如何将键值元组列表转换成字典？

I have a list that looks like:

[('A', 1), ('B', 2), ('C', 3)]

I want to turn it into a dictionary that looks like:

{'A': 1, 'B': 2, 'C': 3}

What’s the best way to go about this?

EDIT: My list of tuples is actually more like:

[(A, 12937012397), (BERA, 2034927830), (CE, 2349057340)]

This gives me the same error as trying to split the list up and zip it. ValueError: dictionary update sequence element #0 has length 1916; 2 is required

THAT is your actual question.

The answer is that the elements of your list are not what you think they are. If you type myList[0] you will find that the first element of your list is not a two-tuple, e.g. ('A', 1), but rather a 1916-length iterable.

Once you actually have a list in the form you stated in your original question (myList = [('A',1),('B',2),...]), all you need to do is dict(myList).

>>> dict([('A', 1), ('B', 2), ('C', 3)])
{'A': 1, 'C': 3, 'B': 2}

Have you tried this?

>>> l=[('A',1), ('B',2), ('C',3)]
>>> d=dict(l)
>>> d
{'A': 1, 'C': 3, 'B': 2}

Here is a way to handle duplicate tuple “keys”:

# An example
l = [('A', 1), ('B', 2), ('C', 3), ('A', 5), ('D', 0), ('D', 9)]

# A solution
d = dict()
[d [t [0]].append(t [1]) if t [0] in list(d.keys()) 
 else d.update({t [0]: [t [1]]}) for t in l]
d

OUTPUT: {'A': [1, 5], 'B': [2], 'C': [3], 'D': [0, 9]}

Another way using dictionary comprehensions,

>>> t = [('A', 1), ('B', 2), ('C', 3)]
>>> d = { i:j for i,j in t }
>>> d
{'A': 1, 'B': 2, 'C': 3}

If Tuple has no key repetitions, it’s Simple.

tup = [("A",0),("B",3),("C",5)]
dic = dict(tup)
print(dic)

If tuple has key repetitions.

tup = [("A",0),("B",3),("C",5),("A",9),("B",4)]
dic = {}
for i, j in tup:
    dic.setdefault(i,[]).append(j)
print(dic)

l=[['A', 1], ['B', 2], ['C', 3]]
d={}
for i,j in l:
d.setdefault(i,j)
print(d)