我有一个元组列表,如下所示：

lst_of_tpls = [(1, 'test2', 3, 4),(11, 'test12', 13, 14),(21, 'test22', 23,24)]

我想将它转换为字典,使它看起来像这样：

mykeys = ['ones', 'text', 'threes', 'fours']
mydict = {'ones': [1,11,21], 'text':['test2','test12','test22'], 
          'threes': [3,13,23], 'fours':[4,14,24]}

我试图枚举lst_of_tplslike所以：

mydict = dict.fromkeys(mykeys, [])
for count, (ones, text, threes, fours) in enumerate(lst_of_tpls):
    mydict['ones'].append(ones)

但是这使得我希望在’ones’中看到的值也在其他“类别”中：

{'ones': [1, 11, 21], 'text': [1, 11, 21], 'threes': [1, 11, 21], 'fours': [1, 11, 21]}

另外,我想保持我的灵活性.

最佳答案 您可以应用两次拉链以找到正确的配对：

lst_of_tpls = [(1, 'test2', 3, 4),(11, 'test12', 13, 14),(21, 'test22', 23,24)]
mykeys = ['ones', 'text', 'threes', 'fours']
new_d = {a:list(b) for a, b in zip(mykeys, zip(*lst_of_tpls))}

输出：

{
 'ones': [1, 11, 21],
 'text': ['test2', 'test12', 'test22'],
 'threes': [3, 13, 23],
 'fours': [4, 14, 24]
}