Python에서 사전의 기본 작업 방법은 무엇입니까?

# 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)

# 使用字典推导式来删除 key a = 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 key a = dict(a=5, b=6, c=7, d=8) keep = remove a = { 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 1 a = 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 Counter counter = 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"] = 5 assert 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] == 10 assert 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 json a = 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.DataFrame df = 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.Series srs = 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}