将列表项列表转换为dict

如何解决将列表项列表转换为dict

我有这样的列表清单

[[],['1.0','1.0 (proxy)','1.1','1.1 (proxy)','2.0','2.0 (proxy)','3.0','3.0 (proxy)'],['2.0','2.0 (proxy)']]

我想将列表列表转换为dict，以使float值成为键并将对应的字符串转换为其值，因此{'1.0': '1.0 (proxy)' ....}

当我尝试使用izip压缩时，它作为迭代器出现，但是在用dict包装时出现错误。

>>> for item in izip(li):
...  print item
... 
([],)
(['1.0',)
(['2.0','2.0 (proxy)'],)
>>> for item in izip(li):
...  if item:
...   print dict(item)
... 
Traceback (most recent call last):
  File "<stdin>",line 3,in <module>
ValueError: dictionary update sequence element #0 has length 0; 2 is required

解决方法

您可以为每个子列表创建一个迭代器，并一次遍历两个元素。这样，您就可以获取键值对，并且当迭代器用尽时，您将转到下一个子列表。这也会跳过空列表

def create_dicts(lst):
    new_list = []

    for item in lst:
        # create new dict
        d = {}

        # iterator over the sub list
        x = iter(item)
        
        while True:
            try:
                # check if anything is left in the iterator
                key,value = next(x),next(x)
            except StopIteration:
                # if not,leave the while loop
                break
            # set the key and value in the dictionary
            d[key] = value

        # if the dictionary has any elements
        if d:
            new_list.append(d)
        else:
            continue
    return new_list


li = [[],['1.0','1.0 (proxy)','1.1','1.1 (proxy)','2.0','2.0 (proxy)','3.0','3.0 (proxy)'],['2.0','2.0 (proxy)']]

a = create_dicts(li)
print(a)

[{'2.0': '2.0 (proxy)','1.0': '1.0 (proxy)','1.1': '1.1 (proxy)','3.0': '3.0 (proxy)'},{'2.0': '2.0 (proxy)',{'2.0': '2.0 (proxy)'}]