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[英]Suppose I have a dictionary. How do I strip out all the keys? Edit: This is a nested dictionary
[英]I have nested lists as entries to a Python dictionary. How do I print the nested list whose first elements is “S”?
我有一个字典,看起来像这样:
my_dict = {(1,0): ['A', 'B'],
(1,1): [[['E'], [['A', 'B'], ['C', 'D']]]],
(1,2): [],
(2,1): [[['E'], [['A', 'B'], ['C', 'F']]], [['S'], [[[['E'], [['A', 'B'], ['C', 'D']]]], [[['G'], [['H', 'J'], [[['E'], [['A', 'B'], ['C', 'F']]]]]]]]]]
}
如何检索我发现的以['S']开头的第一个子列表? 对于以上示例,我希望获得:
answer = [['S'], [[[['E'], [['A', 'B'], ['C', 'D']]]], [[['G'], [['H', 'J'], [[['E'], [['A', 'B'], ['C', 'F']]]]]]]]]
我不知道“ S”嵌套的深度。
编辑:
我尝试如下递归执行:
def recursive_check(longlist):
for a_list in longlist:
if 'S' in a_lis:
return a_lis
else:
rec_check(a_list)
我收到以下错误:
RuntimeError:超过最大递归深度
编辑:列表可能看起来不同,每次都嵌套不同。
def first_s(lst):
if not (isinstance(lst, list) and lst):
return None
if lst[0] == ['S']:
return lst
else:
for x in lst:
y = first_s(x)
if y:
return y
使用您的my_dict
:
>>> print first_s(my_dict.values())
[['S'], [[[['E'], [['A', 'B'], ['C', 'D']]]], [[['G'], [['H', 'J'], [[['E'], [['A', 'B'], ['C', 'F']]]]]]]]]
获取该列表: answer = my_dict[(2,1)][1]
它首先使用键(2, 1)
获取字典值,然后获取该值(这是一个列表)并在索引1处获取其项,这是列表中的第二项(在0之后)。
>>> my_dict = {(1,0): ['A', 'B'],
... (1,1): [[['E'], [['A', 'B'], ['C', 'D']]]],
... (1,2): [],
... (2,1): [[['E'], [['A', 'B'], ['C', 'F']]], [['S'], [[[['E'], [['A', 'B'], ['C', 'D']]]], [[['G'], [['H', 'J'], [[['E'], [['A', 'B'], ['C', 'F']]]]]]]]]]
... }
>>> my_dict[(2,1)][1]
[['S'],
[[[['E'], [['A', 'B'], ['C', 'D']]]],
[[['G'], [['H', 'J'], [[['E'], [['A', 'B'], ['C', 'F']]]]]]]]]
(顺便说一句,在你的例子中,你在(1,2): []
之后错过了一个逗号(1,2): []
...更新:现在已修复:))
您可以打印像my_dict[(2,1)][1][0]
这样的元素。
def nest(a,x): m=False for i in a: if type(i)==list: m=nest(i,x) else: if i==x: return True return m def our_fun(my_dict): for i in my_dict: for j in my_dict[i]: if nest(j,'S')==True: return my_dict[i]
我递归检查了“ S”。
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