1.列表统计

chars = ["a", "b", "a", "c", "a", "d"]

使用count获取单个字符出现次数

chars.count("a")

使用Counter的most_commom获取 出现次数最多的前几位

from collections import Counter
print(Counter(chars).most_common(2)

2.字典键值的集合操作

字典的keys()支持 并集| 交集 & 差集- 等集合操作

dict_a = {"a": 1, "b": 2, "c": 3 }
dict_b = {"a": 1, "c":2, "d": 4}

dict_a.keys() & dict_b.keys()

当字典的values都是字符串(无嵌套)时,字典的items()也支持集合操作

​断言字典a包含字典b​

assertFalse(dict_b.items() - dict_b.items())

3.列表嵌套字典操作

fruits = [{"name": "apple", "price": 4},
{"name": "orange", "price": 5}, {"name": "pear", "price":6} ,{"name": "apple", "price": 5}]

排序

sorted(fruits, key=lambda x: x["price"])

可以使用itemgetter代替lambda表达式

from operator import itemgetter
sorted(fruits, itemgetter("price"))

最小

mim(fruits, key=lambda x: x["price"])

最大

max(fruits, key=lambda x: x["price"])

使用堆获取最大/最小的前几个

'''
学习中遇到问题没人解答？小编创建了一个Python学习交流QQ群：725638078
寻找有志同道合的小伙伴，互帮互助,群里还有不错的视频学习教程和PDF电子书！
'''
import heapq
heapq.nlargest(2, fruits, key=lambda x: x["price"])
heapq.nsmallest(2, fruits, key=lambda x: x["price"]

分组groupby

from itertools import groupby
groups = groupby(fruits, key=lambda x:x["name"])

for name, fruits in groups:
    print(name, len(list(fruits)))