[Date Index]
[Thread Index]
[Author Index]
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
Prev by Date:
**Re: Re: variable "K"? (Really strange behavior . . . )**
Next by Date:
**Re: RE: Mathmatica StyleSheet questions**
Previous by thread:
**Re: Re: generalized foldlist problem - part 2**
Next by thread:
**Re: generalized foldlist problem - part 2**
| |