用昂贵的比较排序？ [英] sorting with expensive compares?
问题描述
大家好。
Python似乎有一个很好的排序方法，但在排序数组元素时
是非常大，因此有非常昂贵的比较，是否有一些已经可用的排序功能，将像元素合并到
a链中，这样他们就赢了''必须多次重新比较？
谢谢！
Hi folks.
Python appears to have a good sort method, but when sorting array elements
that are very large, and hence have very expensive compares, is there some
sort of already-available sort function that will merge like elements into
a chain, so that they won''t have to be recompared as many times?
Thanks!
推荐答案
Dan Stromberg写道：
Dan Stromberg wrote:
嗨伙计们。
Python似乎有一个很好的排序方法，但在排序数组元素时非常大，因此有非常昂贵的比较，是否有一些已经可用的排序功能，将元素合并到链中，这样他们就不必再多次重新比较？
谢谢！
Hi folks.
Python appears to have a good sort method, but when sorting array elements
that are very large, and hence have very expensive compares, is there some
sort of already-available sort function that will merge like elements into
a chain, so that they won''t have to be recompared as many times?
Thanks!
可能更容易记住cmp（），查看在线食谱或
喜欢这个装饰者
http://mail.python.org/pipermail/pyt...er/303035.html
http://aspn.activestate.com/ASPN/Python/Cookbook/
might be simpler to memoize cmp(), look in online cookbook or
something like this decorator
http://mail.python.org/pipermail/pyt...er/303035.html
http://aspn.activestate.com/ASPN/Python/Cookbook/
Dan Stromberg写道：
Dan Stromberg wrote:
嗨伙计。
Python似乎有一个很好的排序方法，但在排序数组元素时
这是非常大的，因此有非常昂贵的比较，是否有一些已经可用的排序功能，将元素合并到一个链，这样他们就不必重新比较多次？
谢谢！
Hi folks.
Python appears to have a good sort method, but when sorting array elements
that are very large, and hence have very expensive compares, is there some
sort of already-available sort function that will merge like elements into
a chain, so that they won''t have to be recompared as many times?
Thanks!
听起来像DSU时间。
[a] - > [（hash（a），a）]
Sounds like DSU time.
[a] -> [ (hash(a), a) ]
bo **** @ gmail.com 写道：
Dan Stromberg写道：
Dan Stromberg wrote:
嗨伙计。
Python似乎有一个很好的排序方法，但是当排序非常大的数组元素，因此有非常昂贵的比较时，是否有一些已经可用的排序函数将类似元素合并到
链条，这样他们就不必多次重新比较了吗？
谢谢！
Hi folks.
Python appears to have a good sort method, but when sorting array elements
that are very large, and hence have very expensive compares, is there some
sort of already-available sort function that will merge like elements into
a chain, so that they won''t have to be recompared as many times?
Thanks!
听起来像DSU时间。
[a] - > [（hash（a），a）]
Sounds like DSU time.
[a] -> [ (hash(a), a) ]
啊哈！或者：记录数组，例如log base 10或其他一些
单调变换和排列顺序索引
http://aspn.activestate.com/ASPN/Coo.../Recipe/306862
Aha! OR: take a log of the array, e.g. log base 10 or some other
monotonic transform and permutation order indexes
http://aspn.activestate.com/ASPN/Coo.../Recipe/306862
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