Re: Coloring curves: not as simple as it sounds..
- To: mathgroup at smc.vnet.net
- Subject: [mg118167] Re: Coloring curves: not as simple as it sounds..
- From: Peter Pein <petsie at dordos.net>
- Date: Fri, 15 Apr 2011 03:55:37 -0400 (EDT)
- References: <io6d17$dpl$1@smc.vnet.net>
Am 14.04.2011 10:59, schrieb Jean-Michel Di Nicola:
> All,
>
> I have a question that looks fairly trivial, but I cannot solve it....
> Please help!
>
> Thanks, JM
>
> Here is a simplified toy example.
> When I type
> Plot[{a x /. {a -> 1}, a x^2 /. {a -> 1}, a x^3 /. {a -> 1}}, {x, 0,
> 2}, PlotStyle -> {Red, Green, Blue}]
> I get 3 curves with different colors.
>
> Well, when I type
> Plot[{a x, a x^2, a x^3} /. {a -> 1}, {x, 0, 2},
> PlotStyle -> {Red, Green, Blue}]
> I get 3 curves but they are all blue, WHY?
>
> However, when I evaluate {a x /. {a -> 1}, a x^2 /. {a -> 1}, a x^3 /.
> {a -> 1} and {a x, a x^2, a x^3} /. {a -> 1}, they both give the same
> result....{x, x^2, x^3}.
>
> Thank you for your help!!!
>
Just to go a little more into detail:
Plot has the Attribute HoldAll:
In[3]:= Attributes@Plot
Out[3]= {HoldAll, Protected}
this means that the arguments are not evaluated before plotting really
starts.
Plot "sees" only one top-level call to ReplaceAll and guesses you've
given one single-valued function:
In[4]:= FullForm@HoldForm[{a x,a x^2,a x^3}/.{a->1}]
Out[4]//FullForm=
HoldForm[ReplaceAll[List[Times[a,x],Times[a,Power[x,2]],Times[a,Power[x,3]]],List[Rule[a,1]]]]
in your first example, you give plot a list with 3 ReplaceAlls and Plot
"knows" that it has to expect three values.
I hope my bad english did not obfuscate the object,
Peter
--
my taylor is rich but my english is poor ;-)