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.
>
>
>
>
>
>