dict 的常用方法¶
提示
- Pycharm 中，点击菜单栏的 View -> Tool Windows -> Structure 可以查看源码中的的类、方法等。
- 在一段代码后直接接个 pass 作为断点， 然后用 debug 模式调试，比较方便。
clear¶
clear的功能是对dict的内容进行清除.没有任何的返回值
copy¶
浅copy¶
a = {"bya1":{"school":"ustc1"},
"bya2": {"school": "ustc2"}
}
new_dict = a.copy()
new_dict["bya1"]["school"] = "ustc3"
print(new_dict)
print(a)
pass
>>>
{'bya1': {'school': 'ustc3'}, 'bya2': {'school': 'ustc2'}}
{'bya1': {'school': 'ustc3'}, 'bya2': {'school': 'ustc2'}}
深copy¶
a = {"bya1":{"school":"ustc1"},
"bya2": {"school": "ustc2"}
}
import copy
new_dict = copy.deepcopy(a)
new_dict["bya1"]["school"] = "ustc3"
print(new_dict)
print(a)
pass
>>>
{'bya1': {'school': 'ustc3'}, 'bya2': {'school': 'ustc2'}}
{'bya1': {'school': 'ustc1'}, 'bya2': {'school': 'ustc2'}}
- dict 内置的 copy 方法返回浅拷贝（shallow copy）,对于嵌套的数据结构，并不会被拷贝
- copy 模块的 deepcopy 方法返回深拷贝（deep copy）, 会拷贝嵌套的数据结构
fromkeys¶
首先这是一个静态方法, 可以直接使用类名调用
@staticmethod # known case
def fromkeys(*args, **kwargs): # real signature unknown
""" Returns a new dict with keys from iterable and values equal to value. """
pass
new_list = ["bya","bya2"]
new_dict = dict.fromkeys(new_list,"school")
print(new_dict)
>>> {'bya': 'school', 'bya2': 'school'}
get¶
使用非常频繁，可以解决 key 没有对应的 value 而报错的问题
a = {"bya1":{"school":"ustc1"},
"bya2": {"school": "ustc2"}
}
# print(a["bya3"]) 会提示 KeyError: 'bya3'
print(a.get("bya3",{})) # a 没有 key bya3，会设置其 value 值为 ｛｝
print(a.get("bya1",{})) # a 有 key bya1， 会取其 value 值 {'school': 'ustc1'}
>>>
{}
{'school': 'ustc1'}
items, keys, values¶
使用 for 循环，进行元组拆包
a = {"bya1":{"school":"ustc1"},
"bya2": {"school": "ustc2"}
}
for key, value in a.items():
print(key, value)
>>>
bya1 {'school': 'ustc1'}
bya2 {'school': 'ustc2'}
pop，popitem¶
删除并返回
setdefault¶
如果key存在使用get方法, 如果key不存在使用set方法
使用 setdefault 方法，比我们自行判断并添加的步骤要少，故而性能更高效
a = {"bya1":{"school":"ustc1"},
"bya2": {"school": "ustc2"}
}
default_value = a.setdefault("bya3", {"school": "ustc3"})
print(default_value)
print(a)
>>>
{'school': 'ustc3'}
{'bya1': {'school': 'ustc1'}, 'bya2': {'school': 'ustc2'}, 'bya3': {'school': 'ustc3'}}
update¶
def update(self, E=None, **F): # known special case of dict.update
"""
D.update([E, ]**F) -> None. Update D from dict/iterable E and F.
If E is present and has a .keys() method, then does: for k in E: D[k] = E[k]
If E is present and lacks a .keys() method, then does: for k, v in E: D[k] = v
In either case, this is followed by: for k in F: D[k] = F[k]
"""
pass
update 方法，中间可以放可迭代的对象，因此 list，dict，tuple 都可以。
a = {"bya1":{"school":"ustc1"},
"bya2": {"school": "ustc2"}
}
a.update({"bya3": {"school": "ustc3"}})
print(a)
# 关键词方式 key=value
a.update(bya4={"school": "ustc4"}) # 同时也可以传递多个 key=value, 使用 , 隔开即可
print(a)
# list 中 放置 tuple 也可以
a.update([("bya5", {"school":"ustc5"})]) # 当然，多个 tuple 也是可以的
print(a)
# tuple 中 放置 tuple 也可以，因为 tuple 也是可迭代的
a.update((("bya6", {"school":"ustc6"}), )) # 注意要有一个 ,
print(a)
>>>
{'bya1': {'school': 'ustc1'}, 'bya2': {'school': 'ustc2'}, 'bya3': {'school': 'ustc3'}}
{'bya1': {'school': 'ustc1'}, 'bya2': {'school': 'ustc2'}, 'bya3': {'school': 'ustc3'}, 'bya4': {'school': 'ustc4'}}
{'bya1': {'school': 'ustc1'}, 'bya2': {'school': 'ustc2'}, 'bya3': {'school': 'ustc3'}, 'bya4': {'school': 'ustc4'}, 'bya5': {'school': 'ustc5'}}
{'bya1': {'school': 'ustc1'}, 'bya2': {'school': 'ustc2'}, 'bya3': {'school': 'ustc3'}, 'bya4': {'school': 'ustc4'}, 'bya5': {'school': 'ustc5'}, 'bya6': {'school': 'ustc6'}}
两个字典合并