让字典节省内存

 旧方法

a = {‘name‘:‘tom‘,‘age‘:18}
aa = {‘job‘:‘student‘,‘hobby‘:‘eat‘}

a.update(aa)
print(a)
>>> {‘name‘: ‘tom‘, ‘age‘: 18, ‘job‘: ‘student‘, ‘hobby‘: ‘eat‘}




新方法

from collections import ChainMap

a = {‘name‘:‘tom‘,‘age‘:18}
aa = {‘job‘:‘student‘,‘hobby‘:‘eat‘}

result = ChainMap(a,aa)
print(result)
>>> ChainMap({‘name‘: ‘tom‘, ‘age‘: 18}, {‘job‘: ‘student‘, ‘hobby‘: ‘eat‘})

a.pop(‘name‘)
result = ChainMap(a,aa)
print(result)
>>> ChainMap({‘age‘: 18}, {‘job‘: ‘student‘, ‘hobby‘: ‘eat‘})

print(result[‘hobby‘])
>>> eat

print(a)
>>> {‘age‘: 18}

总结

新方法类似于列表生成式，不会将所有数据都加载出来，耗费空间，而是要哪个获取哪个

让字典节省内存

原文：https://www.cnblogs.com/zhang-dan/p/11005390.html