Python中字典的基本操作方法有哪些？

# 最常用这种my_object = {"a": 5,"b": 6}# 如果你不喜欢写大括号和双引号：my_object = dict(a=5, b=6)

a = { "a": 5, "b": 5 }b = { "c": 5, "d": 5 }c = { **a, **b } #最简单的方式assert c == { "a": 5, "b": 5, "c": 5, "d": 5 }# 合并后还要修改，可以这样：c = { **a, **b, "a": 10 }assert c == { "a": 10, "b": 5, "c": 5, "d": 5 }b["a"] = 10c = { **a, **b }assert c == { "a": 10, "b": 5, "c": 5, "d": 5 }

# 使用字典推导式来删除 keya = dict(a=5, b=6, c=7, d=8)remove = set(["c", "d"])a = { k: v for k,v in a.items() if k not in remove }# a = { "a": 5, "b": 6 }# 使用字典推导式来保留 keya = dict(a=5, b=6, c=7, d=8)keep = removea = { k: v for k,v in a.items() if k in keep }# a = { "c": 7, "d": 8 }# 使用字典推导式来让所有的 value 加 1a = dict(a=5, b=6, c=7, d=8)a = { k: v+1 for k,v in a.items() }# a = { "a": 6, "b": 7, "c": 8, "d": 9 }

from collections import Countercounter = Counter()#counter 可以统计 list 里面元素的频率counter.update(['a','b','a']#此时 counter = Counter({'a': 2, 'b': 1})#合并计数counter.update({ "a": 10000, "b": 1 })# Counter({'a': 10002, 'b': 2})counter["b"] += 100# Counter({'a': 10002, 'b': 102})print(counter.most_common())#[('a', 10002), ('b', 102)]print(counter.most_common(1)[0][0])# => a

from collections import defaultdict# 如果字典的 value 是 字典a = defaultdict(dict)assert a[5] == {}a[5]["a"] = 5assert a[5] == { "a": 5 }# 如果字典的 value 是列表a = defaultdict(list)assert a[5] == []a[5].append(3)assert a[5] == [3]# 字典的 value 的默认值可以是 lambda 表达式a = defaultdict(lambda: 10)assert a[5] == 10assert a[6] + 1 == 11# 字典里面又是一个字典，不用这个，你要做多少初始化操作？a = defaultdict(lambda: defaultdict(dict))assert a[5][5] == {}

import jsona = dict(a=5, b=6)# 字典转 JSON 字符串json_string = json.dumps(a)# json_string = '{"a": 5, "b": 6}'# JSON 字符串转字典assert a == json.loads(json_string)# 字典转 JSON 字符串保存在文件里with open("dict.json", "w+") as f:json.dump(a, f)# 从 JSON 文件里恢复字典with open("dict.json", "r") as f:assert a == json.load(f)

import pandas as pd# 字典转 pd.DataFramedf = pd.DataFrame([{ "a": 5, "b": 6 },{ "a": 6, "b": 7 }])# df =#ab# 056# 167# DataFrame 转回字典a = df.to_dict(orient="records")# a = [#{ "a": 5, "b": 6 },#{ "a": 6, "b": 7 }# ]# 字典转 pd.Seriessrs = pd.Series({ "a": 5, "b": 6 })# srs =# a5# b6# dtype: int64# pd.Series 转回字典a = srs.to_dict()# a = {'a': 5, 'b': 6}