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‎.gitignore

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.vscode/

‎01-DataSturcture/N_element.py

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'''
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File: N_element.py
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Project: DataSturcture
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===========
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File Created: Monday, 20th July 2020 11:08:11 am
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Author: <<LanLing>> (<<[email protected]>>)
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===========
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Last Modified: Monday, 20th July 2020 11:44:06 am
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Modified By: <<LanLing>> (<<[email protected]>>>)
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===========
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Description: 保留序列中 N 个元素的问题
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Copyright <<2020>> - 2020 Your Company, <<XDU>>
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'''
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import heapq
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from collections import deque
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# 保留最后 N 个元素
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# 先入先出
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q = deque(maxlen=3)
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q.append(1)
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q.append(2)
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q.append(3)
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print(q)
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q.append(6)
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print(q)
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# 队首插入
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q.appendleft(4)
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print(q)
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# 队尾弹出
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a = q.pop()
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print(a, q)
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# 队首弹出
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b = q.popleft()
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print(b, q)
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# 查找最大或最小的 N 个元素
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nums = [1, 2, 3, 4, 5, 6, 6, 6, 1, 1]
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print(heapq.nlargest(3, nums))
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print(heapq.nsmallest(3, nums))
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# 更复杂的数据结构
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portfolio = [
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{'name': 'IBM', 'shares': 100, 'price': 91.1},
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{'name': 'AAPL', 'shares': 50, 'price': 543.22},
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{'name': 'FB', 'shares': 200, 'price': 21.09},
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{'name': 'HPQ', 'shares': 35, 'price': 31.75},
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{'name': 'YHOO', 'shares': 45, 'price': 16.35},
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{'name': 'ACME', 'shares': 75, 'price': 115.65}
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]
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cheap = heapq.nsmallest(3, portfolio, key=lambda s: s['price'])
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expensive = heapq.nlargest(3, portfolio, key=lambda s: s['price'])
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# 会打印全部记录
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print(cheap)
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print(expensive)
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# 当 N 和集合大小差不多时，建议排序后切片

‎01-DataSturcture/anyobj_sort.py

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'''
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File: anyobj_sort.py
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Project: 01-DataSturcture
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===========
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File Created: Tuesday, 21st July 2020 3:16:28 pm
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Author: <<LanLing>> (<<[email protected]>>)
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===========
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Last Modified: Tuesday, 21st July 2020 3:16:32 pm
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Modified By: <<LanLing>> (<<[email protected]>>>)
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===========
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Description: 使对象支持排序
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Copyright <<2020>> - 2020 Your Company, <<XDU>>
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'''
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from operator import attrgetter
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# 要排序的类
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class User:
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def __init__(self, user_id, name):
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self.user_id = user_id
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self.name = name
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def __repr__(self):
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return 'User({})'.format(self.user_id)
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users = [User(23, 'aa'), User(3, 'aa'), User(99, 'bb')]
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# 单属性的排序
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print(sorted(users, key=attrgetter('user_id')))
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# 多属性的排序
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print(sorted(users, key=attrgetter('name', 'user_id')))
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# 最大值
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a = min(users, key=attrgetter('user_id'))
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print(a.user_id, a.name)
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# 最小值
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print(max(users, key=attrgetter('user_id')))

‎01-DataSturcture/dict_add_element.py

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'''
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File: dict_add_element.py
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Project: DataSturcture
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===========
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File Created: Monday, 20th July 2020 12:17:31 pm
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Author: <<LanLing>> (<<[email protected]>>)
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===========
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Last Modified: Monday, 20th July 2020 12:17:37 pm
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Modified By: <<LanLing>> (<<[email protected]>>>)
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===========
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Description: 字典元素的高级添加
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Copyright <<2020>> - 2020 Your Company, <<XDU>>
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'''
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from collections import defaultdict
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# 一个键映射多个值，那么你就需要将这多个值放到另外的容器中
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d = {
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'a' : [1, 2, 3],
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'b' : [4, 5]
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}
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print(d['a'])
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# 保持元素的插入顺序, 使用列表
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d = defaultdict(list)
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d['a'].append(1)
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d['a'].append(1)
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d['b'].append(2)
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print(d)
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# 去掉重复元素, 使用集合, 不关注元素顺序
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d = defaultdict(set)
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d['a'].add(1)
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d['a'].add(1)
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d['b'].add(4)
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print(d)
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# 将数据添加到多值映射的字典
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pairs = {
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('a', 12),
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('a', 13),
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('b', 14)
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}
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# 不用先创建 d['a'] = []
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d = defaultdict(list)
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for key, value in pairs:
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d[key].append(value)
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print(d)

‎01-DataSturcture/dict_calc.py

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'''
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File: dict_calc.py
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Project: 01-DataSturcture
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===========
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File Created: Monday, 20th July 2020 5:32:11 pm
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Author: <<LanLing>> (<<[email protected]>>)
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===========
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Last Modified: Monday, 20th July 2020 5:32:17 pm
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Modified By: <<LanLing>> (<<[email protected]>>>)
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===========
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Description: 字典计算
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Copyright <<2020>> - 2020 Your Company, <<XDU>>
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'''
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# 字典计算，找最小值等
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prices = {
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'ACME': 45.23,
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'AAPL': 612.78,
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'ABM': 205.55,
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'HPQ': 37.20,
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'FB': 10.75
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}
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# 只返回键
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print(min(prices))
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print(max(prices))
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# 返回值和键
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# 按照值排序
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print(min(zip(prices.values(), prices.keys())))
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# 按照键排序
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print(max(zip(prices.keys(), prices.values())))
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# 字典比较，交、并、补等
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a = {
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'x' : 1,
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'y' : 2,
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'z' : 3
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}
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b = {
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'w' : 10,
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'x' : 11,
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'y' : 2
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}
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# 查找共同的键
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print(a.keys() & b.keys())
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# 键在 a，但不在 b
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print(a.keys() - b.keys())
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# 键值都相同
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print(a.items() & b.items())
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# 创建一个新字典
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c = {key:a[key] for key in (a.keys() & b.keys()) - {'z', 'w'}}
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print(c)
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# 取字典元素的最小值
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portfolio = [
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{'name':'GOOG', 'shares': 50},
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{'name':'YHOO', 'shares': 75},
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{'name':'AOL', 'shares': 20},
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{'name':'SCOX', 'shares': 65}
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]
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min_shares = min(s['shares'] for s in portfolio)
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print(min_shares)

‎01-DataSturcture/dict_merge.py

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'''
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File: dict_merge.py
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Project: 01-DataSturcture
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===========
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File Created: Tuesday, 21st July 2020 4:49:05 pm
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Author: <<LanLing>> (<<[email protected]>>)
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===========
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Last Modified: Tuesday, 21st July 2020 4:49:09 pm
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Modified By: <<LanLing>> (<<[email protected]>>>)
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===========
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Description: 字典合并
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Copyright <<2020>> - 2020 Your Company, <<XDU>>
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'''
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from collections import ChainMap
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a = {'x': 1, 'z': 3 }
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b = {'y': 2, 'z': 4 }
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# 合并字典
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# 在内部创建了一个容纳这些字典的列表
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# 相同的键，只保留在第一个字典的
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c = ChainMap(a, b)
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print(c['x'])
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print(c['y'])
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print(c['z'])
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print(list(c.keys()))
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print(list(c.values()))
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# 所有的操作都只会影响第一个字典
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c['z'] = 10
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c['w'] = 20
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print(a, b)
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# 个人感觉 defaultdict 会好一些

‎01-DataSturcture/dict_sort.py

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'''
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File: dict_sort.py
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Project: DataSturcture
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===========
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File Created: Monday, 20th July 2020 5:24:37 pm
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Author: <<LanLing>> (<<[email protected]>>)
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===========
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Last Modified: Monday, 20th July 2020 5:24:58 pm
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Modified By: <<LanLing>> (<<[email protected]>>>)
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===========
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Description: 字典排序
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Copyright <<2020>> - 2020 Your Company, <<XDU>>
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'''
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from collections import OrderedDict
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from operator import itemgetter
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# 保留元素插入时的顺序
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# 大小是一个普通字典的两倍，因为它内部维护着另外一个插入顺序排序的链表。
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d = OrderedDict()
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d['foo'] = 1
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d['spam'] = 3
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d['bar'] = 2
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d['grok'] = 4
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# Outputs "foo 1", "bar 2", "spam 3", "grok 4"
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for key in d:
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print(key, d[key])
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# 按照某一规定排序
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data = [
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{'fname': 'Brian', 'lname': 'Jones', 'uid': 1003},
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{'fname': 'David', 'lname': 'Beazley', 'uid': 1002},
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{'fname': 'John', 'lname': 'Cleese', 'uid': 1001},
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{'fname': 'Big', 'lname': 'Jones', 'uid': 1004}
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]
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# 传入排序的参数
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rows_by_fname = sorted(data, key=itemgetter('fname'))
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rows_by_uid = sorted(data, key=itemgetter('uid'))
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print(rows_by_fname)
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print(rows_by_uid)
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# 传入多个参数，先比较第一个，相同则比较第二个
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rows_by_lfname = sorted(data, key=itemgetter('lname','fname'))
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print(rows_by_lfname)
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# 输出最大值和最小值
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b = min(data, key=itemgetter('uid'))
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a = max(data, key=itemgetter('uid'))
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print(a, '\n', b)

‎01-DataSturcture/element_count.py

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'''
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File: element_count.py
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Project: 01-DataSturcture
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===========
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File Created: Monday, 20th July 2020 10:41:10 pm
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Author: <<LanLing>> (<<[email protected]>>)
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===========
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Last Modified: Monday, 20th July 2020 10:41:15 pm
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Modified By: <<LanLing>> (<<[email protected]>>>)
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===========
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Description: 序列中出现次数最多的元素
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Copyright <<2020>> - 2020 Your Company, <<XDU>>
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'''
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from collections import Counter
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# 出现频率最高的3个单词
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words = [
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'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
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'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
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'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
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'my', 'eyes', "you're", 'under'
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]
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word_counts = Counter(words)
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# 出现频率最高的3个单词
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top_three = word_counts.most_common(3)
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print(top_three)
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# 字典，将元素映射到出现次数中
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print(word_counts['look'])
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# 更新词汇表
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morewords = ['why', 'are', 'you', 'not', 'looking', 'in', 'my', 'eyes']
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word_counts.update(morewords)
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print(word_counts['eyes'])
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# Counter 对象在几乎所有需要制表或者计数数据的场合是非常有用的工具。
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# 在解决这类问题的时候你应该优先选择它，而不是手动的利用字典去实现。
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a = Counter(words)
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b = Counter(morewords)
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# 直接操作
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c = a + b
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d = a - b
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print(c, '\n', d)

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