Re: List manipulation question - 2013

*To*: mathgroup at smc.vnet.net*Subject*: [mg129636] Re: List manipulation question - 2013*From*: Lea Rebanks <lrebanks at gmail.com>*Date*: Fri, 1 Feb 2013 01:15:53 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <20130130092653.09BA46877@smc.vnet.net>

Hi Bob, This is excellent, really impressive & works a dream. However, I did notice that a certain number of combinations of inputs for a & b do not provide a result at all, which was a surprise. For example :- f[19, 8, 6] does not provide a result or any multiple of 8 for b value. Is this TRUE that certain combinations of integers here should not provide a result? Many thanks for your help. Really appreciated. Best regards, Lea... On Fri, Feb 1, 2013 at 9:47 AM, Bob Hanlon <hanlonr357 at gmail.com> wrote: > 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... > >> > > >> > > >> > > > >

**Re: Euclidean distance of all pairwise combinations (redundants)**

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