CV肉饼王

一、*符号

1.解包所有可迭代对象，比如list、tuple、string

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num_list = [1, 2, 3, 4, 5]
num_list_2 = [6, 7, 8, 9, 10]
new_list = [*num_list, *num_list_2]
print(new_list)  # [1,2,3,4,5,6,7,8,9,10]

1.参数的默认值

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def search4letters(phrase: str, letters: str = 'aeiou') -> set:
    """在指定字符串中找到输入的字母"""
    return set(letters) & set(phrase)

一、方法

一、语法

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lambda argument_list:expersion

注意1：lambda函数体只能写一句话

注意2：不支持赋值语句，如下代码报错

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func01 = lambda p: p = 10

文/雨令

有的人在十年前买了好几套房，结果成了千万富翁，有的人在前几年进了股市，结果却一贫如洗。

有的人顺风顺水步步高升一路躺赢，也有的人忙忙碌碌东奔西走依然举步维艰。

每个人一路走来，都在做着这样或那样的选择，你选择上学，他选择出国，你选择上班，他选择创业，那些看似不经意的选择，总是会在往后很长的一段时间里左右着我们的命运起伏。

今天是大年初一，虽然一个多礼拜没洗澡了，但身上的水痘好多了，趁着有大块的空闲时间，来写写年终总结。

Some differences from dict still remain:

• The regular dict was designed to be very good at mapping operations. Tracking insertion order was secondary.
• The OrderedDict was designed to be good at reordering operations. Space efficiency, iteration speed, and the performance of update operations were secondary.
• The OrderedDict algorithm can handle frequent reordering operations better than dict. As shown in the recipes below, this makes it suitable for implementing various kinds of LRU caches.
• The equality operation for OrderedDict checks for matching order.
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一.Counter计数

1.查询某元素出现次数、查询出现此次最多的n个元素

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from collections import Counter

temp = Counter('abcdeabcdabcaba')
print(temp.items())  # dict_items([('a', 5), ('b', 4), ('c', 3), ('d', 2), ('e', 1)])

for item in temp.items():
    print(item)
"""
('a', 5)
('b', 4)
('c', 3)
('d', 2)
('e', 1)
"""

print(temp.most_common(1))  # 统计出现次数最多的一个元素,[('a', 5)]
print(temp.most_common(2))  # 统计出现次数最多个两个元素,[('a', 5), ('b', 4)]

print(temp['a'])  # 查询单个元素出现次数，5
temp['a'] += 1  # 增加单个元素出现次数，6
temp['a'] -= 1  # 增加单个元素出现次数，5
del temp['a']  # 删除,({'b': 4, 'c': 3, 'd': 2, 'e': 1})

1.数据类型

整型int、浮点数float、字符串str、布尔bool(True、False)

2.print格式化输出

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name = input('please enter your name: ')
print('hello,', name)

print('%.1f华氏度 = %.1f摄氏度' % (f, c))
# 或者
print(f'{f:.1f}华氏度 = {c:.1f}摄氏度')