Python的扁平列表（但不是所有的方式）

Question

我列出的清單列表...

A = [ [[1,3]], [[3,5], [4,4], [[5,3]]] ] 

以下功能輸出[1, 3, 3, 5, 4, 4, 5, 3]

def flatten(a): 
    b = [] 
    for c in a: 
     if isinstance(c, list): 
      b.extend(flatten(c)) 
     else: 
      b.append(c) 
    return b 

不過，我想停下來壓扁在最後一級，所以我得到[ [1,3], [3,5], [4,4], [5,3] ]

Answer 1

你可以在扁平化之前測試包含列表：

def flatten(a): 
    b = [] 
    for c in a: 
     if isinstance(c, list) and any(isinstance(i, list) for i in c): 
      b.extend(flatten(c)) 
     else: 
      b.append(c) 
    return b 

演示：

>>> def flatten(a): 
...  b = [] 
...  for c in a: 
...   if isinstance(c, list) and any(isinstance(i, list) for i in c): 
...    b.extend(flatten(c)) 
...   else: 
...    b.append(c) 
...  return b 
... 
>>> A = [ [[1,3]], [[3,5], [4,4], [[5,3]]] ] 
>>> flatten(A) 
[[1, 3], [3, 5], [4, 4], [5, 3]] 

此嘗試是儘可能高效能的情況下; any()只需要測試，直到找到列表，而不是所有元素。

Answer 2

A = [ [[1,3]], [[3,5], [4,4], [[5,3]]] ] 

print [child[0] if isinstance(child[0], list) else child for item in A for child in item] 

輸出

[[1, 3], [3, 5], [4, 4], [5, 3]] 

注：該解決方案是隻用於這個問題。這不是一個通用列表扁平解決方案。

同樣的想法，與itertools.chain

from itertools import chain 
print [item[0] if isinstance(child[0], list) else item for item in chain(*A)] 

輸出

[[1, 3], [3, 5], [4, 4], [5, 3]]