68 lines
1.9 KiB
Lua
68 lines
1.9 KiB
Lua
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--require 'fun' ()
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local utf8 = require 'utf8'
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local ASCII_CHAR_PATTERN = '[\32-\126\009\010\013]'
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local UNICODE_CHAR_PATTERN = '[\01-\127\192-\255][\128-\191]*'
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local function probability_of_ascii_string (str)
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assert(type(str) == 'string')
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-- Find ascii subsequences of the string.
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-- Then find the total number of ascii characters,
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-- and the length of the longest subsequence.
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local len_of_longest_subseq, nr_ascii_chars = 0, 0
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for subseq in str:gmatch(ASCII_CHAR_PATTERN..'+') do
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len_of_longest_subseq = math.max(#subseq, len_of_longest_subseq)
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nr_ascii_chars = nr_ascii_chars + #subseq
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end
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-- Perform probability calculation
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-- This heuristic is based on the observation that large numbers of
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-- ascii characters, and long subsequences are the primary indicators
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-- of ascii strings.
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return (len_of_longest_subseq + nr_ascii_chars) / (2 * #str)
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end
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local function probability_of_utf8_string (str)
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assert(type(str) == 'string')
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-- Find numbers of valid utf8 bytes
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local valid_bytes = 0
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for char, valid in utf8.iterate(str) do
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if valid then valid_bytes = valid_bytes + #char end
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end
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-- Calculate ratio of valid bytes to total number of bytes.
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return valid_bytes / #str
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end
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local function probability_of_utf16_string (str)
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return 0
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end
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local function probability_of_binary_data (str)
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return 2/3
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end
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local str_representations = {
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ascii = probability_of_ascii_string,
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utf8 = probability_of_utf8_string ,
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utf16 = probability_of_utf16_string,
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binary = probability_of_binary_data,
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}
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return function (str)
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local str_info, most_likely, most_likely_prob = {}, 'ascii', 0
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for repr_name, prob_func in pairs(str_representations) do
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local prob = prob_func(str)
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str_info[repr_name..'_prob'] = prob
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if prob >= most_likely_prob then
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most_likely, most_likely_prob = repr_name, prob
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end
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end
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str_info.most_likely = most_likely
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return str_info
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end
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