Python Dict覆盖上一个列表

Question

Temp = {}
Final = {}

OK =['OK','OK']
NOTOK =['NOTOK','NOTOK']

Temp[1] = OK
Final['u1']=Temp

print(Final)

Temp.clear()
Temp[2] = OK
Final['u2']=Temp

print(Final)

Temp.clear()
Temp[3] = NOTOK
Final['u3']=Temp

print(Final)

想知道为什么最后一次打印会覆盖以前的打印，需要帮助。 它与我当前的项目有关，这是我使用的简单术语，即如何使u3 dict（NOTOK）不替换u1和u2内容（确定）。

实际结果

{'u1': {1: ['OK', 'OK']}}
{'u1': {2: ['OK', 'OK']}, 'u2': {2: ['OK', 'OK']}}
{'u1': {3: ['NOTOK', 'NOTOK']}, 'u2': {3: ['NOTOK', 'NOTOK']}, 'u3': {3: ['NOTOK', 'NOTOK']}}

预期结果

{'u1': {1: ['OK', 'OK']}}
{'u1': {2: ['OK', 'OK']}, 'u2': {2: ['OK', 'OK']}}
{'u1': {3: ['OK', 'OK']}, 'u2': {3: ['OK', 'OK']}, 'u3': {3: ['NOTOK', 'NOTOK']}}

Answer 1

使用.copy()

Temp = {}
Final = {}

OK =['OK','OK']
NOTOK =['NOTOK','NOTOK']

Temp[1] = OK.copy()
Final['u1']=Temp.copy()

print(Final)

Temp.clear()
Temp[2] = OK.copy()
Final['u2']=Temp.copy()

print(Final)

Temp.clear()
Temp[3] = NOTOK.copy()
Final['u3']=Temp.copy()

print(Final)