Re: v. 6, third argument to rectangle
- To: mathgroup at smc.vnet.net
- Subject: [mg76786] Re: v. 6, third argument to rectangle
- From: Bill Rowe <readnewsciv at sbcglobal.net>
- Date: Sun, 27 May 2007 04:59:51 -0400 (EDT)
On 5/26/07 at 4:19 AM, dimmechan at yahoo.com (dimitris) wrote:
>Since you are interested in putting two graphs side by side what is
>bad with the following
>In[73]:=
>Block[{$DisplayFunction = Identity}, g1 = Plot[o^(1/2), {o, 1, 2},
>PlotStyle -> Red];
>g2 = Plot[Log[o], {o, 1, 2}, PlotStyle -> Blue]; ]
>Show[GraphicsArray[{g1, g2}], ImageSize -> 600]
>It works in version 5.2 and I think there will not be any problem in
>version 6. (confirmation needed!)
Yes, your code will work in version 6 as is. But, this would not
be the preferred way in version 6 since GraphicsArray is
superseded by GraphicsGrid. So, making this substitution your
code would become
Block[{$DisplayFunction = Identity}, g1 = Plot[o^(1/2), {o, 1, 2},
PlotStyle -> Red];
g2 = Plot[Log[o], {o, 1, 2}, PlotStyle -> Blue]; ]
Show[GraphicsGrid[{{g1, g2}}], ImageSize -> 600]
Also, given the way version 6 does graphics the same result can
be obtained more simply with
GraphicsRow[{Plot[o^(1/2), {o, 1, 2}, PlotStyle -> Red],
Plot[Log[o], {o, 1, 2}, PlotStyle -> Blue]}, ImageSize -> 600]
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