PYTHON合并列表中的单个元素与子列表

Question

I try to make flat list. Now I have list:

L=['aa',['bb','cc']]

and I try:

L=['aa',['bb','cc']]
new=[]

for i in L:
  print i
  new+=i

print new

and I got:

'aa'

['bb','cc']

['a','a','bb','cc']

Why in print i=0 = 'aa' and in new+=ii=0 is only 'a'?

How i could get list ['aa','bb','cc'] ?

Answer 1

In general, meaning when you don't know the depth of the original list, this should work:

L=['aa',['bb','cc', ['dd', 'ee']], 'ff']
new = []
for l_item in L:
    stack = [ l_item ]
    while stack:
        s_item = stack.pop(0)
        if isinstance(s_item, list):
            stack += [ x for x in s_item ]
        else:
            new.append(s_item)
print new 

This gives:

['aa', 'bb', 'cc', 'dd', 'ee', 'ff']

Answer 2

Well, don't forget that strings are iterable in Python.

>>> new = []
>>> new += 'aa'
>>> print new
['a', 'a']

To be sure of adding what you want, you can proceed this way:

>>> L = ['aa',['bb','cc']]
>>> new = []

>>> for e in L:
...     new.extend(e if type(e) == list else (e,))
>>> print new
['aa', 'bb', 'cc']

Seriously,

---

PS You can look at this post ... for more information.

Answer 3

This happens because you iterate over 'aa' , basically treating it like it was ['a', 'a'] .

If you want to avoid iterating over strings, you can look at the type:

for i in L:
    if isinstance(i, list):
        new += i
    else:
        new.append(i)

See this question for more details and how to do it recursively:

Flatten (an irregular) list of lists