从列表创建列表字典

Question

I can create a dict of lists via enumerate() as follows:

def turn_to_dict(*args):
    return {i: v for i, v in enumerate(args)}


lst1 = [1, 2, 3, 4]
lst2 = [3, 4, 6, 7]
lst3 = [5, 8, 9]


x = turn_to_dict(lst1, lst2, lst3)

print(x)

Output:

{0: [1, 2, 3, 4], 
 1: [3, 4, 6, 7], 
 2: [5, 8, 9]
 }

I want the same thing with one change: I want the keys to be the elements of a list:

lst1 = ['a', 'b', 'c']
lst2 = [1, 2, 3, 4]
lst3 = [3, 4, 6, 7]
lst4 = [5, 8, 9]


def turn_to_dict(lst, *args):
    for k in lst:
        return {k: v for v in args}


x = turn_to_dict(lst1, lst2, lst3, lst4)

print(x)

I am getting:

{'a': [5, 8, 9]}

What I want is:

{a: [1, 2, 3, 4], 
 b: [3, 4, 6, 7], 
 c: [5, 8, 9]
 } 

I get no error, just not the output I thought I should be getting.

Answer 1

You need to zip your key-list together with the values:

def turn_to_dict(lst, *args):
    return {k: v for k, v in zip(lst, args)}

This pairs up the elements in lst with each of the lists in args .

Not that you need to use a dictionary comprehension at all here, dict() accepts an iterable of key-value pairs directly:

return dict(zip(lst, args))

You'd only need a dictionary comprehension if either the key or value expressions were more than just a reference to the first and second elements of a 2-value tuple.

Demo:

>>> lst1 = ['a', 'b', 'c']
>>> lst2 = [1, 2, 3, 4]
>>> lst3 = [3, 4, 6, 7]
>>> lst4 = [5, 8, 9]
>>> args = (lst2, lst3, lst4)
>>> {k: v for k, v in zip(lst1, args)}
{'a': [1, 2, 3, 4], 'b': [3, 4, 6, 7], 'c': [5, 8, 9]}
>>> dict(zip(lst1, args))
{'a': [1, 2, 3, 4], 'b': [3, 4, 6, 7], 'c': [5, 8, 9]}

Answer 2

You don't need all this :

dict(zip(lst1, [lst2,lst3, lst4]))

using function :

def turn_to_dict(lst, *args):
    return dict(zip(lst, args))

In :

lst1 = ['a', 'b', 'c']
lst2 = [1, 2, 3, 4]
lst3 = [3, 4, 6, 7]
lst4 = [5, 8, 9]

Output :

>>> turn_to_dict(lst1,lst2,lst3,lst4)
{'a': [1, 2, 3, 4], 'c': [5, 8, 9], 'b': [3, 4, 6, 7]}