"Flatten" list containing lists of lists to list of lists

Question

This question differs from Converting list of lists / nested lists to list of lists without nesting (this yields a very specific set of responses that do not address my situation) and also from the many "flatten list of lists to list" answers out there.

I want to take a list of lists, some of which are in turn lists of lists and "flatten" to a list of lists (not just a list!).

As a concrete example, I want to go from this:

my_list_of_lists = [[1, 2, 3], [[9, 10], [8, 9, 10], [3, 4, 6]], [11, 12, 13]]

to this

target_list_of_lists = [[1, 2, 3], [9, 10], [8, 9, 10], [3, 4, 6], [11, 12, 13]]

(In a visual sense, I want to turn all [[ and ]] inside the outer list into [ and ] respectively.)

I don't think the proposed duplicate is correct, as this question doesn't want to fully flatten the list, just remove some layers of the nesting, while keeping others. — Blckknght
– Blckknght, Commented Jul 21, 2019 at 23:42

Akaisteph7 · Accepted Answer · 2019-07-22 01:49:40Z

4

Here's one way to do it:

def flatten(lst, dh=None):
    # Added so user does not need to pass output list
    if (dh is None):  
        dh = []
    for elem in lst:
        # Will not try to iterate through empty lists
        if elem and (isinstance(elem[0], list)):  
            flatten(elem, dh)
        else:
            dh.append(elem)
    return dh

my_list_of_lists = [[1, 2, 3], [[9, 10], [8, 9, 10], [3, 4, 6]], [11, 12, 13]]
target_list_of_lists = flatten(my_list_of_lists)
print(target_list_of_lists)

Output:

[[1, 2, 3], [9, 10], [8, 9, 10], [3, 4, 6], [11, 12, 13]]

This will work for lists of any length and any depth.

edited Jul 22, 2019 at 1:49

answered Jul 21, 2019 at 23:48

Akaisteph7

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4 Comments

Linuxios Over a year ago

Maybe cleaner to add a default: def flatten(lst, dh = None): #... dh = [] if dh is None else dh and a return dh at the end so that the final user doesn't have to set up the output buffer?

Linuxios Over a year ago

Careful -- you need the ` = None` and the conditional so you start a fresh list with each top level invocation. Otherwise: def test(a = []): a.append(45); return a; test() #=> [45]; test() #=> [45, 45]

Linuxios Over a year ago

:) (We all do... I spent longer than I'd like to admit on one of those the other day...)

Ian Gow Over a year ago

Note that my actual list had [] in the list of lists (for example, my_list_of_lists = [[1, 2, 3], [], [[9, 10], [], [8, 9, 10], [3, 4, 6]], [11, 12, 13]]). This broke the solution, but changing the condition to if elem and (isinstance(elem[0], list)): made it work.

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