在Python中模拟类似Sqlite3的元组排序

我试图对元组列表进行排序，其中某些元组索引不包含任何元组索引。在这种情况下，特定索引位置中无的元组需要优先于元组列表中的其他非无值

例如，如果我有一个如下所示的列表：

[
(3.4, 43, None),
(5345.6, 42, None),
(43.24, 25, None),
(323.4, 433, None),
(5.6, 42, None),
(4.4, 235, None),
(None, None, 'Hello'),
(5.6, None, None),
(None, 235, None),
]

我希望我的最终清单是这样的 （基本上是“按第1列、第3列从表顺序选择*”）

现在，我正在使用

sortedSelectedResult = sorted(selectedResult,key=itemgetter(*indexSortOrderList))

其中，indexSortOrderList是一个列表，其中包含需要对元组列表进行排序的索引顺序列表。但在这种方法中，非索引被随机放置在列表中。我当前的结果如下所示，其中indexSortOrder=[0,2]（即）按第一列排序，然后按第三列排序：

[(3.4, 43, None),
 (5345.6, 42, None),
 (43.24, 25, None),
 (323.4, 433, None),
 (5.6, 42, None),
 (4.4, 235, None),
 (None, None, 'Hello'),
 (5.6, None, None),
 (None, 235, None)]

我尝试查看operator.itemgetter并修改函数以应用于我的用例，

---

但是我也遇到了问题。

您可以添加一个自定义键，将

无

视为0:

sorted(lst, key=lambda x: (isinstance(x[0], str), x[0] or 0,  isinstance(x[2], str) or x[2] or 0))

代码：

lst = [
(3.4, 43, None),
(5345.6, 42, None),
(43.24, 25, None),
(323.4, 433, None),
(5.6, 42, None),
(4.4, 235, None),
(None, None, 'hello'),
(5.6, None, None),
(None, 235, None),
]

print(sorted(lst, key=lambda x: (isinstance(x[0], str), x[0] or 0,  isinstance(x[2], str) or x[2] or 0)))

# [(None, 235, None),
#  (None, None, 'hello'),
#  (3.4, 43, None),
#  (4.4, 235, None),
#  (5.6, 42, None),
#  (5.6, None, None),
#  (43.24, 25, None),
#  (323.4, 433, None),
#  (5345.6, 42, None)]

lst = [
(3.4, 43, None),
(5345.6, 42, None),
(43.24, 25, None),
(323.4, 433, None),
(5.6, 42, None),
(4.4, 235, None),
(None, None, 'hello'),
(5.6, None, None),
(None, 235, None),
]

print(sorted(lst, key=lambda x: (isinstance(x[0], str), x[0] or 0,  isinstance(x[2], str) or x[2] or 0)))

# [(None, 235, None),
#  (None, None, 'hello'),
#  (3.4, 43, None),
#  (4.4, 235, None),
#  (5.6, 42, None),
#  (5.6, None, None),
#  (43.24, 25, None),
#  (323.4, 433, None),
#  (5345.6, 42, None)]