Question

我在守则中工作,我有一份名单:

listA = (
    [ name1 ,  A11,A12,A13 ,  B11,B12,B13 ,  C11,C12,C13 ],
    [ name2 ,  A21,A22 ,  B21,B22 ,  C21,C22 ],
    [ name3 ,  A31,A32,A33,A34,A35 ,  B31,B32,B33,B34,B35 ,  C31,C32,C33,C34,C35  ],
)

我需要:

listA = (
    [ name1 ,  A11 ,  B11 ,  C11 ],
    [ name1 ,  A12 ,  B12 ,  C12 ],
    [ name1 ,  A13 ,  B13 ,  C13 ],
    [ name2 ,  A21 ,  B21 ,  C21 ],
    [ name2 ,  A22 ,  B22 ,  C22 ],
    [ name3 ,  A31 ,  B31 ,  C31 ],
    [ name3 ,  A32 ,  B32 ,  C32 ],
    [ name3 ,  A33 ,  B33 ,  C33 ],
    [ name3 ,  A34 ,  B34 ,  C34 ],
    [ name3 ,  A35 ,  B35 ,  C35 ],
)

please help me, I m staked. Thanks for your time.

Answer 1

list_b = []
for x in list_a:
    i = iter(x)
    name = next(i)
    list_b.extend((name,) + t for t in zip(*(y.split(",") for y in i)))

Answer 2

Kinda ugly, but...

listB = []
for tup in listA:
  tmptup = []
  for elt in tup:
    splt = elt.split( , )
    for n in splt:
      tmptup.append(n)
  listB.append(tmptup)

Answer 3

listA = tuple([name, a, b, c] for (name, aas, bbs, ccs) in listA
                              for (a, b, c) in zip(aas.split( , ), bbs.split( , ), ccs.split( , )))

Answer 4

def uncollapse(L):
    temp = []
    answer = []
    for item in L:
        temp= [item[0]] + [i.split( , ) for i in item]
        for i in range(len(temp[1])):
            answer.append([temp[0]] + [zip(*temp[1:])[i]])
    return answer

测试和工作

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