您可以过滤
values的
dict中
dictcomprehension，那么
Dataframe完美的作品：
print ({k:v[:min_length] for k,v in data_dict.items()}){'b': [1, 2, 3], 'c': [2, 45, 67], 'a': [1, 2, 3]}df = pd.Dataframe({k:v[:min_length] for k,v in data_dict.items()})print (df) a b c0 1 1 21 2 2 452 3 3 67如果可能的话，一些长度可以小于
min_length添加
Series：
data_dict = {'a': [1,2,3,4], 'b': [1,2], 'c': [2,45,67,93,82,92]}min_length = 3df = pd.Dataframe({k:pd.Series(v[:min_length]) for k,v in data_dict.items()})print (df) a b c0 1 1.0 21 2 2.0 452 3 NaN 67时间 ：
In [355]: %timeit (pd.Dataframe({k:v[:min_length] for k,v in data_dict.items()}))The slowest run took 5.32 times longer than the fastest. This could mean that an intermediate result is being cached.1000 loops, best of 3: 520 µs per loopIn [356]: %timeit (pd.Dataframe({k:pd.Series(v[:min_length]) for k,v in data_dict.items()}))The slowest run took 4.50 times longer than the fastest. This could mean that an intermediate result is being cached.1000 loops, best of 3: 937 µs per loop#Allen's solutionIn [357]: %timeit (pd.Dataframe.from_dict(data_dict,orient='index').T.dropna())1 loop, best of 3: 16.7 s per loop计时代码 ：
np.random.seed(123)L = list('ABCDEFGH')N = 500000min_length = 10000data_dict = {k:np.random.randint(10, size=np.random.randint(N)) for k in L}
