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Python在sulbist中搜索项的列表,并返回列表中未找到项的索引_Python_List_Python 2.7_Sublist - Fatal编程技术网

Python在sulbist中搜索项的列表,并返回列表中未找到项的索引

Python在sulbist中搜索项的列表,并返回列表中未找到项的索引,python,list,python-2.7,sublist,Python,List,Python 2.7,Sublist,列表和子列表长度相等,输出的索引应该不匹配,即[3,5]列表的索引号在相应的子列表中找不到我认为正确的答案应该是[3,6]: listA = ['HM', 'GL', 'WD', 'HM', 'WD', 'HM', 'WD'] sublist = [['HM','GL'], ['GL'], ['WD','HM'], ['WD','GL'], ['WD'], ['HM','WD'], ['HM']] 印刷品: listA = ['HM', 'GL', 'WD', 'HM', 'WD', 'HM

列表
子列表
长度相等,输出的索引应该不匹配,即
[3,5]
列表的索引号在相应的
子列表
中找不到
我认为正确的答案应该是
[3,6]

listA = ['HM', 'GL', 'WD', 'HM', 'WD', 'HM', 'WD']

sublist = [['HM','GL'], ['GL'], ['WD','HM'], ['WD','GL'], ['WD'], ['HM','WD'], ['HM']]
印刷品:

listA = ['HM', 'GL', 'WD', 'HM', 'WD', 'HM', 'WD']
sublist = [['HM','GL'], ['GL'], ['WD','HM'], ['WD','GL'], ['WD'], ['HM','WD'], ['HM']]

print([i for i, (a, b) in enumerate(zip(listA, sublist)) if not a in b])

我认为正确的答案应该是
[3,6]

listA = ['HM', 'GL', 'WD', 'HM', 'WD', 'HM', 'WD']

sublist = [['HM','GL'], ['GL'], ['WD','HM'], ['WD','GL'], ['WD'], ['HM','WD'], ['HM']]
印刷品:

listA = ['HM', 'GL', 'WD', 'HM', 'WD', 'HM', 'WD']
sublist = [['HM','GL'], ['GL'], ['WD','HM'], ['WD','GL'], ['WD'], ['HM','WD'], ['HM']]

print([i for i, (a, b) in enumerate(zip(listA, sublist)) if not a in b])