如何在python多个列表中找到给定值的索引？

Question

下面给出了一个示例多个列表，我正在与之合作。现在我需要找到给定值的indices，如果它出现在列表中的任何地方。对于lis[0][2][2][1][0] which is 'span 1 2'，我想把索引作为[0,2,2,1,0]。

lis = [['Root', ['span 1 6'], ['Nucleus', ['span 1 3'], ['Nucleus', ['span 1 2'],  ['Nucleus', ['leaf 1'],  ['text',  "Specific knowledge "]], ['Satellite', ['leaf 2'], ['rel2par Elaboration'], ['text', 'required !_']]], ['Satellite', ['leaf 3'], ['rel2par Elaboration'], ['text', 'The tester is ']]], ['Satellite', ['span 4 6'], ['rel2par Elaboration'], ['Satellite', ['leaf 4'], ['rel2par Attribution'], ['text', 'For instance , the tester is aware!_']], ['Nucleus', ['span 5 6'],  ['Nucleus', ['leaf 5'], ['rel2par Contrast'], ['text', 'that a!_']], ['Nucleus', ['leaf 6'], ['rel2par Contrast'], ['text', 'but ']]]]]]

我尝试了以下方法(从web源代码修改)。

from copy import copy
def scope2(word, list, indexes = None):
    flag = 0
    result = []
    if not indexes:
        indexes = []
    for index, item in enumerate(list):
        try:
            current_index = indexes + [index]
            result.append(current_index + [item.index(word)])
        except ValueError:
            pass
        print item
        print str(current_index) + ":::::" + str(list.index(item))
        for stuff in item:        
            if type(stuff) == type([]):
                flag =1 

    if flag==1:
        indexes.append(index)
        result.extend(scope2(word, item, copy(indexes)))

return result

这里的问题是，同级索引(相同级别的列表)也会被返回，但并不总是这样。一些样本输出如下

for 0,2,3,1 it returns 0,2,3,1,0, similarly for lis[0][2][3][4][3][3] it returns 0,2,3,3,4,3,3等等。可能的问题是什么？

Answer 1

>>> def trail(word, lst):
...     if word in lst:
...         return [lst.index(word)]
...     for i, x in enumerate(lst):
...         if not isinstance(x, list):
...             continue
...         ret = trail(word, x)
...         if ret is not None:
...             return [i] + ret
... 
>>> trail('span 1 2', lis)
[0, 2, 2, 1, 0]
>>> lis[0][2][2][1][0]
'span 1 2'
>>> trail('no such string', lis)
>>> 

Answer 2

这里是深度优先和广度优先搜索的实现，不限于字符串。需要稍加修改才能搜索列表或元组。

>>> l = [['Root', ['span 1 6'], ['Nucleus', ['span 1 3'], ['Nucleus', ['span 1 2'],  ['Nucleus', ['leaf 1'],  ['text',  "Specific knowledge "]], ['Satellite', ['leaf 2'], ['rel2par Elaboration'], ['text', 'required !_']]], ['Satellite', ['leaf 3'], ['rel2par Elaboration'], ['text', 'The tester is ']]], ['Satellite', ['span 4 6'], ['rel2par Elaboration'], ['Satellite', ['leaf 4'], ['rel2par Attribution'], ['text', 'For instance , the tester is aware!_']], ['Nucleus', ['span 5 6'],  ['Nucleus', ['leaf 5'], ['rel2par Contrast'], ['text', 'that a!_']], ['Nucleus', ['leaf 6'], ['rel2par Contrast'], ['text', 'but ']]]]]]
>>> def depth_first(term,data):
...   for i, item in enumerate(data):
...     if isinstance(item,Sequence) and not isinstance(item,basestring):
...       r = depth_first(term,item)
...       if not r is None:
...         return [i] + r
...     else:
...       if item == term:
...         return [i]
...
>>> def breadth_first(term,data):
...   later = []
...   for i, item in enumerate(data):
...     if isinstance(item,Sequence) and not isinstance(item,basestring):
...       later.append((i,item))
...     else:
...       if item == term:
...         return [i]
...   for i, item in later:
...     r = breadth_first(term,item)
...     if not r is None:
...       return [i] + r
>>> depth_first('span 1 2',l)
[0, 2, 2, 1, 0]
>>> breadth_first('span 1 2',l)
[0, 2, 2, 1, 0]