Re: Riddle with Ordering
- To: mathgroup at smc.vnet.net
- Subject: [mg77934] Re: Riddle with Ordering
- From: nazdrovje at gmail.com
- Date: Wed, 20 Jun 2007 05:28:07 -0400 (EDT)
- References: <firstname.lastname@example.org>
I got this reply by Andrzej Kozlowski, which was very close but not what I intended. ****Begin quote Iit's not much of a riddle for anyone who has been reading carefully this forum for a while as it has turned up more than once. It gives you the permutation which will turn Sort[AnyList] into AnyList. In other words: Sort[AnyList][[Ordering[Ordering[AnyList]]]] == Anylist This means, in particular, that if AnyPermuation is any permutation: Ordering[Ordering[AnyPermuation]]==AnyPermutation Moreover, AnyList does not have to be a list of reals, or even a list of numbers, and its elements need not be distinct. *****End quote The intended answer is that the above construct replaces each number with its rank in the list. This is useful for all kinds of statistics involving ranks, such as Spearman rank correlation (although Mathematica has this already built-in). The reason I required the numbers to be unique is that otherwise equal numbers would get unequal (successive) ranks. Usually, a set of equal numbers is mapped to a set of equal ranks (the average of the ranks that would be obtained for the set by the above method). Naz On Jun 19, 1:06 pm, nazdro... at gmail.com wrote: > Just a gem I discovered with Ordering[ ]. > > Anyone have an idea what > > SomeArrayWithUniqueReals // Ordering // Ordering > > does? > > I like this, especially the double use of Ordering. Answer tomorrow.