[英]How do I convert a list of tuples to a dictionary
我有一個元組列表,如下所示:
lst_of_tpls = [(1, 'test2', 3, 4),(11, 'test12', 13, 14),(21, 'test22', 23,24)]
我想將它轉換為字典,使它看起來像這樣:
mykeys = ['ones', 'text', 'threes', 'fours']
mydict = {'ones': [1,11,21], 'text':['test2','test12','test22'],
'threes': [3,13,23], 'fours':[4,14,24]}
我試圖像這樣枚舉lst_of_tpls
:
mydict = dict.fromkeys(mykeys, [])
for count, (ones, text, threes, fours) in enumerate(lst_of_tpls):
mydict['ones'].append(ones)
但是這使得我希望在'ones'中看到的值也在其他“類別”中:
{'ones': [1, 11, 21], 'text': [1, 11, 21], 'threes': [1, 11, 21], 'fours': [1, 11, 21]}
另外,我想保持mykeys
靈活性。
您可以應用兩次zip
以找到正確的配對:
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))}
輸出:
{
'ones': [1, 11, 21],
'text': ['test2', 'test12', 'test22'],
'threes': [3, 13, 23],
'fours': [4, 14, 24]
}
你可以傳遞給(key,value)的dict元組,它比使用字典理解快兩倍
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)))
輸出:
{'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
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