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Re: Re: Variable amount of Buttons in Mathematica
*To*: mathgroup at smc.vnet.net
*Subject*: [mg92882] Re: [mg92851] Re: Variable amount of Buttons in Mathematica
*From*: Fred Simons <f.h.simons at tue.nl>
*Date*: Thu, 16 Oct 2008 05:07:27 -0400 (EDT)
*References*: <gcscrf$8l3$1@smc.vnet.net> <gcv77d$ds9$1@smc.vnet.net> <200810150937.FAA11741@smc.vnet.net>
The problem discussed in this thread is caused by the HoldRest attribute
of Button, and therefore is essentially the same as
In[1]:= Table[Hold[n], {n,1,5}]
Out[1]= {Hold[n],Hold[n],Hold[n],Hold[n],Hold[n]}
returning a list of 5 elements Hold[n].
Since Hold does not evaluate its argument, we have to *substitute*
values for n in Hold[n] rather than assigning values to n.
Here is one solution (you might replace Hold with the Button you are
interested in):
In[2]:= Table[Hold[n] /. n->i, {i,1,5}]
Out[2]= {Hold[1],Hold[2],Hold[3],Hold[4],Hold[5]}
The With-solutions are based on the fact that in the evaluation of
With[{n=something}, argument] the first step is *substitution* of the
value something for n in the expression argument; despite the syntax no
assignments are involved.
In[3]:= Table[With[{n=n},Hold[n]], {n,1,5}]
Out[3]= {Hold[1],Hold[2],Hold[3],Hold[4],Hold[5]}
As can be seen from the syntaxcolouring, the second n in n=n is the
counter of the Table command, while the first n is the vehicle that
takes care of the substitution in the second argument of With.
Still another construction, which I have not seen mentioned, and easier
to understand than the With construction, is based on the fact that
mapping also uses substitution:
In[4]:= Hold[#]& /@ Range[5]
Out[4]= {Hold[1],Hold[2],Hold[3],Hold[4],Hold[5]}
A variant of this is
In[5]:= Array[Hold[#]&, 5]
Out[5]= {Hold[1],Hold[2],Hold[3],Hold[4],Hold[5]}
Kind regards,
Fred Simons
Eindhoven University of Technology
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