提问人:cjg123 提问时间:7/2/2019 最后编辑:cjg123 更新时间:7/2/2019 访问量:271
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
Split list into sublists at every occurrence of element starting with specific substring
问:
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到嵌套列表中,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介于以“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。
所以原来的列表是:
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')]
答:
1赞
Scott Hunter
7/2/2019
#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]
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