我有一个包含一堆字符串的大列表。我需要将原始列表中的元素排序为嵌套列表,由它们在列表中的位置决定。换句话说,我需要将原始列表拆分为子列表,其中每个子列表包含以“ABC”开头的元素之间的所有元素,然后将它们作为嵌套列表连接在一起
原来的清单是:
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')]
您可以直接从原始列表工作:
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