python - 在每次出现以特定子字符串开头的元素时将列表拆分为子列表

Question

我有一个包含一堆字符串的大列表。我需要将原始列表的元素排序为嵌套列表，由它们在列表中的位置确定。换句话说，我需要将原始列表分解为子列表，其中每个子列表包含位于以“ABC”开头的元素之间的所有元素，然后将它们连接在一起作为嵌套列表。

所以原来的名单是：

all_results = ['ABCAccount', 'def = 0', 'gg = 0', 'kec = 0', 'tend = 1234567890', 'ert = abc', 'sed = target', 'id = sadfefsd3g3g24b24b', 'ABCAccount', 'def = 0', 'gg = 0', 'kec = 0', 'tend = NA', 'ert = abc', 'sed = source', 'id = sadfefsd3g3g24b24b', 'ABCAdditional', 'addkey = weds', 'addvalue = false', 'ert = abc', 'sed = target', 'id = sadfefsd3g3g24b24b', 'time_zone = EDT’]

我需要返回：

split_results = [['ABCAccount','def = 0', 'gg = 0', 'kec = 0', 'tend = 1234567890', 'ert = abc', 'sed = target', 'id = sadfefsd3g3g24b24b'],['ABCAccount', 'def = 0', 'gg = 0', 'kec = 0', 'tend = NA', 'ert = abc', 'sed = source', 'id = sadfefsd3g3g24b24b'],['ABCAdditional', 'addkey = weds', 'addvalue = false', 'ert = abc', 'sed = target', 'id = sadfefsd3g3g24b24b', 'time_zone = EDT’]]

我尝试了以下方法：

split_results = [l.split(',') for l in ','.join(all_results).split('ABC')]

标签： pythonpython-2.7listformattingnested-lists

Answer 1

您可以直接从原始列表中工作：

def make_split( lst ):
    if len(lst) == 0:
        return []
    r0 = []
    r1 = []
    for s in lst:
        if s.startswith("ABC"):
            if r1:
                r0.append(r1)
                r1 = []
        r1.append(s)
    return r0 + [r1]