Nit*_*eti 2 ruby algorithm plagiarism-detection rabin-karp
我一直在研究一个小型的抄袭检测引擎,该引擎使用MOSS的 Idea .我需要一个滚动哈希函数,我受到Rabin-Karp算法的启发.
我写的代码 - >
#!/usr/bin/env ruby
#Designing a rolling hash function.
#Inspired from the Rabin-Karp Algorithm
module Myth
module Hasher
#Defining a Hash Structure
#A hash is a integer value + line number where the word for this hash existed in the source file
Struct.new('Hash',:value,:line_number)
#For hashing a piece of text we ned two sets of parameters
#k-->For buildinf units of k grams hashes
#q-->Prime which lets calculations stay within range
def calc_hash(text_to_process,k,q)
text_length=text_to_process.length
radix=26
highorder=(radix**(text_length-1))%q
#Individual hashes for k-grams
text_hash=0
#The entire k-grams hashes list for the document
text_hash_string=""
#Preprocessing
for c in 0...k do
text_hash=(radix*text_hash+text_to_process[c].ord)%q
end
text_hash_string << text_hash.to_s << " "
loop=text_length-k
for c in 0..loop do
puts text_hash
text_hash=(radix*(text_hash-text_to_process[c].ord*highorder)+(text_hash[c+k].ord))%q
text_hash_string << text_hash_string << " "
end
end
end
end
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我正在运行它的值 - > calc_hash(text,5,101)其中text是一个String输入.
代码很慢.我哪里错了?
看看Ruby-Prof,一个Ruby的分析器.使用gem install ruby-prof安装它.
一旦你有一些代码滞后的想法,你可以使用Ruby的Benchmark尝试不同的算法来找到最快的.
在StackOverflow上徘徊,你会看到很多地方,我们将使用Benchmark来测试各种方法,看看哪个是最快的.您还将了解设置测试的不同方法.
例如,查看您的代码,我不确定附加是否<<比使用+或使用字符串插值更好.这是测试它的代码和结果:
require 'benchmark'
include Benchmark
n = 1_000_000
bm(13) do |x|
x.report("interpolate") { n.times { foo = "foo"; bar = "bar"; "#{foo}#{bar}" } }
x.report("concatenate") { n.times { foo = "foo"; bar = "bar"; foo + bar } }
x.report("append") { n.times { foo = "foo"; bar = "bar"; foo << bar } }
end
ruby test.rb; ruby test.rb
user system total real
interpolate 1.090000 0.000000 1.090000 ( 1.093071)
concatenate 0.860000 0.010000 0.870000 ( 0.865982)
append 0.750000 0.000000 0.750000 ( 0.753016)
user system total real
interpolate 1.080000 0.000000 1.080000 ( 1.085537)
concatenate 0.870000 0.000000 0.870000 ( 0.864697)
append 0.750000 0.000000 0.750000 ( 0.750866)
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我想知道根据@ Myth17的评论在下面添加字符串时使用固定与变量的效果:
require 'benchmark'
include Benchmark
n = 1_000_000
bm(13) do |x|
x.report("interpolate") { n.times { foo = "foo"; bar = "bar"; "#{foo}#{bar}" } }
x.report("concatenate") { n.times { foo = "foo"; bar = "bar"; foo + bar } }
x.report("append") { n.times { foo = "foo"; bar = "bar"; foo << bar } }
x.report("append2") { n.times { foo = "foo"; bar = "bar"; "foo" << bar } }
x.report("append3") { n.times { foo = "foo"; bar = "bar"; "foo" << "bar" } }
end
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导致:
ruby test.rb;ruby test.rb
user system total real
interpolate 1.330000 0.000000 1.330000 ( 1.326833)
concatenate 1.080000 0.000000 1.080000 ( 1.084989)
append 0.940000 0.010000 0.950000 ( 0.937635)
append2 1.160000 0.000000 1.160000 ( 1.165974)
append3 1.400000 0.000000 1.400000 ( 1.397616)
user system total real
interpolate 1.320000 0.000000 1.320000 ( 1.325286)
concatenate 1.100000 0.000000 1.100000 ( 1.090585)
append 0.930000 0.000000 0.930000 ( 0.936956)
append2 1.160000 0.000000 1.160000 ( 1.157424)
append3 1.390000 0.000000 1.390000 ( 1.392742)
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这些值与我之前的测试不同,因为代码正在我的笔记本电脑上运行.
附加两个变量比涉及固定字符串时更快,因为存在开销; Ruby必须创建一个中间变量然后附加到它.
这里的一个重要教训是,当我们编写代码时,我们可以做出更明智的决定,因为我们知道什么运行得更快.同时,差异不是很大,因为大多数代码都没有运行1,000,000个循环.您的里程可能有所不同