Re: Re: Re: generalized foldlist problem - part 2

*To*: mathgroup at smc.vnet.net*Subject*: [mg69227] Re: [mg69215] Re: [mg69192] Re: [mg69138] generalized foldlist problem - part 2*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 3 Sep 2006 01:39:16 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

On 1 Sep 2006, at 23:41, Andrzej Kozlowski wrote: > > On 1 Sep 2006, at 11:41, Andrzej Kozlowski wrote: > >> >> On 31 Aug 2006, at 09:38, Arkadiusz Majka wrote: >> >>> Dear All, >>> >>> Thank all of you for your help regarding my "generalized fold list >>> problem", that sounds >>> >>>>> I have two list >>>> >>>> list1=={a,b,c,d,e} >>>> list2=={3,2,5,1,6} >>>> >>>> and I want to apply a modified version of FoldList to list1 in the >>>> following way: list2 indicates that element a appears only 3 times >>>> (if >>>> space enough) beginning from the beginning of the list , element b >>>> appears 2 times, c - 5 times , etc. >>>> >>>> So the output should be >>>> >>>> GeneralizedFoldList[list1,list2]=={a,a+b,a+b+c,c+d,c+e} >>>> >>>> >>> >>> Now, I still need your help. What I want to do now, and what I have >>> problem with, is to add a constraint to the algorithm, i.e every >>> element in my GeneralizedFoldList must be less than one. >>> The following example says what to do if it is not. >>> >>> Lets take two lists >>> >>> list1=={0.9,0.8,0.7} >>> list2=={3,3,3} >>> >>> All your algorithms use PadRight (you pad 0's). So the following >>> matrix >>> is built >>> >>> {{0.9, 0.9, 0.9, 0, 0}, {0, 0.8, 0.8, 0.8, 0}, {0, 0, 0.7, 0.7, >>> 0.7}} >>> >>> and we add elements along colums and obtain {0.9, 1.7, 2.4, 1.5, >>> 0.7} >>> >>> The first element is less than 1 so it's ok. The second is > 1 >>> what I >>> need to avoid. I want to avoid it by shifting the nonzero >>> elements of >>> the second and third row of above matrix of two positions: >>> {0,0,0,0.8,0.8,0.8,0}, {0,0,0,0,0.7,0.7,0.7}. >>> >>> I go on with suming along columns and discover that everything is >>> fine >>> until I have to add 0.8 and 0.7 what is >1. So I repeat the >>> procedure >>> by shfting hte third row of the number of position that is needed to >>> ensure that my sum is <1. >>> >>> Finally I obtain {{0.9, 0.9, 0.9, 0, 0, 0, 0, 0, >>> 0},{0,0,0,0.8,0.8,0.8,0,0,0},{0,0,0,0,0,0,0.7,0.7,0.7}} >>> and Plus@@% is what I desire. >>> >>> Thanks in advance! I hope you will manage :) >>> >>> Arek >>> >> >> >> Here is a (not very pretty) code that should do what you want >> starting with a matrix of the above kind: >> >> >> GF[P_]:==Module[{M==P},Do[M == NestWhile[ >> Join[(PadRight[#1, Length[#1] + 1] & ) /@ >> Take[#1, s], (PadLeft[#1, Length[#1] + 1] & ) /@ >> Take[#1, s - Dimensions[#1][[1]]]] & , M, >> !And @@ Thread[Total[#1[[All,Range[Length[ >> Flatten[Split[#1[[s]]] /. {l___, >> {(0)..}} :> {l}]]]]]] < 1] & ], >> {s, Dimensions[M][[1]]}]; M] >> >> >> Let's verify that it works. First, starting with your original case: >> >> >> In[8]:== >> M=={{0.9,0.9,0.9,0,0},{0,0.8,0.8,0.8,0},{0,0,0.7,0.7,0.7}}; >> >> >> In[9]:== >> GF[M]//InputForm >> >> Out[9]//InputForm== >> {{0.9, 0.9, 0.9, 0, 0, 0, 0, 0, 0}, >> {0, 0, 0, 0.8, 0.8, 0.8, 0, 0, 0}, >> {0, 0, 0, 0, 0, 0, 0.7, 0.7, 0.7}} >> >> Now let's change M somewhat: >> >> In[10]:== >> M=={{0.9,0.9,0.1,0,0},{0,0.1,0.8,0.3,0},{0,0,0.7,0.9,0.7}}; >> >> >> In[11]:== >> GF[M]//InputForm >> >> Out[11]//InputForm== >> {{0.9, 0.9, 0.1, 0, 0, 0, 0, 0}, {0, 0, 0.1, 0.8, 0.3, 0, >> 0, 0}, {0, 0, 0, 0, 0, 0.7, 0.9, 0.7}} >> >> So you only need to combine this with your favourite code from >> part 1. >> >> (PS. In answering part 1 I obviously misunderstood your question, but >> I think that was mainly because I do not think your function is in >> any way a "generalisation of Fold", or even of Fold >> [Plus,element,list]. As far as I can see it does not really perform >> any kind of "folding". >> >> Andrzej Kozlowski >> > > > In fact I still do not fully understand your original question and, > in particular, if the example you originally posted (and which, if > any, of the solutions various people provided) were correct. Just in > case, here is a "solution" to your original question, which I think > is different from any of the ones I have seen: > > GFMatrix[list1_, > list2_] :== Normal[SparseArray[Flatten[Table[If[j =B2 i + list2 > [[i]] - 1, > Rule[{i, j}, list1[[i]]], Rule[{i, > j}, 0]], {i, Length[list2]}, {j, i, Length[list1] + i - > 1}], 1]]] > > GFMatrix builds the matrix, which, I think, you need in the second > part of the question. The code makes no use of PadLeft or PadRight. > However, note that in your original example, the output is not the > same as you gave as the correct one, although the output below seems > to me to fit better your verbal description of what you wanted done: > > > list1=={a,b,c,d,e}; > list2=={3,2,5,1,6}; > > > GFMatrix[list1,list2] > > > {{a, a, a, 0, 0, 0, 0, 0, 0}, > {0, b, b, 0, 0, 0, 0, 0, > 0}, {0, 0, c, c, c, c, c, > 0, 0}, {0, 0, 0, d, 0, 0, > 0, 0, 0}, {0, 0, 0, 0, e, > e, e, e, e}} > > > Total[%] > > > {a,a+b,a+b+c,c+d,c+e,c+e,c+e,e,e} > > This is longer than the output you gave as the right solution, which > was only a part of the above, namely {a,a+b,a+b+c,c+d,c+e}. > > Andrzej Kozlowski > > Oops, the code I copied and pasted in did not come out the way it should have done. It shoudlhave been: GFMatrix[list1_, list2_] := Normal[SparseArray[Flatten[Table[If[j <= i + list2[[i]] - 1, {i, j} -> list1[[i]], {i, j} -> 0], {i, Length [list2]}, {j, i, Length[list1] + i - 1}], 1]]] Andrzej Kozlowski

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**Re: Re: generalized foldlist problem - part 2**

**Re: generalized foldlist problem - part 2**