是否有此代碼一個在線解決方案

Question

我經常使用此代碼unbunch名單和號碼列表|串

def deKnot(someList): 
    l = [] 
    for each in someList: l.extend(each) if (hasattr(each,"__iter__")) else l.append(each) 
    return each 

我希望我可以用一個語法像...

def deKnot(someList): return [extend(each) if (hasattr(each,"__iter__")) else append(each) for each in someList] 

哪一個不行。是否有解決這個問題的單線解決方案？

Answer 1

def deKnot(someList): 
    return [e 
      for each in someList 
      for e in (each if hasattr(each, '__iter__') else [each])] 

打開each爲可迭代，總是循環;而不是.extend()在列表理解中使用嵌套循環。

你可以使用的collections.Iterable而不是屬性檢查：

from collections import Iterable 

def deKnot(someList): 
    return [e 
      for each in someList 
      for e in (each if isinstance(each, Iterable) else [each])]