如何使用3个列表在Python中创建嵌套字典

Let us Suppose, I have created 3 lists and I want to create a dictionary for it. e.g.

让我们假设，我创建了3个列表，我想为它创建一个字典。例如

a= ['A', 'B', 'C', 'D']
b =[1, 2, 3, 4]
c = [9, 8, 7, 6]

Now What I want is to create a dictionary like this:

现在我想要的是创建一个这样的字典：

{'A':{'1' :'9'} , 'B':{'2':'8'}, 'C':{'3':'7'} , 'D':{'4':'6'}}

is it possible, Can Someone Help me on this?

是否有可能，有人可以帮助我吗？

7 个解决方案

#1

You can create the dictionary from zip-ed lists and convert the int values to strings - if I understood your question proper

你可以从zip-ed列表创建字典，并将int值转换为字符串 - 如果我理解你的问题

dct = {x: {str(y): str(z)} for x, y, z in zip(a,b,c)}

Output:

输出：

{'A': {'1': '9'}, 'C': {'3': '7'}, 'B': {'2': '8'}, 'D': {'4': '6'}}

#2

a = ['A', 'B', 'C', 'D'] # don't forget the quotation marks
b = [1, 2, 3, 4]
c = [9, 8, 7, 6]

res = dict()
for i, index_a in enumerate(a):
    res[index_a] = {str(b[i]): c[i]}

Edit: Alternatively with list comprehension (mainly for the voters in here, as it's advanced python and harder to understand):

编辑：或者使用列表理解（主要针对这里的选民，因为它是高级python并且更难理解）：

res = dict((a[i], {str(b[i]): c[i]}) for i in range(len(a)))

#3

You can also use map() here:

你也可以在这里使用map（）：

a = ['A', 'B', 'C', 'D']
b = [1, 2, 3, 4]
c = [9, 8, 7, 6]

dct = dict(map(lambda x, y, z : (x, {str(y): str(z)}), a, b, c))

print(dct)

Which outputs:

哪个输出：

{'A': {'1': '9'}, 'B': {'2': '8'}, 'C': {'3': '7'}, 'D': {'4': '6'}}

#4

Assuming what you want is to have a be keys in the outer dictionary, and b and c the key and value element of the inner dicts:

假设你想要的是在外部字典中有一个be键，而b和c是内部dicts的键和值元素：

d = {k: {x: y} for k, x, y in zip(a, b, c)}

Update: However, in your example x and y are strings, so if that's what you want:

更新：但是，在您的示例中，x和y是字符串，所以如果这是您想要的：

d = {k: {str(x): str(y)} for k, x, y in zip(a, b, c)}

#5

{ a[x]: {b[x]: c[x]} for x in range(len(a))}

or if you really mean it:

或者如果你真的是这个意思：

{ a[x]: {str(b[x]): str(c[x])} for x in range(len(a))}

#6

Are you looking for something like this ?

你在找这样的东西吗？

a= ['A', 'B', 'C', 'D']
b =[1, 2, 3, 4]
c = [9, 8, 7, 6]

new_dict={}
set(map(lambda x,y,z:(new_dict.__setitem__(x,{y,z})),a,b,c))
print(new_dict)

output:

输出：

{'D': {4, 6}, 'A': {9, 1}, 'B': {8, 2}, 'C': {3, 7}}

#7

You can try this:

你可以试试这个：

a= ['A', 'B', 'C', 'D']
b =[1, 2, 3, 4]
c = [9, 8, 7, 6]
new_data = dict([[a, dict([map(str, i)])] for a, i in zip(a, zip(b, c))])

Output:

输出：

{'A': {'1': '9'}, 'C': {'3': '7'}, 'B': {'2': '8'}, 'D': {'4': '6'}}

Or

要么

new_data = dict(zip(a, map(lambda x:dict([x]), zip(map(str, b), map(str, c)))))