Re: Length and Drop
- To: mathgroup at smc.vnet.net
- Subject: [mg32193] Re: Length and Drop
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Sat, 5 Jan 2002 00:10:31 -0500 (EST)
- References: <a13vdg$dgm$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Steve, Drop[timeexp,1] evaluates by the Drop[{{0,0},{7,8}},1 ] {{7,8}} It does nont redefine timeexp. To get this you would have to evaluate timeexp = Drop[timeexp,1] -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Steve Gray" <stevebg at adelphia.net> wrote in message news:a13vdg$dgm$1 at smc.vnet.net... > The following sequence seems wrong to me, in that > after dropping some elements from a list, its Length is > still the same. I copied this directly from the notebook > but removed the In[] and Out[] statements. > > > timexp={{0,0}} -> {{0,0}} [this is ok] > > timexp=Append[timexp,{7,8}] -> {{0,0},{7,8}} [this is ok] > > Length[timexp] -> 2 [this is ok] > > Drop[timexp,1] -> {{7,8}} [this is ok] > > Length[timexp] -> 2 [ this is what I did not expect!] > > Should not Drop reduce the Length of the list by the amount > dropped? > >