Question

我的代码在这里

str = "Early in his first term in office, Obama signed into law economic stimulus legislation in response"
arr= str.split(" ")
set_element= arr.each_cons(2).to_a
sub_str = set_element.map {|i| i.join(' ')}

如果我有一个很大的字符串，那么这个过程需要6.50秒因为我想要这种类型的结果

sub_str= ["Early in", "in his", "his first", "first term", "term in", "in office,", "office, Obama", "Obama signed", "signed into", "into law", "law economic", "economic stimulus", "stimulus legislation", "legislation in", "in response"]

是否有可能采用其他有效方式

Answer 1

使用scan而不是split，您可以直接获取单词对。

s.scan(/\S+(?:\s+\S+)?/)

编辑：只是为了向自己保证这是相对有效的，我做了a little micro-benchmark。以下是迄今为止所见答案的结果：

ruby 1.9.3p125 (2012-02-16 revision 34643) [x86_64-linux]
10 times on string of size 2284879
                 user     system      total        real
original     4.180000   0.070000   4.250000 (  4.272856)
sergio       2.090000   0.000000   2.090000 (  2.102469)
dbenhur      1.050000   0.000000   1.050000 (  1.042167)

Answer 2

set_element = arr.each_cons(2).to_a

上面的行会创建大量您不需要的临时对象。试试这个，应该更快：

str = "Early in his first term in office, Obama signed into law economic stimulus legislation in response"
arr = str.split(" ")
sub_str = arr.each_with_object([]).with_index do |(el, memo), idx|
  if idx % 2 == 0
    memo << el
  else
    memo.last << ' ' << el
  end

end

sub_str # => ["Early in", "his first", "term in", "office, Obama", "signed into", "law economic", "stimulus legislation", "in response"]

Answer 3

你可以试试这个。少一步：）

arr= str.scan(/\S+/)
s = []
arr.each_with_index { |x, i| s << (x + " " + arr[i + 1]) if arr[i+1] }

以有效的方式将字符串拆分为多个部分

3 个答案: