python中关于深嵌元组转列表的小问题

1.下面是提的一个问题：如何将aaa=[[(‘s1’,)], [(‘s1’,)], [(‘s1’,)], [(‘s1’, ‘s2’)], [(‘s1’, ‘s2’)], [(‘s1’, ‘s2’)], [(‘s1’, ‘s2’)]] 变成bbb= [[‘s1’], [‘s1’], [s1’], [‘s1’,’s2’], [‘s1’, ‘s2’], [‘s1’, ‘s2’], [‘s1’, ‘s2’]]

2.下面一位网友叫乔布斯给出的答案

aaa=[[('s1',)], [('s1',)], [('s1',)], [('s1', 's2')], [('s1', 's2')], [('s1', 's2')], [('s1', 's2')]]
bbb=[]
for each in aaa:
for ec in each:
bbb.append(list(ec))
print bbb

3.下面的我当时写的代码，其中dataSetM就是aaa,返回的就是bbb

def getAprioriSet(apri_dataSetM):
aprioriD = []
# print apri_dataSetM
for apri1 in apri_dataSetM:
kkk=[]
for apri2 in apri1:
apri2 = apri2
if (len(apri2)==1):
kkk.extend(list(apri2))
else:
sss = []
for apri3 in xrange(len(apri2)):
sss.extend([apri2[apri3]])
kkk.extend(sss)
aprioriD.append(kkk)
return aprioriD

4.很明显就是网友给出的代码效率高