairs(patterns) do
    str = string.gsub(str, v, "")
end

This will require many invocations of string.gsub(), but the code should be much more maintainable than having a lot of string.gsub() calls.

回答3:

You could do it in a simple function, that way you do not need to write the code each time per string, or just put string.gsub, and the replacement value for the string you need

Function:

local large_name = "B.A.Pass 2013 Hindi 720p DvDRip CROPPED AAC x264 RickyKT"

function clean_name(str)
  local v = string.gsub(str, "(.-)%s([%(%[']?%d%d%d?%d?[%)%]]?)%s*(.*)", "%1")
  return v
end

print(clean_name(large_name))

Only string.gsub for value

local large_name = "B.A.Pass 2013 Hindi 720p DvDRip CROPPED AAC x264 RickyKT"
local clean_name = string.gsub(large_name, "(.-)%s([%(%[']?%d%d%d?%d?[%)%]]?)%s*(.*)", "%1")

print(clean_name)

The replacement pattern places the first value (name of the movie) separated by a space and prints it, also identifies the year as the second value, to avoid error in the titles, so it is not necessary to place all the values that can exist within the name of the movie and will avoid many false positives

I add a testing function to test different movie names

local testing = {"Whiplash 2014 [1080p]",
"Anon (2018) [WEBRip] [1080p] [YTS.AM]",
"Maze Runner The Death Cure 2018 [WEBRip] [1080p] [YTS.AM]",
"12 Strong [2018] [WEBRip] [1080p] [YTS.AM]",
"Kingsman The Secret Service (2014) [1080p]",
"The Equalizer [2014] [1080p]",
"Annihilation 2018 [WEBRip] [1080p] [YTS.AM]",
"The Shawshank Redemption '94",
"Assassin's Creed 2016 HC 720p HDRip 850 MB - iExTV",
"Captain Marvel (2019) [WEBRip] [1080p] [YTS.AM]",}

for k,v in pairs(testing) do
  local result = string.gsub(v, "(.-)%s([%(%[']?%d%d%d?%d?[%)%]]?)%s*(.*)", "%1")
  print(result)
end

Output:

Whiplash
Anon
Maze Runner The Death Cure
12 Strong
Kingsman The Secret Service
The Equalizer
Annihilation
The Shawshank Redemption
Assassin's Creed
Captain Marvel

回答4:

To avoid to write keys and values on a table for every new entry, i'd write a function to handle a numerically indexed table (the patterns being the values).

This way I dont need to write {["pattern_n"] = ""} for every new pattern.

Ex:

PATTERNS = {"2013", "Hindi", "720p", "DvDRip", "CROPPED", "AAC", "x264", "RickyKT"}
function replace(match)
    local ret = nil
    for i, v in ipairs(PATTERNS) do
        if v:find(match) then
            ret = ""
        end
    end
    return ret
end


local movie = "B.A.Pass 2013 Hindi 720p DvDRip CROPPED AAC x264 RickyKT"
movie = movie:gsub("%S+", replace)

print(movie)

来源：https://stackoverflow.com/questions/25280374/lua-string-gsub-with-multiple-patterns

标签

string

lua

lua-patterns