假设所有键始终存在于所有字典中:
ds = [d1, d2]
d = {}
for k in d1.iterkeys():
d[k] = tuple(d[k] for d in ds)
注意:在Python 3.x中,使用以下代码:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = tuple(d[k] for d in ds)
并且如果dic包含numpy数组:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = np.concatenate(list(d[k] for d in ds))