[Python] Outermost Brackets

[ki4jgt]

[[a][b]][[a][b][c]]

If you can't tell, there are two sets of outer brackets. How would I place the contents of ONLY the two outermost brackets into a list, or a couple of strings?

[NovaAesa]

I'm having trouble understanding your question. Is "[[a][b]][[a][b][c]]" being input as a string or in a list or what? Could you give an example of input and corresponding output?

[MadCow108]

you seem to be looking for the flatten operation, which is not available natively in python
the closest thing available I know is itertools.chain.from_iterable:

In [3]: [i for i in itertools.chain.from_iterable([[1],["ab"],[3,3]])]
Out[3]: [1, 'ab', 3, 3]
http://docs.python.org/library/itertools.html

[Lux Perpetua]

You do this using a counter: it starts at zero, and every time you see a "[", you increment the counter, and for every "]", you decrement it.


[ [ a ] [ b ] ] [ [ a ] [ b ] [ c ] ]
0 1 2 1 2 1 0 1 2 1 2 1 2 1 0

(Notice that the two pieces you want of the top row are demarcated by the zeros in the bottom row.)

[ki4jgt]

You do this using a counter: it starts at zero, and every time you see a "[", you increment the counter, and for every "]", you decrement it.


[ [ a ] [ b ] ] [ [ a ] [ b ] [ c ] ]
0 1 2 1 2 1 0 1 2 1 2 1 2 1 0

(Notice that the two pieces you want of the top row are demarcated by the zeros in the bottom row.)

I actually thought of this one before I posted this, but I ran into some kind of problem where the going up and coming down intermingled with each other and the entire idea was a fail in my head. I don't exactly know how it went now, so maybe I'll give it another shot.

EDIT: I know it had something to do with the fact that the 1 was on the start bracket and 0 was on the close but anyway.

[nvteighen]

A weird Lisp-ish Python implementation for checking whether brackets are balanced or not:




def check_brackets(string, open_bracket = '[', close_bracket = ']'):
# Yay, Lisp-like Python: recursion and lexical closures in a
# "named-let"-like idiom!

def recurse_string_for_brackets(string, counter):
if (string == "") and (counter == 0):
return True # Match made in heaven!
elif (string == "") and (counter != 0):
return False # Mismatch

head = string[0]
tail = string[1:]

if head == open_bracket:
counter += 1 # Breaking FP-style...
elif head == close_bracket:
counter -= 1

return recurse_string_for_brackets(tail, counter)


return recurse_string_for_brackets(string, 0)


EDIT: Great... I completely missed the thread's topic. Forget this post.