Question

我正在使用代码，我有一个列表：

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' ],
)

我需要得到：

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'],
)

请帮助我，我已经赌了。谢谢你的时间。

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

有点难看，但是......

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

经过测试和工作

如何在python中使用逗号分隔列解开列表？

4 个答案: