Python CookBook(self report)

Python CookBook

中文版:https://python3-cookbook.readthedocs.io/zh_CN/latest/copyright.html

英文版:https://d.cxcore.net/Python/Python_Cookbook_3rd_Edition.pdf

Data Structures and Algorithms

start expressions

items = [1, 10, 7, 4, 5, 9] 
def sum(items):
... head, *tail = items
... return head + sum(tail) if tail else head
...
>>> head, *tail = items
>>> head
1
>>> tail
[10, 7, 4, 5, 9]
>>> sum(items)
36

Deque

Using deque(maxlen=N) creates a fixed-sized queue. When new items are added and the queue is full, the oldest item is automatically removed. 

from collections import deque
>>> q = deque(maxlen=3) >>> q.append(1) >>> q.append(2) >>> q.append(3) >>> q deque([1, 2, 3], maxlen=3) >>> q.append(4) >>> q

deque([1, 2, 3])
>>> q.appendleft(4)
>>> q deque([4, 1, 2, 3])
>>> q.pop() 3
>>> q deque([4, 1, 2])
>>> q.popleft() 4  

★ Adding or popping items from either end of a queue has O(1) complexity. This is unlike a list where inserting or removing items from the front of the list is O(N).

Heapq

The heapq module has two functions—nlargest() and nsmallest()—that do exactly what you want. For example:
import heapq
nums = [1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
print(heapq.nlargest(3, nums)) # Prints [42, 37, 23]
print(heapq.nsmallest(3, nums)) # Prints [-4, 1, 2]
Both functions also accept a key parameter that allows them to be used with more
complicated data structures. For example:
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'])

 

 

 

 

 

 

 

 

 

 

 

 

声明:如果涉及侵权请联系本人‘删除’,谢谢~
posted @ 2019-11-11 18:46   ̄□ ̄  阅读(389)  评论(0编辑  收藏  举报