如何在Python中将数组（二维列表）分配给字典？

Question

我想将一个数组（列表中的列表）分配给一个字典。

为此，我已经尝试了fromkeys()方法。但这并不像我期望的那样工作。

# a list with 3 string elements
list_of_Strings = ['index', 'substance', 'container']
# a list with 3 lists, each list has a few values
index_list = [0,1,2]
substance_list = ['water','oil','acetone']
container_list = ['B100','B200','B300']
list_of_lists = [index_list, substance_list, container_list]

# this line sets all for all dict keys the value to 0
dict_dummy_1 = dict.fromkeys(list_of_Strings,0)

# when I try this line, in every "key value" are all 3 lists from list_of_lists:
# Key -> Value
# index -> index_list, substance_list, container_list
# substance -> index_list, substance_list, container_list
# container -> index_list, substance_list, container_list
dict_dummy_1 = dict.fromkeys(list_of_Strings,list_of_lists)

# but I want to have it like this:
# Key -> Value
# index -> index_list
# substance -> substance_list
# container -> container_list

# with a loop it is possible:
dict_dummy_2 = dict.fromkeys(list_of_Strings,0)
for i in range(len(list_of_Strings)):
    dict_dummy_2[list_of_Strings[i]] = list_of_lists[i]

有没有办法避免循环解决方案？有没有相应的函数呢？ 非常感谢您的提前帮助。 :)

- DerDressing

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你能添加你的最终输出吗？ - sahasrara62

Answer 1

以下是使用zip函数的简单一行代码：

dict(zip(list_of_Strings,list_of_lists))

- yukashima huksay

Answer 2

您可以使用字典推导式：

{list_of_Strings[i]: list_of_lists[i] for i in range(len(list_of_Strings))}

- Carsten

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{key: value for key, value in zip(list_of_Strings, list_of_lists)}： - h4z3

Answer 3

# a list with 3 string elements
list_of_Strings = ['index', 'substance', 'container']
# a list with 3 lists, each list has a few values
index_list = [0,1,2]
substance_list = ['water','oil','acetone']
container_list = ['B100','B200','B300']
list_of_lists = [index_list, substance_list, container_list]


res = {i:j for i,j in zip(list_of_Strings, list_of_lists)}

print(res)

输出

{'index': [0, 1, 2],
 'substance': ['water', 'oil', 'acetone'],
 'container': ['B100', 'B200', 'B300']}

- sahasrara62