Re: Ordering function weird?

*To*: mathgroup at smc.vnet.net*Subject*: [mg82691] Re: Ordering function weird?*From*: "David Park" <djmpark at comcast.net>*Date*: Sun, 28 Oct 2007 04:08:58 -0500 (EST)*References*: <ffv29v$aph$1@smc.vnet.net>

Claus, Read the Help for Ordering carefully. y = {1, 2, 7, 8, 9, 1, 2} Sort[y] Ordering[y] The identity between Ordering and Sort is as follows: Part[y, Ordering[y]] == Sort[y] This spells it out in more detail: Column@MapThread[ Row[{"Element number ", #1, " \[Equal] ", #2, " in Sort[y] is in position ", #3, " \[Equal] ", Part[y, #3], " in y"}] &, {Range[Length[y]], Sort[y], Ordering[y]}] The advantage of Ordering is that you could now reorder another equal length list, say the x list, in the same way that Sort[y] reordered the y list. -- David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ "Claus" <claus.haslauer at web.de> wrote in message news:ffv29v$aph$1 at smc.vnet.net... > Hi, > say I've got two sets of number, x and y, which I want to rank. See the > example below. I totally expect and want the result of Ordering[x]. But > I neiter understand nor expect the result of Ordering[y]. Both Sort[x] > and Sort[y] are ok. > Can anybody explain to me Ordering[y]? > Thanks, > Claus > > > In[3]:= x = {1, 2, 3, 6, 10, 3, 4} > y = {1, 2, 7, 8, 9, 1, 2} > > Out[3]= {1, 2, 3, 6, 10, 3, 4} > > Out[4]= {1, 2, 7, 8, 9, 1, 2} > > In[5]:= Sort[x] > Sort[y] > > Out[5]= {1, 2, 3, 3, 4, 6, 10} > > Out[6]= {1, 1, 2, 2, 7, 8, 9} > > In[7]:= Ordering[x] > Ordering[y] > > Out[7]= {1, 2, 3, 6, 7, 4, 5} > > Out[8]= {1, 6, 2, 7, 3, 4, 5} >