從一個列表創建可變長度列表

Question

我想從單個列表中創建多個列表。 如果有來自用戶的數據，列表將具有1個值或3個。 但是，由於數據附加到特定用戶，因此必須維護訂單。

例如：

data = ['no data', 'choice 1', 'choice 4', 'choice 2', 'no data', 'choice 1', ...etc]

我希望輸出為：

list1 = ['no data']
list2 = ['choice 1', 'choice4', 'choice 2']
list3 = ['no data']
list4 = ['choice 1'...]

我怎么做到這一點？

Answer 1

不幸的是，如果你需要將非'no-data'運行到3個組中，或者每個'no-data'單個數據塊，那么itertools.groupby將無效。 所以，這是一個拋在一起的東西：

In [40]: def group_lists(data):
    ...:     final = []
    ...:     contains_data = False
    ...:     temp = []
    ...:     for sub in data:
    ...:         if sub == 'no data':
    ...:             if contains_data:
    ...:                 final.append(temp)
    ...:                 temp = []
    ...:                 final.append([sub])
    ...:                 contains_data = False
    ...:             else:
    ...:                 final.append([sub])
    ...:         else:
    ...:             contains_data = True
    ...:             if len(temp) < 3:
    ...:                 temp.append(sub)
    ...:             else:
    ...:                 final.append(temp)
    ...:                 temp = []
    ...:                 contains_data = False
    ...:     return final
    ...:

In [41]: data = ['choice 1', 'choice 2', 'choice 1', 'choice 1', 'choice 4', 'choice 1', 'no data']

In [42]: group_lists(data)
Out[42]: [['choice 1', 'choice 2', 'choice 1'], ['choice 4', 'choice 1'], ['no data']]

In [43]: data2 = ['choice 1', 'choice 2', 'choice 1', 'choice 1', 'choice 4', 'choice 1', 'choice1', 'no data']

In [44]: group_lists(data2)
Out[44]:
[['choice 1', 'choice 2', 'choice 1'],
 ['choice 4', 'choice 1', 'choice1'],
 ['no data']]

原始答案

使用itertools.groupby ：

In [15]: data = ['no data', 'choice 1', 'choice 4', 'choice 2', 'no data', 'choice 1']
    ...:

In [16]: import itertools

In [17]: grouped = [list(g) for _, g in itertools.groupby(data, lambda s: s == 'no data')]

In [18]: grouped[0]
Out[18]: ['no data']

In [19]: grouped[1]
Out[19]: ['choice 1', 'choice 4', 'choice 2']

In [20]: grouped[2]
Out[20]: ['no data']

In [21]: grouped[3]
Out[21]: ['choice 1']

拆開那些相當笨重的單線：

In [26]: lists = []
    ...: for _, g in itertools.groupby(data, lambda s: s == 'no data'):
    ...:     lists.append(list(g))
    ...:

In [27]: lists[0]
Out[27]: ['no data']

In [28]: lists[1]
Out[28]: ['choice 1', 'choice 4', 'choice 2']

In [29]: lists[2]
Out[29]: ['no data']

In [30]: lists[3]
Out[30]: ['choice 1']

因此，結果是列表列表：

In [32]: lists
Out[32]: [['no data'], ['choice 1', 'choice 4', 'choice 2'], ['no data'], ['choice 1']]

Answer 2

這似乎有效並且相當容易理解。 它不會創建單獨的變量，而是創建名為lists 。

data = ['no data', 'choice 1', 'choice 4', 'choice 2', 'no data', 'choice 1', '...etc']

lists = []
tmp = []
for elem in data:
    if elem != 'no data':
        tmp.append(elem)
    else:
        if tmp:
            lists.append(tmp)
        lists.append([elem])
        tmp = []

if tmp:
    lists.append(tmp)

for i, sublist in enumerate(lists, start=1):
    print('list{}: {}'.format(i, sublist))

輸出：

list1: ['no data']
list2: ['choice 1', 'choice 4', 'choice 2']
list3: ['no data']
list4: ['choice 1', '...etc']