重要知识点
- 生成式(推导式)的用法
prices = {
‘AAPL’: 191.88,
‘GOOG’: 1186.96,
‘IBM’: 149.24,
‘ORCL’: 48.44,
‘ACN’: 166.89,
‘FB’: 208.09,
‘SYMC’: 21.29
}
# 用股票价格大于100元的股票构造一个新的字典
prices2 = {key: value for key, value in prices.items() if value > 100}
print(prices2)
说明:生成式(推导式)可以用来生成列表、集合和字典。
嵌套的列表的坑
names = [‘关羽’, ‘张飞’, ‘赵云’, ‘马超’, ‘黄忠’]
courses = [‘语文’, ‘数学’, ‘英语’]
# 录入五个学生三门课程的成绩
# 错误 - 参考http://pythontutor.com/visualize.html#mode=edit
# scores = [[None] len(courses)] len(names)
scores = [[None] * len(courses) for _ in range(len(names))]
for row, name in enumerate(names):
for col, course in enumerate(courses):
scores[row][col] = float(input(f‘请输入{name}的{course}成绩: ‘))
print(scores)
Python Tutor - VISUALIZE CODE AND GET LIVE HELP
heapq
模块(堆排序)“””
从列表中找出最大的或最小的N个元素
堆结构(大根堆/小根堆)
“””
import heapq
list1 = [34, 25, 12, 99, 87, 63, 58, 78, 88, 92]
list2 = [
{‘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}
]
print(heapq.nlargest(3, list1))
print(heapq.nsmallest(3, list1))
print(heapq.nlargest(2, list2, key=lambda x: x[‘price’]))
print(heapq.nlargest(2, list2, key=lambda x: x[‘shares’]))
itertools
模块“””
迭代工具模块
“””
import itertools
# 产生ABCD的全排列
itertools.permutations(‘ABCD’)
# 产生ABCDE的五选三组合
itertools.combinations(‘ABCDE’, 3)
# 产生ABCD和123的笛卡尔积
itertools.product(‘ABCD’, ‘123’)
# 产生ABC的无限循环序列
itertools.cycle((‘A’, ‘B’, ‘C’))
collections
模块常用的工具类:
namedtuple
:命令元组,它是一个类工厂,接受类型的名称和属性列表来创建一个类。deque
:双端队列,是列表的替代实现。Python中的列表底层是基于数组来实现的,而deque底层是双向链表,因此当你需要在头尾添加和删除元素是,deque会表现出更好的性能,渐近时间复杂度为$O(1)$。Counter
:dict
的子类,键是元素,值是元素的计数,它的most_common()
方法可以帮助我们获取出现频率最高的元素。Counter
和dict
的继承关系我认为是值得商榷的,按照CARP原则,Counter
跟dict
的关系应该设计为关联关系更为合理。OrderedDict
:dict
的子类,它记录了键值对插入的顺序,看起来既有字典的行为,也有链表的行为。defaultdict
:类似于字典类型,但是可以通过默认的工厂函数来获得键对应的默认值,相比字典中的setdefault()
方法,这种做法更加高效。
“””
找出序列中出现次数最多的元素
“””
from collections import Counter
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’
]
counter = Counter(words)
print(counter.most_common(3))