基于另一个较短列表的列表排序

你可以用 in 要检测元素是否在short\列表中,请使用三元组返回基于该元素的元组:

>>> long_list = ['y', 'z', 'x', 'c', 'a', 'b']
>>> short_list = ['b', 'c', 'a']
>>> sorted(long_list, key=lambda e: (short_list.index(e),e) if e in short_list  else (len(short_list),e))
['b', 'c', 'a', 'x', 'y', 'z']

(index_of_the_element, element) 属于 (len(short_list), element) 为了实现这种改变。

>>> sorted(long_list, key=lambda e: (short_list.index(e),e) if e in short_list  else (len(short_list),))
['b', 'c', 'a', 'y', 'z', 'x']

def position(value):
    try:
        return short_list.index(value)
    except ValueError:
        return len(short_list)

long_list.sort(key=position)

Sorting is guaranteed to be stable ,所以使用 len(short_list)

我会先在短名单中搜索,必要时在长名单中搜索:

>>> def index(list1, list2):
...  def inner(value):
...   try:
...    return list1.index(value)
...   except ValueError:
...    return list2.index(value)
... 
>>> long_list = ['x', 'y', 'z', 'a', 'b', 'c']
>>> short_list = ['a', 'b', 'c']
>>> long_list.sort(key=index(short_list, long_list))
>>> long_list
['a', 'b', 'c', 'x', 'y', 'z']

pointed out ,此解决方案并不总是有效。将两种解决方案结合起来:

>>> def index(list1, list2):
...  def inner(value, l=len(list1)):
...   try:
...    return list1.index(value)
...   except ValueError:
...    return l
...  return inner
... 
>>> long_list = ['x', 'y', 'z', 'a', 'b', 'c']
>>> short_list = ['a', 'b', 'c', 'y']
>>> sorted(long_list, key=index(short_list, long_list))
['a', 'b', 'c', 'y', 'x', 'z']
>>>