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MathGroup Archive 2000

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Re: Re: Set in Scan

  • To: mathgroup at smc.vnet.net
  • Subject: [mg22008] Re: [mg21900] Re: [mg21856] Set in Scan
  • From: "Tomas Garza" <tgarza at mail.internet.com.mx>
  • Date: Thu, 10 Feb 2000 02:25:50 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

It looks as if the suggested approach of Hartmut Wolf's - as well as Jens
Peer Kuska's - to this problem works fine and is extremely efficient.
However, suppose that instead of having a few possibly very long lists {v1,
v2, v3} one has a large number N - perhaps of the order of tens of
thousands - of such lists, which need not be too long. The procedure would
have to be modified in order to use some symbol like allv = {v1, v2,..., vN}
in Scan, instead of writing the complete list with all individual v's, as
used by Hartmut and Jens:

Scan[Function[sym, sym[[3]] = "-XXX-", {HoldFirst}],
   Unevaluated[{v1, v2, v3}]]

However, if I write

Scan[Function[sym, sym[[3]] = "-XXX-", {HoldFirst}],
   Unevaluated[allv]]

nothing happens. And the second suggestion by Hartmut is unusable in case
the assignment may be something which depends on the value of the element to
be replaced, e.g. Function[sym,If[sym[[3]] < 0.5, "-XXX-","-YYY-"]].

Any suggestions?

Tomas Garza
Mexico Cit

Hartmut Wolf wrote:

> Johannes Ludsteck schrieb:
> >
> > Dear MathGroup members
> >
> > I have to do replacements in very long lists. An efficient way to do
> > this is to use
> >
> > x[[index]]=newElement.
> >
> > This, however, doesn't work if I want to do replacements for a set of
> > lists. If I try to replace the third element in the lists v1,
> v2 and v3 by
> > typing
> >
> > Scan[#[[3]] = newElement &, {v1,v2,v3}]
> > I get the error message
> > Set::setps: #1 in assignment of part is not a symbol.
> >
>
> Hallo Johannes,
>
> the problem with your expression is to get the symbols v1,...
> unevaluated into the lhs of Set. There are two points where evaluation
> occurs, first at the rhs of Scan (or Map) where the list {v1,v2,v3} and
> then recursively v1,... are evaluated. You can suppress that by wrapping
> the list with Unevaluated. The second is the evaluation of #1 before
> Function inserts the argument at the lhs of Set. This can be avoided by
> specifing the argument HoldFirst for evaluation of Function:
>
> In[8]:= allv = {v1, v2, v3} = Table[Random[], {3}, {4}]
> Out[8]=
> {{0.308624, 0.0113816, 0.452114, 0.100156},
>  {0.368705, 0.44588, 0.878839, 0.736246},
>  {0.71764, 0.18652, 0.667573, 0.338795}}
>
> In[9]:=
> Scan[Function[sym, sym[[3]] = "-XXX-", {HoldFirst}],
>   Unevaluated[{v1, v2, v3}]]
>
> In[10]:= {v1, v2, v3}
> Out[10]=
> {{0.308624, 0.0113816, "-XXX-", 0.100156},
>  {0.368705, 0.44588, "-XXX-", 0.736246},
>  {0.71764, 0.18652, "-XXX-", 0.338795}}
>
> Yet another idea would be to use the compound object allv (see above),
> and then


> In[11]:= allv[[All, 2]] = "-YYY-"
> Out[11]= "-YYY-"
>
> In[14]:= allv
> Out[14]=
> {{0.308624, "-YYY-", 0.452114, 0.100156},
>  {0.368705, "-YYY-", 0.878839, 0.736246},
>  {0.71764, "-YYY-", 0.667573, 0.338795}}
>
> Of course the values of v1,... are not affected, but perhaps you may
> dispense with them alltogether, or use them only for access or display
> purposes:
>
> In[15]:= Clear[v1, v2, v3]
> In[16]:=
> v1 := allv[[1]]; v2 := allv[[2]]; v3 := allv[[3]]
>
> In[17]:= {v1, v2, v3}
> Out[17]=
> {{0.308624, "-YYY-", 0.452114, 0.100156},
>  {0.368705, "-YYY-", 0.878839, 0.736246},
>  {0.71764, "-YYY-", 0.667573, 0.338795}}



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