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MathGroup Archive 2001

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Re: Q: Recursion on a list

  • To: mathgroup at smc.vnet.net
  • Subject: [mg28903] Re: [mg28886] Q: Recursion on a list
  • From: BobHanlon at aol.com
  • Date: Fri, 18 May 2001 01:13:06 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

The elements of a list are designated by r[[n]], i.e., with Part[r, n], 
not r[n], which is a function r evaluated for the argument n.

To generate the complete new list

f[r_List] := FoldList[#1*(1+#2)&, 1, r];

However, to calculate an element of the new list without 
calculating and storing the whole new list

g[r_List, n_Integer?NonNegative] := 
    Times @@ (1+Take[r, n]);

r = {r1, r2, r3, r4, r5};

f[r]

{1, r1 + 1, (r1 + 1)*(r2 + 1), 
  (r1 + 1)*(r2 + 1)*(r3 + 1), 
  (r1 + 1)*(r2 + 1)*(r3 + 1)*
   (r4 + 1), (r1 + 1)*(r2 + 1)*
   (r3 + 1)*(r4 + 1)*(r5 + 1)}

g[r, 4]

(r1 + 1)*(r2 + 1)*(r3 + 1)*(r4 + 1)

f[r] == Table[g[r, k], {k, 0, Length[r]}]

True

For a large list

r = Table[2Random[]-1, {10000}];

Last[f[r]] == g[r, Length[r]]//Timing

{0.9166666666665151*Second, True}


Bob Hanlon

In a message dated 2001/5/17 5:13:35 AM, mscmsc at mediaone.net writes:

>I'm looking for an efficient method (i.e., with the use of an explicit
>loop)
>for the following problem. I am working with large lists (several thousand
>elements) of data. Denote one such list r. I would like to define a new
>list
>x such that:
>
>
>x[0] = 1.0
>
>x[t] = x[t-1]*(1.0 + r[t])
>
>Can anyone suggest an efficient way?
>


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