展平包含元组，字母和整数列表的列表

Question

I need to flatten a list recursively:

Before the list looks like:

L=[1,[2,[‘a’,(3,’b’)]],(5,6),([11,22])] 

After:

Lflat=[1,2,’a’,(3,’b’),(5,6),([11,22])]

I've come across an issue with my code (lst1 is an empty lst1)

def list_flatten(lst,lst1):
    for item in lst:
        if type(item) == tuple:
            print(item)
            lst1.append(item)
        elif type(item) == list:
            list_flatten(item,lst1)
        else:
            lst1.append(item)
    return lst1

This returns the following:

OUTPUT : [1, 2, 'a', (3, 'b'), (5, 6), 11, 22]

Which lead me to find out that ([]) is considered a list, not a tuple.

Now my questions are as following:

1. Say I were to define lst1=[] inside the main program. How do I make it so the recursion doesn't empty the list out every iteration?
2. Why is ([]) considered a list?

Answer 1

Your list_flatten function mutates the lst1 argument, so you don't really need to return anything. You can call it like this:

L = [1,[2,['a',(3,'b')]],(5,6),([11,22])] 

def list_flatten(lst, lst1):
    for item in lst:
        if isinstance(item, list):
            list_flatten(item, lst1)
        else:
            lst1.append(item)

Lflat = []
list_flatten(L, Lflat)
print(Lflat)

output

[1, 2, 'a', (3, 'b'), (5, 6), 11, 22]

It's recommended to use isinstance rather than type because that makes the code more versatile: it will also work with objects derived from list .

We can re-write the function so that you don't need to pass in lst1 :

def list_flatten(lst, lst1=None):
    if lst1 is None:
        lst1 = []
    for item in lst:
        if isinstance(item, list):
            list_flatten(item, lst1)
        else:
            lst1.append(item)
    return lst1

Lflat = list_flatten(L)
print(Lflat)

We give lst1 a default value of None and on the top level of the recursion we re-bind the name lst1 to an empty list to collect the results.

We can't give lst1 a default value of [] . That's because default args are created when the function is compiled, not when the function is called, and if we gave lst1 a default value of [] that same list would get used on every call. It would look like it does what we want the first time we used list_flatten , but it would not behave as desired on subsequent calls. Here's a short demo.

L = [1,[2,['a',(3,'b')]],(5,6),([11,22])] 

def list_flatten(lst, lst1=[]):
    for item in lst:
        if isinstance(item, list):
            list_flatten(item, lst1)
        else:
            lst1.append(item)
    return lst1

Lflat = list_flatten(L)
print(Lflat)
Lflat = list_flatten(L)
print(Lflat)

output

[1, 2, 'a', (3, 'b'), (5, 6), 11, 22]
[1, 2, 'a', (3, 'b'), (5, 6), 11, 22, 1, 2, 'a', (3, 'b'), (5, 6), 11, 22]

As you can see, lst1 has retained its contents from the first call. For more info on this important topic, please see “Least Astonishment” and the Mutable Default Argument . There are times when this behviour is desirable, but in such cases it's wise to add a comment to your code that you're intentionally using a mutable default argument.

---

Yet another way is to make list_flatten into a generator, and collect its output into a list:

def list_flatten(lst):
    for item in lst:
        if isinstance(item, list):
            yield from list_flatten(item)
        else:
            yield item

Lflat = list(list_flatten(L))
print(Lflat)

In recent versions of Python you can replace list(list_flatten(L)) with [*list_flatten(L)] .

Python 2 doesn't have yield from , but you can replace that line with:

for u in list_flatten(item):
    yield u

If you don't actually need the list you can call the generator like this:

for u in list_flatten(L):
    print(u)

output

1
2
a
(3, 'b')
(5, 6)
11
22

Answer 2

You can note that what you need is:

• if the list is empty, return it unchanged
• if it has one single element and that element is not a list, return the list unchanged
• else flatten first element, flatten the end of the list and concatenate the two sublists

In Python code, it leads to:

def flatten(L):
    if len(L) == 0: return L
    elif len(L) == 1 and not isinstance(L[0], list): return L
    else:
        return (flatten(L[0] if isinstance(L[0], list)
                else [L[0]]) + flatten(L[1:]))

it gives as expected:

>>> L = [1, 2, 'a', (3, 'b'), (5, 6), ([11, 22],)]
>>> flatten(L)
[1, 2, 'a', (3, 'b'), (5, 6), ([11, 22],)]