Question

我有一个元组列表，像这样：

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

我想将其转换为字典，使其看起来像这样：

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

我试图像这样枚举lst_of_tpls：

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

但这会将我想在“一个”中看到的值也放入其他“类别”中：

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

此外，我想保持mykeys的灵活性。

Answer 1

您可以两次应用zip来找到正确的配对：

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]
}

Answer 2

您可以传递（key，value）的dict元组，它比使用字典理解的速度快两倍

lst_of_tpls = [(1, "test2", 3, 4), (11, "test12", 13, 14), (21, "test22", 23, 24)]
mykeys = ["ones", "text", "threes", "fours"]
my_dict = dict(zip(mykeys, zip(*lst_of_tpls)))

输出：

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

Profiler示例：

lst_of_tpls = [(1, "test2", 3, 4), (11, "test12", 13, 14), (21, "test22", 23, 24)]
mykeys = ["ones", "text", "threes", "fours"]


def dict_comprehension():
    return {a: list(b) for a, b in zip(mykeys, zip(*lst_of_tpls))}


def dict_generator():
    return dict(zip(mykeys, zip(*lst_of_tpls)))


if __name__ == "__main__":
    import timeit

    funcs = (dict_comprehension, dict_generator)
    for f in funcs:
        result = timeit.timeit(f, number=10000, globals=globals())
        print(f"{f.__name__}: {result:.5f}")


dict_comprehension: 0.05009 
dict_generator: 0.02468

如何将元组列表转换为字典

2 个答案: