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 But, this would not be the preferred way in version 6 since GraphicsArray is superseded by GraphicsGrid.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] -- To reply via email subtract one hundred and four