粗略估计字符串比较Python所需的时间

2 回答

• 3

  我会将你的5500个字符串数组转换为一个集合，并使用一个集合交集 .

  >>> paragraph = "five hundred to one hundred words separated by spaces"
  >>> array_of_strings = set(['hundred', 'spaces', ])  # make a set..
  
  >>> print set(paragraph.split()).intersection(array_of_strings)
  set(['hundred', 'spaces'])
  

  这是你如何计时的 .

  阅读timeit模块 . 这是另一个教程：http://diveintopython.net/performance_tuning/timeit.html

  import timeit
  s = """paragraph = "five hundred to one hundred words separated by spaces"
  array_of_strings = set(['hundred', 'spaces', ])  # make a set..
  
  set(paragraph.split()).intersection(array_of_strings)
  """
  t = timeit.Timer(stmt=s)
  print "%.2f usec/pass" % (1000000 * t.timeit(number=100000)/100000)
  

  回复于 2024-05-02T14:16:32+08:00
• 1

  如果使用列表，请先对其进行排序并使用二进制搜索 .

  但是使用字典可能会更好;）

  import time
  
  def timeo(fun, n=1000): 
      def void(  ): pass 
      start = time.clock(  ) 
      for i in range(n): void(  ) 
      stend = time.clock(  ) 
      overhead = stend - start 
      start = time.clock(  ) 
      for i in range(n): fun(  ) 
      stend = time.clock(  ) 
      fulltime = stend-start 
      return fun.__name__, fulltime-overhead 
  
  for f in solution1, solution2, solution3:
      print "%s: %.2f" % timeo(f)
  

  回复于 2024-05-02T14:16:32+08:00