Python collections类

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c = Counter(‘abcdeabcdabcaba‘) 

import collections
c = collections.Counter(‘abcdeabcdabcaba‘)
print (c)

结果：Counter({‘a‘: 5, ‘b‘: 4, ‘c‘: 3, ‘d‘: 2, ‘e‘: 1})
结论：统计字符串出现的次数

import collections
c = collections.Counter(‘abcdeabcdabcaba‘)
print (c)

ret = c.most_common(2)
print (ret)

结果：[(‘a‘, 5), (‘b‘, 4)]
结论：统计出现次数最多的两个字符串

import collections
c = collections.Counter(‘abcdeabcdabcaba‘)
print (c)
print (sorted(c))
结果：[‘a‘, ‘b‘, ‘c‘, ‘d‘, ‘e‘]
结论：列出所有唯一的元素

import collections
c = collections.Counter(‘abcdeabcdabcaba‘)
print (c)
for k,v in c.items():
    print (k,v)
结果：

　Counter({‘a‘: 5, ‘b‘: 4, ‘c‘: 3, ‘d‘: 2, ‘e‘: 1})
　a 5
　c 3
　e 1
　d 2
　b 4

　结论：统计各个字符串出现的次数

OrderedDict

import collections
dic = collections.OrderedDict()
dic[‘k3‘] = ‘v3‘
dic[‘k1‘] = ‘v1‘
dic[‘k2‘] = ‘v2‘
print (dic)
结果：OrderedDict([(‘k3‘, ‘v3‘), (‘k1‘, ‘v1‘), (‘k2‘, ‘v2‘)])
结论：用列表来维系有序字典

defaultdict

import collections
values = [11,22,33,44,55,66,77,88,99]
dic = collections.defaultdict(list)   #设置缺省为list类型
for value in values:
    if value <66:
        dic[‘k1‘].append(value)
    else:
        dic[‘k2‘].append(value)
print (dic[‘k1‘])
print (dic[‘k2‘])

不使用缺省字典

values = [11,22,33,44,55,66,77,88,99]
dic1 = {‘k1‘:[]}
dic2 = {‘k2‘:[]}

for value in values:
    if value < 66:
            dic1[‘k1‘].append(value)
    else:
            dic2[‘k2‘].append(value)
print (dic1)
print (dic2)
结果：

　　{‘k1‘: [11, 22, 33, 44, 55]}
　　{‘k2‘: [66, 77, 88, 99]}

　结论：必须制定dic1,dic2的值为list空表

 　namedtuple(可命名的元祖)

import collections

Test = collections.namedtuple(‘Test‘,(‘x‘,‘y‘,‘z‘))

obj = Test(11,22,33)
print (obj.x)

结果：11
结论：针对元祖，要访问数据只能通过下标来访问，但是对于可命名的元祖，可以通过构建新的一个类（譬如Test），然后调用方法来实现

 双向队列

import collections

dq = collections.deque()

单向队列

import queue
q = queue.Queue()
q.put("123")
r = q.get()
print (r)

Python collections类

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原文地址：http://www.cnblogs.com/python-study/p/5447939.html