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Re: List manipulation question - 2013

• To: mathgroup at smc.vnet.net
• Subject: [mg129632] Re: List manipulation question - 2013
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Thu, 31 Jan 2013 20:47:21 -0500 (EST)
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• Delivered-to: l-mathgroup@wolfram.com
• Delivered-to: mathgroup-newout@smc.vnet.net
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• References: <20130130092653.09BA46877@smc.vnet.net>

Clear[f]

f[a_, b_, outputLen_: (-1), c_: 2000] :=
  Module[{list1, list2},
   list1 = Table[24 + (a*n1), {n1, c}];
   list2 = Select[list1, IntegerQ[#/24] &];
   list3 = Last /@
     (Select[
        Solve[{12 + (b*n2) == #,
            Element[n2, Integers]}, n2] & /@
         list2,
        Length[#] > 0 &] // Flatten);
   If[outputLen == -1,
    list3,
    list3[[1 ;; Min[outputLen, Length[list3]]]]]];

f[15, 3, 6]

{44, 84, 124, 164, 204, 244}

f[13, 7, 6]

{180, 492, 804, 1116, 1428, 1740}


Bob Hanlon


On Thu, Jan 31, 2013 at 2:56 AM, Lea Rebanks <lrebanks at gmail.com> wrote:
> Hi Bob,
>
> Many thanks for your excellent reply.
> Your coding for lists 1,2,3 is exactly what I want.
> And your statement - "Apparently you want list2 to consist of the members of
> list1 that are
> integer multiples of 24 rather than the integers in list1/24." is correct.
>
> But I tried changing the input values from the example given
> From 24 + (15*n1)   to    24 + (13*n1) .....15 to 13
> From 12 + (3*n2)     to    12 + (7*n2) .....3 to 7
> This immediately caused me error messages, some of which I was able to
> correct by expanding the list range of the Take function.
> But I was unable to calculate the last part for list3 :-
>
> list1 = Table[24 + (13*n1), {n1, 2000}];
> list2 = Take[Select[list1, IntegerQ[#/24] &], 40];
> list3 = n2 /.
>     Solve[{12 + (7*n2) == #, Element[n2, Integers]}, n2][[1]] & /@
>   list2
>
> Hopefully the result should look something like :-
> {1272, 3456, 5640, 7824, 10008, 12192}
> These values above being the first & then following repetitions thereafter
> of equal points of repetitions of n1 & n2 in list2 & list3 respectively.
>
> Please could you rewrite the enclosed code so that :-
> (i) The above new values work & provide a similar result to that which I
> have shown.
> (ii)  Also, if possible, allow in the rewrite the flexibility within the
> function to accept much larger & challenging input values for list1 & 2.
> This would include a domain allowance relative to the size of input values
> as in this case its values are 13 & 7, but they could be integer values of
> 127 & 56 for list1 & list3 respectively.
> (Note higher value first.) Actually I will probably want to use much much
> higher values later.
> Ideally I am only looking for about 6 elements in the final answer list like
> - {1272, 3456, 5640, 7824, 10008, 12192}
>
> Many thanks for your help & attention.
> Really appreciated.
> Best regards,
> Lea...
>
>
>
>
> On Thu, Jan 31, 2013 at 11:05 AM, Bob Hanlon <hanlonr357 at gmail.com> wrote:
>>
>> list1 = Table[24 + (15*n1), {n1, 2000}];
>>
>> Take[Select[list1/24, IntegerQ], 9]
>>
>> {6, 11, 16, 21, 26, 31, 36, 41, 46}
>>
>> Take[Select[list1, IntegerQ[#/24] &], 9]
>>
>> {144, 264, 384, 504, 624, 744, 864, 984, 1104}
>>
>> Apparently you want list2 to consist of the members of list1 that are
>> integer multiples of 24 rather than the integers in list1/24.
>>
>> list2 = Select[list1, IntegerQ[#/24] &];
>>
>> list3 = n2 /. Solve[
>>        {12 + (3*n2) == #, Element[n2, Integers]},
>>        n2][[1]] & /@ list2;
>>
>> Length[list2] == Length[list3]
>>
>> True
>>
>>
>> Bob Hanlon
>>
>> Consequently, there is an integer n2 for every element of list2.
>>
>> On Wed, Jan 30, 2013 at 4:26 AM, Lea Rebanks <lrebanks at gmail.com> wrote:
>> >
>> > Dear All,
>> >
>> > I have the follow problem with the combination of a few lists.
>> > I shall outline the problem or what I am trying to do.
>> > I know that to solve this problem requires a number of processes, but I
>> > am
>> > not sure how to setup the coding to achieve my desired result.
>> > Please could someone show me the coding for this problem.
>> >
>> > The problem:-
>> > 1 - I have a list1 created from 24+(15*n1)        All n1 are integers
>> > 1,2,3,4.....to a large number of integers.
>> > 2 - I want to divide the list1 by 24 and create another list2 with only
>> > the
>> > integer results in list2.
>> >      Table[24 + 15*i, {i, 100}]/24  ...    so that integer results in
>> > list2
>> > = { 144, 264, 384, 504, 624, 744, 864, 984, 1104 }
>> > 3 - I have another list3 created from 12+(3*n2)   All n2 are integers
>> > 1,2,3,4.....to a large number of integers.
>> > 4 - With list3 I want to find  :-
>> >      (i) The integer number of n2 that either equates 12+(3*n2) = 144 or
>> > the next FIRST available number in list2 that meets this equality.
>> >  then also    (ii)The integer number(s) of n2 that equates 12+(3*n2) =
>> > 24+(15*n1) after the first equal value for say 6 repetitions.
>> >  then also     (iii)The integer number(s) of n1 that equates 12+(3*n2) =
>> > 24+(15*n1) after the first equal value for say 6 repetitions.
>> >
>> > The above example is quite simple, but I am hoping to setup the coding
>> > to
>> > work with other integer values instead of 3 & 15 which will present more
>> > of
>> > a challenge.
>> >
>> > Any help or advice greatly appreciated.
>> > Best regards,
>> > Lea...
>> >
>> >
>>
>

• References:
  • List manipulation question - 2013
    • From: Lea Rebanks <lrebanks@gmail.com>