python标准库：collections和heapq模块

以你之姓@ 2022-08-02 11:53 211阅读 0赞

http://[blog.csdn.net/pipisorry/article/details/46947833][blog.csdn.net_pipisorry_article_details_46947833]

python额外的数据类型。collections模块和heapq模块的主要内容。

## 集合库collection ##

### collections模块介绍 ###

Python拥有一些内置的数据类型，比如str, int, list, tuple, dict等， collections模块在这些内置数据类型的基础上，提供了几个额外的数据类型：  
1.namedtuple(): 生成可以使用名字来访问元素内容的tuple子类  
2.deque: 双端队列，可以快速的从另外一侧追加和推出对象  
3.Counter: 计数器，主要用来计数  
4.OrderedDict: 有序字典  
5.defaultdict: 带有默认值的字典

## Named tuples元素命名的元组(collections.namedtuple) ##

>>> Point = collections.namedtuple('Point', ['x', 'y'])
    >>> p = Point(x=1.0, y=2.0)
    >>> p
    Point(x=1.0, y=2.0)
    >>> p.x
    1.0
    >>> p.y
    2.0

namedtuple主要用来产生可以使用名称来访问元素的数据对象，通常用来增强代码的可读性， 在访问一些tuple类型的数据时尤其好用。

### 举个例子 ###

比如我们用户拥有一个这样的数据结构，每一个对象是拥有三个元素的tuple。使用namedtuple方法就可以方便的通过tuple来生成可读性更高也更好用的数据结构。  
"""  
from collections import namedtuple  
websites = \[  
('Sohu', 'http://www.google.com/', u'张朝阳'),  
('Sina', 'http://www.sina.com.cn/', u'王志东'),  
('163', 'http://www.jbxue.com/', u'丁磊')  
\]  
Website = namedtuple('Website', \['name', 'url', 'founder'\])  
for website in websites:  
website = Website.\_make(website)  
print website  
\# Result:  
Website(name='Sohu', url='http://www.google.com/', founder=u'\\u5f20\\u671d\\u9633')  
Website(name='Sina', url='http://www.sina.com.cn/', founder=u'\\u738b\\u5fd7\\u4e1c')

Website(name='163', url='http://www.jbxue.com/', founder=u'\\u4e01\\u78ca')

### 继承命名tuples ###

>>> class Point(collections.namedtuple("PointBase", \["x", "y"\])):

... \_\_slots\_\_ = ()

... def \_\_add\_\_(self, other):

... return Point(x=self.x + other.x, y=self.y + other.y)

...

>>> p = Point(x=4.0, y=2.0)

>>> q = Point(x=2.0, y=3.0)

>>> p + q

Point(x=6.0, y=5.0)

## deque双端队列 ##

deque其实是 double-ended queue 的缩写，翻译过来就是双端队列，它最大的好处就是实现了从队列 头部快速增加和取出对象: .popleft(), .appendleft() 。  
你可能会说，原生的list也可以从头部添加和取出对象啊？就像这样：  
l.insert(0, v)  
l.pop(0)

list对象的这两种用法的时间复杂度是 O(n) ，也就是说随着元素数量的增加耗时呈 线性上升。而使用deque对象则是 O(1) 的复杂度，所以当你的代码有这样的需求的时候， 一定要记得使用deque。  
作为一个双端队列，deque还提供了一些其他的好用方法，比如 rotate 等。  
举个栗子  
**使用deque的rotate方法来实现了一个无限循环的加载动画**  
import sys  
import time  
from collections import deque  
fancy\_loading = deque('>--------------------')  
while True:  
print '\\r%s' % ''.join(fancy\_loading),  
fancy\_loading.rotate(1)  
sys.stdout.flush()  
time.sleep(0.08)  
\# Result:  
\# 一个无尽循环的跑马灯  
\------------->-------

**`双向队列`操作**

`>>> Q``=``collections.deque()`

`>>> Q.append(``1``)`

`>>> Q.appendleft(``2``)`

`>>> Q.extend([``3``,``4``])`

`>>> Q.extendleft([``5``,``6``])`

`>>> Q`

`deque([``6``,``5``,``2``,``1``,``3``,``4``])`

`>>> Q.pop()`

`4`

`>>> Q.popleft()`

`6`

`>>> Q`

`deque([``5``,``2``,``1``,``3``])`

`>>> Q.rotate(``3``)`

`>>> Q`

`deque([``2``,``1``,``3``,``5``])`

`>>> Q.rotate(``-``3``)`

`>>> Q`

`deque([``5``,``2``,``1``,``3``]) 限制长度的双向队列`

`>>> last_three``=` `collections.deque(maxlen``=``3``)`

`>>>``for``i``in``xrange``(``10``):`

`... last_three.append(i)`

`...``print``", "``.join(``str``(x)``for``x``in``last_three)`

`...`

`0`

`0``,``1`

`0``,``1``,``2`

`1``,``2``,``3`

`2``,``3``,``4`

[皮皮Blog][Blog]

## Counter计数器 - Multisets运算 ##

A Counter is a dict subclass for counting hashable objects.It is an unordered collection where elements are stored as dictionary keysand their counts are stored as dictionary values. Counts are allowed to beany integer value including zero or negative counts. The Counter class is similar to bags or multisets in other languages.

**统计所有字符出现次数**  
from collections import Counter  
counter = Counter()  
a = '隐蔽式喷淋头盖板面板装饰板 外壳 whats app'  
counter.update(a)  
print(counter)

Counter(\{'板': 3, ' ': 3, 'a': 2, 'p': 2, '隐': 1, '蔽': 1, '式': 1, '喷': 1, '淋': 1, '头': 1, '盖': 1, '面': 1, '装': 1, '饰': 1, '外': 1, '壳': 1, 'w': 1, 'h': 1, 't': 1, 's': 1\})

**统计字符串出现次数**

counter.update(\[a\])

### Counter方法 ###

counter还有几个自己独有的方法如下：

**elements()**

Return an iterator over elements repeating each as many times as itscount. Elements are returned in arbitrary order. If an element’s countis less than one, elements() will ignore it.

>>> c = Counter(a=4, b=2, c=0, d=-2)
    >>> sorted(c.elements())
    ['a', 'a', 'a', 'a', 'b', 'b']

**most\_common(\[n\]) 获取出现频率最高的n个/按频率降序的字符**

Return a list of the n most common elements and their counts from themost common to the least. If n is omitted or None,most\_common() returns all elements in the counter.Elements with equal counts are ordered arbitrarily:

>>> Counter('abracadabra').most_common(3)  
    [('a', 5), ('r', 2), ('b', 2)]

**subtract(\[iterable-or-mapping\])**

Elements are subtracted from an iterable or from another mapping(or counter). Like dict.update() but subtracts counts insteadof replacing them. Both inputs and outputs may be zero or negative.

>>> c = Counter(a=4, b=2, c=0, d=-2)
    >>> d = Counter(a=1, b=2, c=3, d=4)
    >>> c.subtract(d)
    >>> c
    Counter({'a': 3, 'b': 0, 'c': -3, 'd': -6})

除了fromkeys和update外，counter支持所有dict的其它函数。如返回所有的值counter.values()。The usual dictionary methods are available for Counter objects except for two which work differently for counters.

fromkeys(iterable)

This class method is not implemented for Counter objects.

update(\[iterable-or-mapping\])

Elements are counted from an iterable or added-in from anothermapping (or counter). Like dict.update() but adds countsinstead of replacing them. Also, the iterable is expected to be asequence of elements, not a sequence of (key, value) pairs.

counter = Counter()
    for c in c_type_list:
      counter.update(c)

Another, maybe better, example might be counting all the different lines that appear in a file. It becomes very simple

with open('/some/file', 'r') as f:
    
        line_count = Counter(f)

**Multisets运算**

`>>> A``=``collections.Counter([``1``,``2``,``2``])`

`>>> B``=``collections.Counter([``2``,``2``,``3``])`

`>>> A`

`Counter({``2``:``2``,``1``:``1``})`

`>>> B`

`Counter({``2``:``2``,``3``:``1``})`

`>>> A | B`

`Counter({``2``:``2``,``1``:``1``,``3``:``1``})`

`>>> A & B`

`Counter({``2``:``2``})`

`>>> A``+``B`

`Counter({``2``:``4``,``1``:``1``,``3``:``1``})`

`>>> A``-``B`

`Counter({``1``:``1``})`

`>>> B``-``A`

`Counter({``3``:``1``})`  
\[[Counter][] objects\]

## OrderedDict有序字典 ##

在Python中，dict这个数据结构由于hash的特性，是无序的，这在有的时候会给我们带来一些麻烦， 幸运的是，collections模块为我们提供了OrderedDict，当你要获得一个有序的字典对象时，用它就对了。

>>> m = dict((str(x), x) for x in range(10))
    >>> print ', '.join(m.keys())
    1, 0, 3, 2, 5, 4, 7, 6, 9, 8
    >>> m = collections.OrderedDict((str(x), x) for x in range(10))
    >>> print ', '.join(m.keys())
    0, 1, 2, 3, 4, 5, 6, 7, 8, 9
    >>> m = collections.OrderedDict((str(x), x) for x in range(10, 0, -1))
    >>> print ', '.join(m.keys())
    10, 9, 8, 7, 6, 5, 4, 3, 2, 1

**举个栗子**  
from collections import OrderedDict  
items = (  
('A', 1),  
('B', 2),  
('C', 3)  
)  
regular\_dict = dict(items)  
ordered\_dict = OrderedDict(items)  
print 'Regular Dict:'  
for k, v in regular\_dict.items():  
print k, v  
print 'Ordered Dict:'  
for k, v in ordered\_dict.items():  
print k, v  
\# Result:  
Regular Dict:  
A 1  
C 3  
B 2  
Ordered Dict:  
A 1  
B 2  
C 3

**Note**:直接输出的OrderedDict形式如下：print(OrderedDict(reguale\_slist))

OrderedDict(\[('GET', 'http://www.jobbole.com/login/?redirect=http://www.jobbole.com/ HTTP/1.1'), ('Host', 'www.jobbole.com'), ('Connection', 'keep-alive'),  ('Cookie', 'wordpress\_test\_cookie=WP+Cookie+check')...\])

而不是dict的形式:print(dict(reguale\_slist))

\{'Referer': 'http://www.jobbole.com/', 'Host': 'www.jobbole.com', 'Connection': 'keep-alive', 'Accept-Encoding': 'gzip, deflate, sdch', ...\}

## defaultdict默认字典 ##

我们都知道，在使用Python原生的数据结构dict的时候，如果用 d\[key\] 这样的方式访问， 当指定的key不存在时，是会抛出KeyError异常的。  
但是，如果使用defaultdict，只要你传入一个默认的工厂方法，那么请求一个不存在的key时， 便会调用这个工厂方法使用其结果来作为这个key的默认值。

### defaultdict字典元素类型初始化 ###

collections.defaultdict(\[default\_factory\[,...\]\])

default\_factory指定字典的value类型

from collections import defaultdict  
>>> s = \[('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)\]  
>>> d = defaultdict(list)  
>>> for k, v in s:  
...     d\[k\].append(v)...  
>>> d.items()

\[('blue', \[2, 4\]), ('red', \[1\]), ('yellow', \[1, 3\])\]

**如果给default\_dict传入int，则可以用来计数：**  
>>> s = 'mississippi'  
>>> d = defaultdict(int)  
>>> for k in s:  
...     d\[k\] += 1  
>>> d.items()

\[('i', 4), ('p', 2), ('s', 4), ('m', 1)\]

Note:defaultdict()参数设置成int，那么就设置字典值的初始值为0了

### 嵌套defaultdict ###

from collections import defaultdict
    u_lt_ts_dict = defaultdict(lambda: defaultdict(list))

但是这个对象是不可序列化的，因为有lamba函数。也可以这样

from collections import defaultdict
    from functools import partial
    u_lt_ts_dict = defaultdict(partial(defaultdict, list))

这样就可以pickle了。

\[[Multiple levels of 'collection.defaultdict' in Python][Multiple levels of _collection.defaultdict_ in Python]\]

### Using default dictionaries to represent simple trees ###

>>> import json
    >>> tree = lambda: collections.defaultdict(tree)
    >>> root = tree()
    >>> root['menu']['id'] = 'file'
    >>> root['menu']['value'] = 'File'
    >>> root['menu']['menuitems']['new']['value'] = 'New'
    >>> root['menu']['menuitems']['new']['onclick'] = 'new();'
    >>> root['menu']['menuitems']['open']['value'] = 'Open'
    >>> root['menu']['menuitems']['open']['onclick'] = 'open();'
    >>> root['menu']['menuitems']['close']['value'] = 'Close'
    >>> root['menu']['menuitems']['close']['onclick'] = 'close();'
    >>> print json.dumps(root, sort_keys=True, indent=4, separators=(',', ': '))
    {
        "menu": {
            "id": "file",
            "menuitems": {
                "close": {
                    "onclick": "close();",
                    "value": "Close"
                },
                "new": {
                    "onclick": "new();",
                    "value": "New"
                },
                "open": {
                    "onclick": "open();",
                    "value": "Open"
                }
            },
            "value": "File"
        }
    }

(See [https://gist.github.com/hrldcpr/2012250][https_gist.github.com_hrldcpr_2012250] for more on this.)

\[[一行Python代码实现树结构][Python]\]

### Mapping objects to unique counting numbers (collections.defaultdict) ###

>>> import itertools, collections
    >>> value_to_numeric_map = collections.defaultdict(itertools.count().next)
    >>> value_to_numeric_map['a']
    0
    >>> value_to_numeric_map['b']
    1
    >>> value_to_numeric_map['c']
    2
    >>> value_to_numeric_map['a']
    0

**举个栗子**  
from collections import defaultdict  
members = \[  
\# Age, name  
\['male', 'John'\],  
\['male', 'Jack'\],  
\['female', 'Lily'\],  
\['male', 'Pony'\],  
\['female', 'Lucy'\],  
\]  
result = defaultdict(list)  
for sex, name in members:  
result\[sex\].append(name)  
print result  
\# Result:  
defaultdict(<type 'list'>, \{'male': \['John', 'Jack', 'Pony'\], 'female': \['Lily', 'Lucy'\]\})

\[[Python collections模块实例][Python collections]\]

**常用操作**

>>> m =dict()

>>> m\["a"\]

Traceback (most recent call last):

File"<stdin>", line1,in<module>

KeyError: "a"

>>>

>>> m =collections.defaultdict(int)

>>> m\["a"\]

>>> m\["b"\]

>>> m =collections.defaultdict(str)

>>> m\["a"\]

>>> m\["b"\]+="a"

>>> m\["b"\]

"a"

>>> m =collections.defaultdict(lambda:"\[default value\]")

>>> m\["a"\]

"\[default value\]"

>>> m\["b"\]

"\[default value\]"  
`使用defaultdict代表tree`

>>> importjson

>>> tree = lambda: collections.defaultdict(tree)

>>> root = tree()

>>> root\["menu"\]\["id"\]="file"

>>> root\["menu"\]\["value"\]="File"

>>> root\["menu"\]\["menuitems"\]\["new"\]\["value"\]="New"

>>> root\["menu"\]\["menuitems"\]\["new"\]\["onclick"\]="new();"

>>> root\["menu"\]\["menuitems"\]\["open"\]\["value"\]="Open"

>>> root\["menu"\]\["menuitems"\]\["open"\]\["onclick"\]="open();"

>>> root\["menu"\]\["menuitems"\]\["close"\]\["value"\]="Close"

>>> root\["menu"\]\["menuitems"\]\["close"\]\["onclick"\]="close();"

>>> printjson.dumps(root, sort\_keys=True, indent=4, separators=(",",": "))

"menu": \{

"id":"file",

"menuitems": \{

"close": \{

"onclick":"close();",

"value":"Close"

\},

"new": \{

"onclick":"new();",

"value":"New"

\},

"open": \{

"onclick":"open();",

"value":"Open"

\},

"value":"File"

`[查看更多] 映射对象到唯一的计数数字`

>>> importitertools, collections

>>> value\_to\_numeric\_map=collections.defaultdict(itertools.count().next)

>>> value\_to\_numeric\_map\["a"\]

>>> value\_to\_numeric\_map\["b"\]

>>> value\_to\_numeric\_map\["c"\]

>>> value\_to\_numeric\_map\["a"\]

>>> value\_to\_numeric\_map\["b"\]

1  
[皮皮blog][Blog]

# 堆队列heapq — Heap queue algorithm #

这个模块提供了堆队列算法的实现，也称为优先队列算法。

默认是个最小堆。堆是一个二叉树，它的每个父节点的值都只会小于或等于所有孩子节点（的值）。 它使用了数组来实现：从零开始计数，对于所有的 *k* ，都有 `heap[k] <= heap[2*k+1]` 和 `heap[k] <= heap[2*k+2]`。 为了便于比较，不存在的元素被认为是无限大。 堆最有趣的特性在于最小的元素总是在根结点：`heap[0]`。

这个API与教材的堆算法实现有所不同，具体区别有两方面：（a）我们使用了从零开始的索引。这使得节点和其孩子节点索引之间的关系不太直观但更加适合，因为 Python 使用从零开始的索引。 （b）我们的 pop 方法返回最小的项而不是最大的项（这在教材中称为“最小堆”；而“最大堆”在教材中更为常见，因为它更适用于原地排序）。

基于这两方面，把堆看作原生的Python list也没什么奇怪的： `heap[0]` 表示最小的元素，同时 `heap.sort()` 维护了堆的不变性！

heap = \[\]

**heapq.heappush(heap, item)**

将 item 的值加入 heap 中，保持堆的不变性。Push the value item onto the heap, maintaining the heap invariant.

Note: item可以是tuple类型带卫星数据，默认使用item\[0\]进行排序，如

**`heapq.``heapify`(*x*)[¶][Link 1]**

将list *x* 转换成堆，原地，线性时间内。

**heapq.heappop(heap)**

弹出并返回 heap 的最小的元素，保持堆的不变性。如果堆为空，抛出 IndexError 。使用 heap\[0\] ，可以只访问最小的元素而不弹出它。Pop and return the smallest item from the heap, maintaining the heapinvariant. If the heap is empty, IndexError is raised. To access thesmallest item without popping it, use heap\[0\].

**heapq.heappushpop(heap, item)**

Push item on the heap, then pop and return the smallest item from the heap. The combined action runs more efficiently than heappush()followed by a separate call to heappop().  
**heapq.heapreplace(heap, item)**

弹出并返回 heap 中最小的一项，同时推入新的 item。 堆的大小不变。 如果堆为空则引发 IndexError。

这个单步骤操作比 heappop() 加 heappush() 更高效，并且在使用固定大小的堆时更为适宜。 pop/push 组合总是会从堆中返回一个元素并将其替换为 item。

返回的值可能会比添加的 item 更大。 如果不希望如此，可考虑改用 heappushpop()。 它的 push/pop 组合会返回两个值中较小的一个，将较大的值留在堆中。

**`heapq.``merge`(*\*iterables*, *key=None*, *reverse=False*)**

将多个已排序的输入合并为一个已排序的输出（例如，合并来自多个日志文件的带时间戳的条目）。 返回已排序值的 [iterator][]。

**heapq.nlargest(n, iterable, key=None)**

Return a list with the n largest elements from the dataset defined byiterable. key, if provided, specifies a function of one argument that isused to extract a comparison key from each element in the iterable:key=str.lower Equivalent to: sorted(iterable, key=key,reverse=True)\[:n\]

**heapq.nsmallest(n, iterable, key=None)**

Return a list with the n smallest elements from the dataset defined byiterable. key, if provided, specifies a function of one argument that isused to extract a comparison key from each element in the iterable:key=str.lower Equivalent to: sorted(iterable, key=key)\[:n\]

\[[`heapq`][heapq] \--- 堆队列算法[¶][heapq]\]\*\*

## 示例 ##

>>> h = \[\]  
>>> heappush(h, (5, 'write code'))  
>>> heappush(h, (1, 'write spec'))  
>>> heappush(h, (3, 'create tests'))  
>>> heappop(h)  
(1, 'write spec')

### 维护top n的最小堆（最大的k个元素） ###

def push(self, x):
        """Pushes a new element."""
        assert self._data is not None
        if len(self._data) < self._n:
            heapq.heappush(self._data, x)
        else:
            heapq.heappushpop(self._data, x)

### 第k大的元素 ###

import heapq
    class Solution:
        def findKthLargest(self, nums: List[int], k: int) -> int:
            maxk_l = []
            for i in range(k):
                heapq.heappush(maxk_l, nums[i])
            
            for i in nums[k:]:
                min0 = maxk_l[0]
                if min0 < i:
                    heapq.heappushpop(maxk_l, i)
            return maxk_l[0]

return maxk\_l即最大的k个元素。

\[[https://leetcode-cn.com/problems/kth-largest-element-in-an-array/submissions/][https_leetcode-cn.com_problems_kth-largest-element-in-an-array_submissions]\]

解法2：

from typing import List
    import heapq
    
    
    class Solution:
    
        # 使用容量为 k 的小顶堆
        # 元素个数小于 k 的时候，放进去就是了
        # 元素个数大于 k 的时候，小于等于堆顶元素，就扔掉，大于堆顶元素，就替换
    
        def findKthLargest(self, nums: List[int], k: int) -> int:
            size = len(nums)
            if k > size:
                raise Exception('程序出错')
    
            L = []
            for index in range(k):
                # heapq 默认就是小顶堆
                heapq.heappush(L, nums[index])
    
            for index in range(k, size):
                top = L[0]
                if nums[index] > top:
                    # 看一看堆顶的元素，只要比堆顶元素大，就替换堆顶元素
                    heapq.heapreplace(L, nums[index])
            # 最后堆顶中的元素就是堆中最小的，整个数组中的第 k 大元素
            return L[0]

\[[https://leetcode-cn.com/problems/kth-largest-element-in-an-array/solution/partitionfen-er-zhi-zhi-you-xian-dui-lie-java-dai-/][https_leetcode-cn.com_problems_kth-largest-element-in-an-array_solution_partitionfen-er-zhi-zhi-you-xian-dui-lie-java-dai-]\]

### 最大 & 最小的k个元素(heapq.nlargest and heapq.nsmallest) ###

>>> a =\[random.randint(0,100)for\_\_inxrange(100)\]

>>> heapq.nsmallest(5, a)

\[3,3,5,6,8\]

>>> heapq.nlargest(5, a)

\[100,100,99,98,98\]  
\[其具体实现算法可以参考：[编程之美之2.5 寻找最大的K个数 ][2.5 _K_]\]  
\[[A fast way to find the largest N elements in an numpy array][]\]  
\[[How to get indices of N maximum values in a numpy array?][How to get indices of N maximum values in a numpy array]\]  
\[[`heapq` — Heap queue algorithm][heapq_ _ Heap queue algorithm]\]

from:[http://blog.csdn.net/pipisorry/article/details/46947833][blog.csdn.net_pipisorry_article_details_46947833]

ref: \[[`collections`][collections] — Container datatypes\]

\[[Data Types][]\]

[blog.csdn.net_pipisorry_article_details_46947833]: http://blog.csdn.net/pipisorry/article/details/46947833
[Blog]: http://blog.csdn.net/pipisorry
[Counter]: https://docs.python.org/3/library/collections.html#collections.Counter
[Multiple levels of _collection.defaultdict_ in Python]: http://stackoverflow.com/questions/2600790/multiple-levels-of-collection-defaultdict-in-python
[https_gist.github.com_hrldcpr_2012250]: https://gist.github.com/hrldcpr/2012250
[Python]: http://codingpy.com/article/one-line-tree-in-python/
[Python collections]: http://www.cnblogs.com/yes123/p/3651172.html
[Link 1]: https://docs.python.org/zh-cn/3/library/heapq.html#heapq.heapify
[iterator]: https://docs.python.org/zh-cn/3/glossary.html#term-iterator
[heapq]: https://docs.python.org/zh-cn/3/library/heapq.html#module-heapq
[https_leetcode-cn.com_problems_kth-largest-element-in-an-array_submissions]: https://leetcode-cn.com/problems/kth-largest-element-in-an-array/submissions/
[https_leetcode-cn.com_problems_kth-largest-element-in-an-array_solution_partitionfen-er-zhi-zhi-you-xian-dui-lie-java-dai-]: https://leetcode-cn.com/problems/kth-largest-element-in-an-array/solution/partitionfen-er-zhi-zhi-you-xian-dui-lie-java-dai-/
[2.5 _K_]: http://blog.csdn.net/sunnyyoona/article/details/26380473
[A fast way to find the largest N elements in an numpy array]: http://stackoverflow.com/questions/10337533/a-fast-way-to-find-the-largest-n-elements-in-an-numpy-array
[How to get indices of N maximum values in a numpy array]: http://stackoverflow.com/questions/6910641/how-to-get-indices-of-n-maximum-values-in-a-numpy-array
[heapq_ _ Heap queue algorithm]: https://docs.python.org/3/library/heapq.html
[collections]: https://docs.python.org/3/library/collections.html#module-collections
[Data Types]: https://docs.python.org/3/library/datatypes.html