Re: ListPlot: corner problem ??
- To: mathgroup at smc.vnet.net
- Subject: [mg19060] Re: ListPlot: corner problem ??
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Thu, 5 Aug 1999 01:34:50 -0400
- References: <7o5fq5$rhj@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Drago, The range of the axes that is shown, but not the axes origin, is determine by PlotRange Its safer to keep the AxesOrigin seq = Table[1/n, {n, 4}];; ListPlot [seq, AxesOrigin -> {0, 0}, PlotRange -> {{-2, 4}, {-2, 1}}] Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 Drago Ganic <drago.ganic at in2.hr> wrote in message news:7o5fq5$rhj at smc.vnet.net... > Hi there, > > When I tried to plot a sequence ... > > seq = Table[1/n, {n, 4}]; > ListPlot [seq, AxesOrigin ->{0,0}] > > .. I didn't saw the lower left corner of the picture. Why ? > I use Mathematica 3.0 and run it from the CD. > > I used AxesOrigin because I didn't become the desired picture with: > ListPlot[seq] > > > I can get rid of the "corner-problem" if I use PlotRange: > > ListPlot > [ > seq, > PlotRange->{{0,Length[seq]},{0,1}} > ] > > It's interesting that I don't have to use AxesOrigin now to becamo the > "desired" picture. > > > > > >