python字符串过滤性能比较5种方法
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 random
import time
import os
import string
base = string.digits+string.punctuation
total = 100000
def loop(ss):
"""循环"""
rt = ''
for c in ss:
if c in '0123456789':
rt = rt + c
return rt
def regular(ss):
"""正则表达式"""
import re
rt = re.sub(r'\D', '', ss)
return rt
def 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 ss
if __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
感谢阅读，希望能帮助到大家，谢谢大家对本站的支持！
