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MathGroup Archive 2005

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Re: enumerating list items

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55986] Re: [mg55976] enumerating list items
  • From: Søren Merser <merser at image.dk>
  • Date: Wed, 13 Apr 2005 01:10:10 -0400 (EDT)
  • References: <4FDA877403669E48A2CB7F1122A7948045C6BA@dassne02.da.gei>
  • Sender: owner-wri-mathgroup at wolfram.com

thanks for your reply
what i really want is:
i'm getting data from a database link, including column headers
assigning  1 : ncols to these names  should enable call column[data, 
colum-index-by-name]
(or better: assign each data column to it's header name)
regards søren

----- Original Message ----- 
From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
To: mathgroup at smc.vnet.net
Subject: [mg55986] Re: [mg55976] enumerating list items



>-----Original Message-----
>From: Søren Merser [mailto:merser at image.dk]
To: mathgroup at smc.vnet.net
>Sent: Tuesday, April 12, 2005 11:27 AM
>To: mathgroup at smc.vnet.net
>Subject: [mg55986] [mg55976] enumerating list items
>
>hi
>
>is there a way to make each item in the list 'assignable',
>i.e. i want to
>enumerate the items like id=1, gender=2 etc.
>
>a={"id", "gender", "status", "odag"}
>
>b=Range@Length@a
>
>a=b
>
>regards soren
>
>
>

You have to convert the strings to symbols, before you can assign to them. 
This works, provided the symbols already don't exist with values:

In[1]:= a={"id","gender","status","odag"};

In[2]:= MapIndexed[(Evaluate[Symbol[#1]]=First[#2])&,a];

In[3]:= ?status
        Global`status
        status = 3

Or

In[4]:= Clear/@ a;    (* to repeat that, we first have to clear the symbols 
*)

In[5]:= MapThread[Set,{Symbol/@a , Range[Length[a]]}];

In[6]:= ?odag
        Global`odag
        odag = 4



The question as how to proceed when you want to change values, and you only 
have the names at hand, is not trivial but a FAQ, see the archive, if you 
are interested.

Good advice to you depends on what you really want to do. If you just want 
to associate your "names" with values you might better proceed differently 
(assoc here is the name of your association):

In[7]:= Thread[Evaluate[assoc /@ a] = Range[Length[a]]]
Out[7]= {1, 2, 3, 4}

This retrieves a "value":

In[8]:= assoc["gender"]
Out[8]= 2


This re-assigns a "value":

In[9]:= assoc["gender"] = 999
Out[9]= 999

--
Hartmut Wolf




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