用多个条件对元组列表进行排序

由 hrerum发布于 2020-04-09 06:05:46

我有一个k元素的元组列表。我想对元素0进行排序，然后对元素1进行排序，依此类推。我用谷歌搜索，但我仍然不知道该怎么做。会是这样吗？

list.sort(key = lambda x : (x[0], x[1], ...., x[k-1])

特别是，我想使用不同的条件进行排序，例如，从元素0降序，从元素1升序，依此类推。

海绵宝宝

2020-04-09 06:05:47

Since python's sort is stable for versions after 2.2 (or perhaps 2.3), the easiest implementation I can think of is a serial repetition of sort using a series of index, reverse_value tuples:

# Specify the index, and whether reverse should be True/False
sort_spec = ((0, True), (1, False), (2, False), (3, True))

# Sort repeatedly from last tuple to the first, to have final output be
# sorted by first tuple, and ties sorted by second tuple etc
for index, reverse_value in sort_spec[::-1]:
    list_of_tuples.sort(key = lambda x: x[index], reverse=reverse_value)

这会进行多次，因此就恒定的时间成本而言可能效率低下，但就渐进复杂度而言仍为O（nlogn）。

If the sort order for indices is truly 0, 1... n-1, n for a list of n-sized tuples as shown in your example, then all you need is a sequence of True and False to denote whether you want reverse or not, and you can use enumerate to add the index.

sort_spec = (True, False, False, True)
for index, reverse_value in list(enumerate(sort_spec))[::-1]:
    list_of_tuples.sort(key = lambda x: x[index], reverse=reverse_value)

虽然原始代码允许按索引的任何顺序进行排序的灵活性。

Incidentally, this "sequence of sorts" method is recommended in the Python Sorting HOWTO with minor modifications.

编辑如果您不需要按某些索引升序排序，而按其他索引降序排序，则

from operator import itemgetter
list_of_tuples.sort(key = itemgetter(1, 3, 5))

将按索引1排序，然后按索引3排序关系，按索引5进一步排序关系。但是，更改每个索引的升序/降序并不容易。

pnon

2020-04-09 06:05:47

因此，我假设您要对tuple_0进行升序排序，然后对tuple_1进行降序排序，依此类推。有点冗长，但这是您可能正在寻找的：

ctr = 0
for i in range(list_of_tuples):
    if ctr%2 == 0:
        list_of_tuples[0] = sorted(list_of_tuples[0])
    else:
        list_of_tuples[0] = sorted(list_of_tuples[0], reverse=True)
    ctr+=1

print list_of_tuples

钱海秋

2020-04-09 06:05:47

 list.sort(key = lambda x : (x[0], x[1], ...., x[k-1])

This is actually using the tuple as its own sort key. In other words, the same thing as calling sort() with no argument.

如果我假设您简化了问题，并且实际元素实际上并不是按照您想要的排序顺序排序（例如，最后一个值的优先级最高），则可以使用相同的技术，但是可以对排序基于优先级的密钥：

 list.sort(key = lambda x : (x[k-1], x[1], ...., x[0])

通常，这是一个非常方便的技巧，即使在C ++之类的其他语言中（如果您使用的是库）：当您要按优先级不同的多个成员对对象列表进行排序时，也可以通过创建一个按照优先级顺序包含所有相关成员的元组。

Final trick (this one is off topic, but it may help you at some point): When using a library that doesn't support the idea of "sort by" keys, you can usually get the same effect by building a list that contains the sort-key. So, instead of sorting a list of Obj, you would construct then sort a list of tuples: (ObjSortKey, Obj). Also, just inserting the objects into a sorted set will work, if they sort key is unique. (The sort key would be the index, in that case.)

hrerum

这家伙很懒，什么都没留下

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