python字符串过滤性能比较5种方法

Python  /  管理员 发布于 7年前   158

python字符串过滤性能比较5种方法比较

总共比较5种方法。直接看代码：

import randomimport timeimport osimport stringbase = string.digits+string.punctuationtotal = 100000def loop(ss):  """循环"""  rt = ''  for c in ss:    if c in '0123456789':      rt = rt + c  return rtdef regular(ss):  """正则表达式"""  import re  rt = re.sub(r'\D', '', ss)  return rtdef filter_mt(ss):  """函数式"""  return filter(lambda c:c.isdigit(), ss)def list_com(ss):  """列表生成式"""  isdigit = {'0': 1, '1': 1, '2': 1, '3': 1, '4': 1,'5':1, '6':1, '7':1, '8':1, '9':1}.has_key  return ''.join([x for x in ss if isdigit(x)])def str_tran(ss):  """string.translate()"""  table = string.maketrans('', '')  ss = ss.translate(table,string.punctuation)  return ssif __name__ == '__main__':  lst = []  for i in xrange(total):    num = random.randrange(10, 50)    ss = ''    for j in xrange(num):      ss = ss + random.choice(base)    lst.append(ss)  s1 = time.time()  map(loop,lst)  print "loop: ",time.time() - s1  print '*'*20  s1 = time.time()  map(regular, lst)  print "regular: ", time.time() - s1  print '*' * 20  s1 = time.time()  map(str_tran, lst)  print "str_tran: ", time.time() - s1  print '*' * 20  s1 = time.time()  map(filter_mt, lst)  print "filter_mt: ", time.time() - s1  print '*' * 20  s1 = time.time()  map(list_com, lst)  print "list_com: ", time.time() - s1

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