I have a dictionary that looks like this:
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']]]]]]]]]]
}
How do I retrieve the first sublist I find that begins with ['S']? For the example above I would like to get:
answer = [['S'], [[[['E'], [['A', 'B'], ['C', 'D']]]], [[['G'], [['H', 'J'], [[['E'], [['A', 'B'], ['C', 'F']]]]]]]]]
I don't know how deep the 'S' is nested.
EDIT:
I attempted to do it recursively as follows:
def recursive_check(longlist):
for a_list in longlist:
if 'S' in a_lis:
return a_lis
else:
rec_check(a_list)
I received the following error:
RuntimeError: maximum recursion depth exceeded
EDIT: The list may look different and be nested differently every time.
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
Using your my_dict
:
>>> print first_s(my_dict.values())
[['S'], [[[['E'], [['A', 'B'], ['C', 'D']]]], [[['G'], [['H', 'J'], [[['E'], [['A', 'B'], ['C', 'F']]]]]]]]]
To get that list: answer = my_dict[(2,1)][1]
It first gets the dictionary value with key of (2, 1)
, then takes that value (which is a list) and gets its item at index 1, which is the second item in the list (after 0).
>>> 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']]]]]]]]]
(By the way, in your example, you are missing a comma after (1,2): []
...Update: which has now been fixed :) )
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]
I checked for 'S' recursively.
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