Python3-cookbook- 笔记1 - 数据结构和算法

系统管理员 2021-04-19 14:49 519阅读 0赞

## 元祖或者数组赋值 ##

data = [ 'ACME', 50, 91.1, (2012, 12, 21) ]
    name , shares , price , (year,month,day) = data
    print(name , shares , price ,year,month,day)
    # ACME 50 91.1 2012 12 21

### 查找最大或者最小的N个元素 ###

import heapq
    
    nums = [1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
    print(heapq.nlargest(3,nums)) # [42, 37, 23] 
    print(heapq.nsmallest(3,nums)) # [-4, 1, 2]
    # 两个函数都能接受一个关键字参数，用于更复杂的数据结构中：
    
    portfolio = [
        {
        'name': 'IBM', 'shares': 100, 'price': 91.1},
        {
        'name': 'AAPL', 'shares': 50, 'price': 543.22},
        {
        'name': 'FB', 'shares': 200, 'price': 21.09},
        {
        'name': 'HPQ', 'shares': 35, 'price': 31.75},
        {
        'name': 'YHOO', 'shares': 45, 'price': 16.35},
        {
        'name': 'ACME', 'shares': 75, 'price': 115.65}
    ]
    
    cheap = heapq.nsmallest(3, portfolio, key=lambda s: s['price'])
    expensive = heapq.nlargest(3, portfolio, key=lambda s: s['price'])
    
    print(cheap)
    print(expensive)
    # [{'name': 'YHOO', 'shares': 45, 'price': 16.35}, {'name': 'FB', 'shares': 200, 'price': 21.09}, {'name': 'HPQ', 'shares': 35, 'price': 31.75}]
    # [{'name': 'AAPL', 'shares': 50, 'price': 543.22}, {'name': 'ACME', 'shares': 75, 'price': 115.65}, {'name': 'IBM', 'shares': 100, 'price': 91.1}]

### 下面的类利用 heapq 模块实现了一个简单的优先级队列 ###

class PriorityQueue:
    	def __init__(self):
    		self._queue = []
    		self._index = 0
    
    	def push(self,item,priority):
    		heapq.heappush(self._queue,(-priority,self._index,item))
    		self._index += 1
    
    	def pop(self):
    		return heapq.heappop(self._queue)[-1]
    
    	def __len__(self):
    		return len(self._queue)
    
    	def pop_last(self,last):
    		return heapq.heappop(self._queue)[-last]
    
    
    class Item:
    	def __init__(self,name):
    		self.name = name
    
    	def __repr__(self):
    		return 'Item({!r})'.format(self.name)
    # 使用
    q = PriorityQueue()
    q.push(Item('foo'),1)
    q.push(Item('bar'),5)
    q.push(Item('spam'),4)
    q.push(Item('grok'),1)
    
    print(q.pop())
    print(q.pop())
    print(q.pop())
    print(q.pop())
    
    # Item('bar')
    # Item('spam')
    # Item('foo')
    # Item('grok')

### 字典排序 ###

from collections import OrderedDict
    
    d = OrderedDict()
    d['a'] = 1
    d['b'] = 2
    d['c'] = 3
    
    for k in d:
    	print(k , d[k])
    
    # a 1
    # b 2
    # c 3

### 字典的运算 ###

> 通常需要使用 zip() 函数先将键和值反转过来

min_price = min(zip(prices.values(), prices.keys()))
    # min_price is (10.75, 'FB')
    max_price = max(zip(prices.values(), prices.keys()))
    # max_price is (612.78, 'AAPL')
    
    prices_sorted = sorted(zip(prices.values(), prices.keys()))
    # prices_sorted is [(10.75, 'FB'), (37.2, 'HPQ'),
    # (45.23, 'ACME'), (205.55, 'IBM'),
    # (612.78, 'AAPL')]

### 命名切片 ###

> 如果你的程序包含了大量无法直视的硬编码切片

record = '....................100 .......513.25 ..........'
    cost = int(record[20:23]) * float(record[31:37])
    SHARES = slice(20, 23)
    PRICE = slice(31, 37)
    cost = int(record[SHARES]) * float(record[PRICE])

### 序列中出现次数最多的元素 ###

words = [
        'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
        'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
        'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
        'my', 'eyes', "you're", 'under'
    ]
    from collections import Counter
    word_counts = Counter(words)
    
    
    # 出现频率最高的3个单词
    top_three = word_counts.most_common(3)
    print(top_three)
    
    # [('eyes', 8), ('the', 5), ('look', 4)]

### 通过某个关键字排序一个字典列表 ###

from operator import itemgetter
    
    rows = [
        {
        'fname': 'Brian', 'lname': 'Jones', 'uid': 1003},
        {
        'fname': 'David', 'lname': 'Beazley', 'uid': 1002},
        {
        'fname': 'John', 'lname': 'Cleese', 'uid': 1001},
        {
        'fname': 'Big', 'lname': 'Jones', 'uid': 1004}
    ]
    
    
    rows_by_fname = sorted(rows, key=itemgetter('fname'))
    rows_by_uid = sorted(rows, key=itemgetter('uid'))
    print(rows_by_fname)
    print(rows_by_uid)
    
    # [{'lname': 'Jones', 'uid': 1004, 'fname': 'Big'}, {'lname': 'Jones', 'uid': 1003, 'fname': 'Brian'}, {'lname': 'Beazley', 'uid': 1002, 'fname': 'David'}, {'lname': 'Cleese', 'uid': 1001, 'fname': 'John'}]
    # [{'lname': 'Cleese', 'uid': 1001, 'fname': 'John'}, {'lname': 'Beazley', 'uid': 1002, 'fname': 'David'}, {'lname': 'Jones', 'uid': 1003, 'fname': 'Brian'}, {'lname': 'Jones', 'uid': 1004, 'fname': 'Big'}]
    
    # itemgetter() 函数也支持多个 keys，比如下面的代码
    
    rows_by_lfname = sorted(rows, key=itemgetter('lname','fname'))
    print(rows_by_lfname)
    # [{'lname': 'Beazley', 'fname': 'David', 'uid': 1002}, {'lname': 'Cleese', 'fname': 'John', 'uid': 1001}, {'lname': 'Jones', 'fname': 'Big', 'uid': 1004}, {'lname': 'Jones', 'fname': 'Brian', 'uid': 1003}]

### 排序不支持原生比较的对象 ###

from operator import attrgetter
    
    class User:
        def __init__(self, user_id):
            self.user_id = user_id
    
        def __repr__(self):
            return 'User({})'.format(self.user_id)
    
    users = [User(23), User(3), User(99)]
    sorted(users, key=attrgetter('user_id'))

### 通过某个字段将记录分组 ###

from operator import itemgetter
    from itertools import groupby
    
    rows = [
        {
        'address': '5412 N CLARK', 'date': '07/01/2012'},
        {
        'address': '5148 N CLARK', 'date': '07/04/2012'},
        {
        'address': '5800 E 58TH', 'date': '07/02/2012'},
        {
        'address': '2122 N CLARK', 'date': '07/03/2012'},
        {
        'address': '5645 N RAVENSWOOD', 'date': '07/02/2012'},
        {
        'address': '1060 W ADDISON', 'date': '07/02/2012'},
        {
        'address': '4801 N BROADWAY', 'date': '07/01/2012'},
        {
        'address': '1039 W GRANVILLE', 'date': '07/04/2012'},
    ]
    
    # 现在假设你想在按 date 分组后的数据块上进行迭代。为了这样做，你首先需要按照指定的字段(这里就是 date )排序， 然后调用 itertools.groupby() 函数：
    # Sort by the desired field first
    rows.sort(key=itemgetter('date'))
    
    for date, items in groupby(rows, key=itemgetter('date')):
        print(date)
        for i in items:
            print(' ', i)
    # 07/01/2012
    # {'address': '5412 N CLARK', 'date': '07/01/2012'}
    # {'address': '4801 N BROADWAY', 'date': '07/01/2012'}
    # 07/02/2012
    # {'address': '5800 E 58TH', 'date': '07/02/2012'}
    # {'address': '5645 N RAVENSWOOD', 'date': '07/02/2012'}
    # {'address': '1060 W ADDISON', 'date': '07/02/2012'}
    # 07/03/2012
    # {'address': '2122 N CLARK', 'date': '07/03/2012'}
    # 07/04/2012
    # {'address': '5148 N CLARK', 'date': '07/04/2012'}
    # {'address': '1039 W GRANVILLE', 'date': '07/04/2012'}

### 从字典中提取子集 ###

prices = {
        'ACME': 45.23,
        'AAPL': 612.78,
        'IBM': 205.55,
        'HPQ': 37.20,
        'FB': 10.75
    }
    # Make a dictionary of all prices over 200
    p1 = {key: value for key, value in prices.items() if value > 200}
    # Make a dictionary of tech stocks
    tech_names = {
        'AAPL', 'IBM', 'HPQ', 'MSFT'}
    p2 = {key: value for key, value in prices.items() if key in tech_names}

### 映射名称到序列元素 ###

from collections import namedtuple
    
    Subscriber = namedtuple('Subscriber', ['addr', 'joined'])
    sub = Subscriber('jonesy@example.com', '2012-10-19')
    print(sub.addr)
    print(sub.joined)