[英]how to update the value of dictionary in a dictionary using python
我有:
new_dict['a1']['Road_Type']=[0,0,0,0,0,0,0]
new_dict['a2']['Road_Type']=[0,0,0,0,0,0,0]
new_dict['a3']['Road_Type']=[0,0,0,0,0,0,0]
new_dict['a4']['Road_Type']=[0,0,0,0,0,0,0]
现在更新后我希望它是:
new_dict['a1']['Road_Type']=[0,0,0,0,0,0,0]
new_dict['a2']['Road_Type']=[5,0,0,0,0,0,0]
new_dict['a3']['Road_Type']=[0,0,0,0,0,0,0]
new_dict['a4']['Road_Type']=[0,0,0,0,0,0,0]
我的代码:
kk=new_dict['a2']['Road_Type']
kk[0]=5
new_dict['a2']['Road_Type']=kk
但结果是:
new_dict['a1']['Road_Type']=[5,0,0,0,0,0,0]
new_dict['a2']['Road_Type']=[5,0,0,0,0,0,0]
new_dict['a3']['Road_Type']=[5,0,0,0,0,0,0]
new_dict['a4']['Road_Type']=[5,0,0,0,0,0,0]
所有值都在更新,那么我如何更新特定值。
根据您对问题的评论,您由于不知道 Python 的工作原理而犯了一个错误。 我会在示例中使其更简单,但是当您拥有new_dict[a][b]...[n]
时,这也代表您的情况。
以下是您生成字典的方式:
lst = [0, 0, 0, 0, 0, 0, 0]
new_dict = []
for p in range(N):
new_dict[p] = lst
然而,这new_dict[p]
每个new_dict[p]
绑定到相同的lst
,因为p=0,...,N
,即每个new_dict[p]
值引用相同的list
实例。
您必须为每个new_dict[p]
生成新列表。
以下是您应该如何生成它:
new_dict = {}
for p in range(N):
new_dict[p] = [0, 0, 0, 0, 0, 0, 0]
填充字典后,您可以用一行对其进行编辑:
new_dict['a1']['RoadType'][0] = 5
只需单独更新a2
:
new_dict = {
'a1': {'Road_Type': [0,0,0,0,0,0,0]},
'a2': {'Road_Type': [0,0,0,0,0,0,0]},
'a3': {'Road_Type': [0,0,0,0,0,0,0]},
'a4': {'Road_Type': [0,0,0,0,0,0,0]},
}
new_dict['a2']['Road_Type'][0] = 5
print(new_dict)
输出:
{'a1': {'Road_Type': [0, 0, 0, 0, 0, 0, 0]},
'a2': {'Road_Type': [5, 0, 0, 0, 0, 0, 0]},
'a3': {'Road_Type': [0, 0, 0, 0, 0, 0, 0]},
'a4': {'Road_Type': [0, 0, 0, 0, 0, 0, 0]}}
试试这个代码:你的代码有一个小错误,你更新'a2'的索引0,然后分配指向'a2'的new_dict的'kk'
程序 :
new_dict = {}
new_dict['a1']= {'Road_Type': [0,0,0,0,0,0,0]}
new_dict['a2']= {'Road_Type': [0,0,0,0,0,0,0]}
new_dict['a3']= {'Road_Type': [0,0,0,0,0,0,0]}
new_dict['a4']= {'Road_Type': [0,0,0,0,0,0,0]}
kk=new_dict['a2']['Road_Type']
kk[0]=5
print(new_dict)
输出 :
{'a2': {'Road_Type': [5, 0, 0, 0, 0, 0, 0]},
'a3': {'Road_Type': [0, 0, 0, 0, 0, 0, 0]},
'a4': {'Road_Type': [0, 0, 0, 0, 0, 0, 0]},
'a1': {'Road_Type': [0, 0, 0, 0, 0, 0, 0]}}
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