# -*- coding: utf-8 -*-
#使用Counter（）计数
from collections import Counter
breakfast = ['spam','spam','egg','spam']
breakfast_counter = Counter(breakfast)
print(breakfast_counter)
#函数most_commom()以降序返回所有元素，或者给定一个数字，会返回该数字前的元素
print(breakfast_counter.most_common())
print(breakfast_counter.most_common(1))
#新建一个列表lunch和一个计数器lunch_counter
lunch = ['egg','egg','bacon']
lunch_counter = Counter(lunch)
print(lunch_counter)
#组合计数器
print(breakfast_counter + lunch_counter)
print(breakfast_counter - lunch_counter)
print(breakfast_counter & lunch_counter)
print(breakfast_counter | lunch_counter)
#使用有序字典orderdDict()按键排序
#一个字典的键的顺序是不可知的
quotes = {
'moe':'a wise guy huh?',
'larry':'ow!',
'curly':'nyuk nyuk!',
}
for stooge in quotes:
print(stooge)
#使用OrderdDict()记忆字典添加的顺序
from collections import OrderedDict
quotes = OrderedDict([('moe','a wise guy,huh?'),
('larry','ow'),
('curly','nyuk nyuk'),
])
for stooge in quotes:
print(stooge)
#双端队列：栈+队列
def palindrome(word):
from collections import deque
dq = deque(word)
while len(dq) > 1:
if dq.popleft() != dq.pop():
return False
return True
print(palindrome('a'))
print(palindrome('raeccar'))
#快速判断回文的程序
def another_palindrome(word):
return word == word[::-1]
print(another_palindrome('radar'))
#使用迭代器
import itertools
for item in itertools.chain([1,2],[3,4]): #chain()的参数是单个迭代对象
print(item)
import itertools
for item in itertools.cycle([1,2]): #无线循环迭代
print(item)
from itertools import accumulate
for item in itertools.accumulate([1,2,3,4]): #累加,python2中没有
print(item)
import itertools
def multiply(a,b):
return a*b
for item in itertools.accumulate([1,2,3,4],accumulate):
print(item)
#使用pprint()友好输出
from pprint import pprint
quotes = OrderedDict([('moe', 'a wise guy,huh?'),
('larry', 'ow'),
('curly', 'nyuk nyuk'),
])
pprint(quotes)
print(quotes)
