Question

Possible Duplicate:
Flatten (an irregular) list of lists in Python

Example 1:

Lets say I have a list: [[1,2], [3,4]]. I could use two for loops to be able to print out: 1, 2, 3, 4.

Example 2:

So, now let’s suppose that I’m given an output and I don’t know how many nested lists there within list1:

list1 = [1, [1, 2, [3, 5, 6,[ .. ], …, ] ] ] ] ]

So, my question is how would I be able to print out every individual number in the same format as the first example. I’m working with something right now that gives me nested lists as a result, but different inputs to the function will give me different amount of nested lists.

What I can think of is to do this, but I don’t know what to do after the isinstance part:

c = 0
for i in list1:
   while c < len(list1):
         if isinstance(i, list):

         else:
              print i
         c += 1

Thanks

First Edit

If there is also a way to deconstruct all the nested lists to become a single list that would work as well for me, but I’m curious to know answers to both these problems.

Answer 1

The itertools documentation has some pretty good examples of iterating over lists and such, so it’s always a good place to start when faced with such a task.

I would recomend using a generator, which avoid creating many levels of lists:

def flatten_all(iterable):
    for elem in iterable:
        if not isinstance(elem, list):
            yield elem
        else:
            for x in flatten_all(elem):
                yield x
            # in Python 3.3 just: yield from flatten_all(elem)

Application:

for x in flatten_all([1, [2, [3]]]):
    print(x)

# or if you need a list:
my_lst = list(flatten_all([1, [2, [3]]])
assert my_lst == [1, 2, 3]

Edit: Non-recursive linear version

def flatten_all(iterable):
    stack = [iter(iterable)]
    while stack:
        try:
            elem  = stack[-1].next()
            if not isinstance(elem, list):
                yield elem
            else:
                stack.append(iter(elem))
        except StopIteration:
            stack.pop()