python3数据结构与算法

python内置的数据结构包括：列表（list）、集合（set）、字典（dictionary），一般情况下我们可以直接使用这些数据结构，但通常我们还需要考虑比如搜索、排序、排列以及赛选等一些常见的问题。

如何巧妙的使用数据结构和同数据有关的算法，在collections模块中包含了针对各种数据结构的解决方法。

1、序列分解为变量

In [5]: a = (4,5,6)
In [6]: x,y,z = a
In [7]: x
Out[7]: 4
In [8]: z
Out[8]: 6
In [9]: y
Out[9]: 5
In [10]: b = ['python',222,(2018,9,30)]  #嵌套分解变量
In [11]: p,n,(year,mon,day) = b
In [12]: p
Out[12]: 'python'
In [13]: n
Out[13]: 222
In [14]: year
Out[14]: 2018
In [15]: day
Out[15]: 30
#可以分解的对象只要是可迭代对象如字符串、文件、迭代器和生成器
In [16]: s = 'py'
In [17]: x,y = s
In [18]: x
Out[18]: 'p'
#忽略某个值使用下划线代替
In [19]: data = 'python'
In [20]: x,_,_,y,_,_ = data
In [21]: x
Out[21]: 'p'
In [22]: y
Out[22]: 'h'

2、任意长度对象分解元素

要从某个可迭代对象中分解出N个元素，可以使用python的“*表达式”来代表多个

列1：在作业成绩中去掉最高和最低后取平均分

In [47]: grades = (68,98,85,78,84,79,88)
In [48]: def drop_first_last(grades):
    ...:     first,*middle,last = grades
    ...:     return sum(middle) / len(middle)
    ...:
    ...: 

In [49]: drop_first_last(sorted(list(grades),reverse=True))
Out[49]: 82.8

列2：在嵌套元组中*式语法的分解应用

records = [
    ('foo',1,2,3),
    ('bar',11,22,33),
    ('foo',4,5,6),
    ('bar',44,55,66),
]

def do_foo(x,y,z):
    print('foo',x,y,z)

def do_bar(a,b,c):
    print('bar',a,b,c)

for tag,*args in records:  #分解元组打印
    if tag == 'foo':
        do_foo(*args)
    elif tag == 'bar':
        do_bar(*args)

#outing
foo 1 2 3
bar 11 22 33
foo 4 5 6
bar 44 55 66

列3：通过split拆分分解元素

In [52]: passwd = 'root:x:0:0:root:/root:/bin/bash'

In [53]: username,*_,homedir,sh = passwd.split(":")

In [54]: username
Out[54]: 'root'

In [55]: homedir
Out[55]: '/root'

In [56]: sh
Out[56]: '/bin/bash'

3、保存最后N个元素

列1：使用collections.deque保存有限的历史纪录，deque用来创建一个固定长度的队列

In [61]: from collections import deque
#创建队列长度对象
In [62]: q = deque(maxlen=3)
#加入数据到队列
In [63]: q.append(1)

In [64]: q.append(2)

In [65]: q.append(3)

In [66]: q
Out[66]: deque([1, 2, 3])

In [67]: q.append(4)

In [68]: q
Out[68]: deque([2, 3, 4])
#从左边加入数据到队列
In [69]: q.appendleft(5)

In [70]: q
Out[70]: deque([5, 2, 3])
#从末尾取出一个数据
In [71]: q.pop()
Out[71]: 3

In [72]: q
Out[72]: deque([5, 2])

In [73]: q.popleft()
Out[73]: 5

In [74]: q
Out[74]: deque([2])

4、找到最大或最小的N个元素

在heapq模块中有两个函数nlargest()从最大的值开始取，nsmallest()从最小的值开始取

In [75]: import heapq

In [76]: numbers = [1,3,4,9,11,34,55,232,445,9812,321,45,67,434,555]
#取三个最大的值
In [77]: heapq.nlargest(3,numbers)
Out[77]: [9812, 555, 445]
#取三个最小的值
In [78]: heapq.nsmallest(3,numbers)
Out[78]: [1, 3, 4]

5、python堆排序peapq模块

hepaq模块实现了python中的推排序，并提供了很多方法，让用python实现排序算法有了简单快捷的方式

In [1]: import heapq
In [2]: date = [19,1,9,3,11,21]
In [3]: heap = []
#heappush方法会插入一个元素到堆中，并按从小到大排序
In [4]: for i in date:
   ...:     heapq.heappush(heap,i)
   ...:
In [5]: heap
Out[5]: [1, 3, 9, 19, 11, 21]
In [6]: date
Out[6]: [19, 1, 9, 3, 11, 21]
#heapify方法会重新排序整个列表
In [7]: heapq.heapify(date)
In [8]: date
Out[8]: [1, 3, 9, 19, 11, 21]
#heappop()方法会取出第一个元素，并将剩下的元素堆排序
In [10]: date
Out[10]: [19, 1, 9, 3, 11, 21]
In [11]: heapq.heappop(date)
Out[11]: 19
In [12]: date
Out[12]: [1, 3, 9, 21, 11]
#heapreplace()的作用是在堆中取第一个元素并插入一个元素
In [27]: date = [11,8,3,78,35]
In [28]: heapq.heapreplace(date,1)
Out[28]: 11
In [29]: date
Out[29]: [1, 8, 3, 78, 35]
#在集合中找出最大或者最小的N个元素，可以使用nlargest()和nsmallest()
In [30]: date = [3,88,32,97,56]
In [31]: heapq.nlargest(2,date)
Out[31]: [97, 88]
In [33]: heapq.nsmallest(2,date)
Out[33]: [3, 32]
#nlargest()和nsmallest()还可以接受一个key参数来实现复杂的数据结构上的取值，如根据字典的值取值
In [34]: port = [
    ...: {'name':'dhcp','port':67},
    ...: {'name':'mysql','port':3306},
    ...: {'name':'memcached','port':11211},
    ...: {'name':'nginx','port':80},
    ...: {'name':'ssh','port':22},]

In [35]: heapq.nlargest(3,port,key=lambda x:x['port'])
Out[35]:
[{'name': 'memcached', 'port': 11211},
 {'name': 'mysql', 'port': 3306},
 {'name': 'nginx', 'port': 80}]

In [36]: heapq.nsmallest(3,port,key=lambda x:x['port'])
Out[36]:
[{'name': 'ssh', 'port': 22},
 {'name': 'dhcp', 'port': 67},
 {'name': 'nginx', 'port': 80}]

实现优先级队列实例：

import heapq
class priorityqueue(object):
    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 listt(self):
        return self._queue

q = priorityqueue()
q.push('python',44)
q.push('java',2)
q.push('c++',4)
q.push('c#',8)
q.push('goo',88)
q.push('perl',1)
date1 = q.listt()
print(date1)
print(q.pop())
print(q.listt())

#output
[(-88, 4, 'goo'), (-44, 0, 'python'), (-4, 2, 'c++'), (-2, 1, 'java'), (-8, 3, 'c#'), (-1, 5, 'perl')]
goo
[(-44, 0, 'python'), (-8, 3, 'c#'), (-4, 2, 'c++'), (-2, 1, 'java'), (-1, 5, 'perl')]

6、在字典中将键映射到多个值上

可以使用列表、元组、集合来创建多个值的字典键值

dictlist = {
    'a':[1,2],
    'b':[3,4],
    'c':[5,6],
}
dictset = {
    'as':{7,8},
    'bs':{9,0},
}

在collection模块中的defaultdict类，它可以自动初始化第一个值，只需要添加元素即可

In [1]: from collections import defaultdict

In [2]: d = defaultdict(list)

In [3]: d
Out[3]: defaultdict(list, {})

In [4]: d['a'].append(1)

In [5]: d['a'].append(2)

In [6]: d
Out[6]: defaultdict(list, {'a': [1, 2]})

7、让字典保持有序

要控制字典中元素的顺序，可以使用collections模块中的OrderedDict类，当对字典做迭代时，它会严格按照元素初始添加的顺序进行迭代

from collections import OrderedDict

d = OrderedDict()
d['one'] = 1
d['two'] = 2
d['three'] = 3
d['four'] = 4

for key,value in d.items():
    print(key,value)

#output
one 1
two 2
three 3
four 4

当我们先精确控制字典中各字段的顺序然后序列化时，只需要在序列化前使用OrderdDist来构建字典数据，OrderedDict内部维护了一个双向链表，它会根据元素加入的顺序来排列键的位置，第一个新加入的元素被放置在链表的末尾，以后对已存在的键做修改也不会改变键的顺序，由于它额外创建了链表所占用的空间会是普通字典的2倍

from collections import OrderedDict
import json
d = OrderedDict()
d['one'] = 1
d['two'] = 2
d['three'] = 3
d['four'] = 4
jsd = json.dumps(d)
d1 = json.loads(jsd)
print(jsd)
print(d1)

#
{"one": 1, "two": 2, "three": 3, "four": 4}
{'one': 1, 'two': 2, 'three': 3, 'four': 4}

8、字典中的计算（求最大值、最小值和排序）

prices = {
    'ACME':45.23,
    'AAPL':612.78,
    'IBM':205.55,
    'HPQ':10.75,
    'FB':10.75
}

print(min(zip(prices.values(),prices.keys())))
print(max(zip(prices.values(),prices.keys())))
print(sorted(zip(prices.values(),prices.keys())))

#使用zip()将字典中的值映射为元组的迭代器，但zip()只能被使用一次
#如果对比的值相同，则选择键的排序大小
#
(10.75, 'FB')
(612.78, 'AAPL')
[(10.75, 'FB'), (10.75, 'HPQ'), (45.23, 'ACME'), (205.55, 'IBM'), (612.78, 'AAPL')]

9、在两个字典中寻找相同点

a = {'x':1,'y':2,'z':3}
b = {'w':10,'x':11,'y':2}
print(a.keys() & b.keys())  #a和b中同时都有的key
print(a.keys() - b.keys())  #a中的键不在b中出现的key
print(a.items() & b.items())  #a和b中键值都相同的元素

#
{'x', 'y'}
{'z'}
{('y', 2)}

In [1]: a = {'a':11,'b':22,'c':44,'d':99,'f':101}
#推倒式排除键新建字典
In [2]: c = {key:a[key] for key in a.keys() - {'b','d'}}

In [3]: c
Out[3]: {'f': 101, 'a': 11, 'c': 44}

10、从序列中移除重复项并保持元素顺序

dic = [{'x':1,'y':3},{'x':3,'y':8},{'x':1,'y':11},{'x':1,'y':3}]
def dedupe(items,key=None):
    seen = set()
    for item in items:
        val = item if key is None else key(item)
        if val not in seen:
            yield item
            seen.add(val)

#key传递函数将序列中的元素转换为可哈希值，来去除重复项
date = list(dedupe(dic,key=lambda d:(d['x'],d['y'])))
print(date)
date1 = list(dedupe(dic,key=lambda x:(x['x'])))
print(date1)

#
[{'x': 1, 'y': 3}, {'x': 3, 'y': 8}, {'x': 1, 'y': 11}]
[{'x': 1, 'y': 3}, {'x': 3, 'y': 8}]

11、对切片命名

使用内置函数slice来创建切片对象

In [4]: li = [1,2,3,4,5,6,7,8,9,0]

In [5]: cost = li[slice(2,8)]

In [6]: cost
Out[6]: [3, 4, 5, 6, 7, 8]

In [8]: li[slice(2,9,2)]
Out[8]: [3, 5, 7, 9]

12、找出序列中出现次数最多的元素

collections模块中的Counter类可以直接统计每个元素出现的次数，它会以字典的形式映射每个元素出现的次数，其中的most_common()方法可以直接显示结果，可传参数为显示的元素个数

In [15]: date = [1,23,4,3,2,5,23,123,553,23,1,3,4,5,2,3,423,12,3,4,23,412,43]

In [16]: from collections import Counter

In [17]: Counter(date)
Out[17]:
Counter({1: 2,
         23: 4,
         4: 3,
         3: 4,
         2: 2,
         5: 2,
         123: 1,
         553: 1,
         423: 1,
         12: 1,
         412: 1,
         43: 1})
In [18]: Counter(date).most_common()
Out[18]:
[(23, 4),
 (3, 4),
 (4, 3),
 (1, 2),
 (2, 2),
 (5, 2),
 (123, 1),
 (553, 1),
 (423, 1),
 (12, 1),
 (412, 1),
 (43, 1)]

In [19]: Counter(date).most_common(2)
Out[19]: [(23, 4), (3, 4)]

In [20]: Counter(date).most_common(4)
Out[20]: [(23, 4), (3, 4), (4, 3), (1, 2)]

In [27]: date1 = ['a','b','c','a','a','b']

In [28]: from collections import Counter
#生成一个Counter对象，为字典映射的统计值
In [29]: counts = Counter(date1)

In [30]: counts['a']
Out[30]: 3
#创建第二个序列
In [31]: date2 = ['b','b','a']
#先统计元素出现的次数
In [32]: counts.most_common()
Out[32]: [('a', 3), ('b', 2), ('c', 1)]
#使用update()方法来手动更新counts对象
In [33]: counts.update(date2)
#查看结果
In [34]: counts.most_common()
Out[34]: [('a', 4), ('b', 4), ('c', 1)]
#创建第二个counter对象
In [35]: counts1 = Counter(date2)
#counter对象可以用加减来运算
In [36]: counts - counts1
Out[36]: Counter({'a': 3, 'b': 2, 'c': 1})

13、字典列表排序

operator模块中的itemgetter函数可以对嵌套数据结构的排序会非常简单且运行很快

from operator import itemgetter
date1 = [
    {'fname':'Brian','lname':'Jones','uid':1003},
    {'fname':'David','lname':'Beazley','uid':1002},
    {'fname':'John','lname':'Cleese','uid':1001},
    {'fname':'Big','lname':'Jones','uid':1004},
]

print(sorted(date1,key=itemgetter('uid')))
print(sorted(date1,key=itemgetter('uid'),reverse=True)) #反向排序
print(sorted(date1,key=itemgetter('uid','fname')))  #通过多个公共键排序

print(sorted(date1,key=lambda x:x['uid']))  #也可以使用匿名函数来代替，但速度没有itemgetter()函数快

print(min(date1,key=itemgetter('uid'))) #itemgetter()也可以用在去最大或最小值上
print(max(date1,key=itemgetter('uid')))

#
[{'fname': 'John', 'lname': 'Cleese', 'uid': 1001}, {'fname': 'David', 'lname': 'Beazley', 'uid': 1002}, {'fname': 'Brian', 'lname': 'Jones', 'uid': 1003}, {'fname': 'Big', 'lname': 'Jones', 'uid': 1004}]
[{'fname': 'Big', 'lname': 'Jones', 'uid': 1004}, {'fname': 'Brian', 'lname': 'Jones', 'uid': 1003}, {'fname': 'David', 'lname': 'Beazley', 'uid': 1002}, {'fname': 'John', 'lname': 'Cleese', 'uid': 1001}]
[{'fname': 'John', 'lname': 'Cleese', 'uid': 1001}, {'fname': 'David', 'lname': 'Beazley', 'uid': 1002}, {'fname': 'Brian', 'lname': 'Jones', 'uid': 1003}, {'fname': 'Big', 'lname': 'Jones', 'uid': 1004}]
[{'fname': 'John', 'lname': 'Cleese', 'uid': 1001}, {'fname': 'David', 'lname': 'Beazley', 'uid': 1002}, {'fname': 'Brian', 'lname': 'Jones', 'uid': 1003}, {'fname': 'Big', 'lname': 'Jones', 'uid': 1004}]
{'fname': 'John', 'lname': 'Cleese', 'uid': 1001}
{'fname': 'Big', 'lname': 'Jones', 'uid': 1004}

对原生不支持比较操作的对象排序

from operator import attrgetter
class user(object):
    def __init__(self,user_id):
        self.user_id = user_id
    def __repr__(self):
        return 'user({})'.format(self.user_id)
users = [user(11),user(22),user(3)]
print(users)
print(sorted(users,key=lambda x:x.user_id))
print(sorted(users,key=attrgetter('user_id')))  #使用attrgetter()函数来对实例化对象的参数值排序

14、根据字段将记录分组

itertools模块中的函数groupby()可以通过扫描序列找出拥有相同值或是参数key指定的函数所返回的值的序列项，并将它们分组，groupby()创建一个迭代器，而每次迭代时都回返回一个值，和一个子迭代器，这个子迭代器可以产生所有在该分组内具有该值的项。

from operator import itemgetter
from itertools import groupby
rows = [
    {'a':'python','date':'07/01/2012'},
    {'a':'java','date':'08/11/2015'},
    {'a':'c++','date':'09/12/2018'},
    {'a':'perl','date':'17/06/2017'},
]
rows.sort(key=itemgetter('date'))
print(rows)
for date,items in groupby(rows,key=itemgetter('date')):
    print(date)
    for i in items:
        print(' ',i)

#
[{'a': 'python', 'date': '07/01/2012'}, {'a': 'java', 'date': '08/11/2015'}, {'a': 'c++', 'date': '09/12/2018'}, {'a': 'perl', 'date': '17/06/2017'}]
07/01/2012
  {'a': 'python', 'date': '07/01/2012'}
08/11/2015
  {'a': 'java', 'date': '08/11/2015'}
09/12/2018
  {'a': 'c++', 'date': '09/12/2018'}
17/06/2017
  {'a': 'perl', 'date': '17/06/2017'}

from collections import defaultdict
#根据数据分组来构建一个一键多值的字典
rows_date = defaultdict(list)
for row in rows:
    rows_date[row['date']].append(row)

print(rows_date)

#
defaultdict(<class 'list'>, {'07/01/2012': [{'a': 'python', 'date': '07/01/2012'}], '08/11/2015': [{'a': 'java', 'date': '08/11/2015'}], '09/12/2018': [{'a': 'c++', 'date': '09/12/2018'}], '17/06/2017': [{'a': 'perl', 'date': '17/06/2017'}]})

15、筛选序列中的元素

#使用列表推到式来赛选列表中符合要求的值
In [37]: mylist = [1,2,-5,10,-8,3,-1]

In [38]: list(i for i in mylist if i > 0)
Out[38]: [1, 2, 10, 3]

In [39]: list(i for i in mylist if i < 0)
Out[39]: [-5, -8, -1]

#如果输入的值非常多，可以先生成生成器然后筛选结果值
In [43]: pos = (i for i in mylist if i > 0)

In [46]: for i in pos:
    ...:     print(i)
    ...:
1
2
10
3

如果碰到筛选不标准的值如包含字符和数字，只筛选出数字呢？

In [47]: values = [1,'','-4','-',88,'N/A','python','']

In [48]: def is_int(val):
    ...:     try:
    ...:         x = int(val)
    ...:         return True
    ...:     except ValueError:
    ...:         return False
    ...:
#在筛选不规则的值式使用函数来过滤异常然后使用filter函数处理
In [49]: list(filter(is_int,values))
Out[49]: [1, '', '-4', 88, '']

#用新值替换掉筛选不和规定的值
In [50]: mylist = [1,4,-5,10,-7,2,3,-1]

In [51]: list(i if i > 0 else 0 for i in mylist)
Out[51]: [1, 4, 0, 10, 0, 2, 3, 0]

In [52]: list(i if i < 0 else 0 for i in mylist)
Out[52]: [0, 0, -5, 0, -7, 0, 0, -1]

#还可以使用itertools.compress()来构建一个布尔选择器序列来赛选数据
In [53]: addresses = ['one','two','three','four','five']

In [54]: from itertools import compress

In [55]: counts = [1,3,5,6,3]

In [56]: more1 = [i > 3 for i in counts]

In [57]: more1
Out[57]: [False, False, True, True, False]

In [58]: list(compress(addresses,more1))
Out[58]: ['three', 'four']

16、从字典中提取子集

prices = {
    'ACME':45.23,
    'AAPL':612.78,
    'IBM':205.55,
    'HPQ':37.20,
    'FB':10.75,
}
#推到式创建值大于30的字典集合
P1 = {key:value for key,value in prices.items() if value > 30}
print(P1)
#推倒式创建在tech中有的键的字典集合
tech = {'ACME','IBM','HPQ','FB'}
P2 = {key:value for key,value in prices.items() if key in tech}
print(P2)

#
{'ACME': 45.23, 'AAPL': 612.78, 'IBM': 205.55, 'HPQ': 37.2}
{'ACME': 45.23, 'IBM': 205.55, 'HPQ': 37.2, 'FB': 10.75}

#使用dict()函数来创建会更加清晰，效率会是上面的两倍
P3 = dict((key,value) for key,value in prices.items() if value > 100)
print(P3)

#
{'AAPL': 612.78, 'IBM': 205.55}

#多种实现方式，但这种方法会慢很多
p4 ={key:prices[key] for key in prices.keys() & tech}
print(p4)

#
{'ACME': 45.23, 'IBM': 205.55, 'FB': 10.75, 'HPQ': 37.2}

17、将名称映射到序列的元素中

collections.namedtuple()模块定义命名元组

In [1]: from collections import namedtuple
In [2]: subject = namedtuple('subject',['one','two','three'])
In [3]: sub = subject(1,2,3)
In [4]: sub
Out[4]: subject(one=1, two=2, three=3)
In [7]: sub.index(2)
Out[7]: 1
In [9]: sub.one
Out[9]: 1
In [10]: sub.two
Out[10]: 2
In [11]: sub.three
Out[11]: 3
In [12]: len(sub)
Out[12]: 3
In [13]: a,b,c=sub
In [14]: a
Out[14]: 1
In [15]: c
Out[15]: 3

#如果要修改某个值可以使用_replace()方法
In [17]: sub._replace(one=88)
Out[17]: subject(one=88, two=2, three=3)

18、同时对数据做转换和换算

#使用生成器表达式将数据转换和换算
In [19]: numbers = [1,2,3,4,5]
In [20]: s = sum(x * x for x in numbers)
In [21]: s
Out[21]: 55
In [26]: portfolio = [{'name':'GOOG','shares':50},{'name':'YHOO','shares':75},{'name':'AOL','shares':20},{'name':'SCOX','shares':65}]
#对所有商品求和
In [27]: sumnmber = sum(i['shares'] for i in portfolio)
In [28]: sumnmber
Out[28]: 210
In [29]: minmber = min(i['shares'] for i in portfolio)
In [30]: minmber
Out[30]: 20
In [31]: maxmber = max(i['shares'] for i in portfolio)
In [32]: maxmber
Out[32]: 75
#也可以使用key参数来换算
In [33]: min(portfolio,key=lambda s:s['shares'])
Out[33]: {'name': 'AOL', 'shares': 20}

19、将多个映射合并为单个映射

In [34]: a = {'x':11,'z':33}

In [35]: b = {'y':22,'z':44}
#利用collections模块中的ChainMap类来实现多个字典的合并检查
In [36]: from collections import ChainMap
In [37]: c = ChainMap(a,b)
In [38]: c
Out[38]: ChainMap({'x': 11, 'z': 33}, {'y': 22, 'z': 44})
In [39]: c['z'] = 33
In [40]: c
Out[40]: ChainMap({'x': 11, 'z': 33}, {'y': 22, 'z': 44})
In [41]: c['z'] = 55
In [42]: c
Out[42]: ChainMap({'x': 11, 'z': 55}, {'y': 22, 'z': 44})
In [43]: values = ChainMap()
In [44]: values['x'] = 100
In [45]: values = values.new_child()
In [46]: values['x'] = 200
In [47]: values
Out[47]: ChainMap({'x': 200}, {'x': 100})
In [48]: values = values.new_child()
In [50]: values['x'] = 50
In [51]: values
Out[51]: ChainMap({'x': 50}, {'x': 200}, {'x': 100})
In [52]: values['x']
Out[52]: 50

#利用字典的update()方法将多个字典合并一起，它会重新构建一个完整的字典
In [58]: a = {'x':1,'z':3}
In [59]: b = {'y':2,'z':4}
In [60]: merged = dict(b)
In [61]: merged.update(a)
In [62]: merged
Out[62]: {'y': 2, 'z': 3, 'x': 1}
#ChainMap使用的是原始的字典，对原始数据的更改会映射到新建的对象上
In [63]: a = {'x':1,'z':3}
In [64]: b = {'y':2,'z':4}
In [65]: merged = ChainMap(a,b)
In [66]: merged
Out[66]: ChainMap({'x': 1, 'z': 3}, {'y': 2, 'z': 4})
In [67]: merged['x']
Out[67]: 1
In [68]: a['x'] = 100
In [69]: merged['x']
Out[69]: 100
In [70]: merged
Out[70]: ChainMap({'x': 100, 'z': 3}, {'y': 2, 'z': 4})