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I am trying to separate/split an expression (consisting of polynomials) to a list. However I want to keep the exponents intact. Here is an example:

k1=((x+y)^(2+e) (a+b)^(-1-e) (k - z) )//Expand
k2=k1 /. Times | Plus  | Power -> List

output:

(a + b)^(-1 - e) k (x + y)^(2 + e) - (a + b)^(-1 - e) (x + y)^(2 + e) z
{{{{a, b}, {-1, {-1, e}}}, k, {{x, y}, {2, e}}}, {-1, {{a, b}, {-1, {-1, e}}}, {{x, y}, {2, e}}, z}}

The problem in this approach is it makes list in which it is not apparant to distinguish -e and -1+e .

 -e /. Times | Plus | Power -> List
 -1+e /. Times | Plus | Power -> List

leads to same {-1, e}. For each - sign it creates a list which is unwanted.

In this example,

How do I prevent applying Plus -> List to the exponent?

The reason is I want to finally get back to the following form from each term of the list at the end, i.e.

k2[[1]], k2[[2]] etc.

where

 k2[[1]][[1]] = (a + b)^(-1 - e) 
 k2[[1]][[2]] = (x + y)^(2 + e)
 k2[[1]][[3]] = k

 k2[[2]][[1]] = (a + b)^(-1 - e) 
 k2[[2]][[2]] = (x + y)^(2 + e) 
 k2[[2]][[3]] = -z

EDIT::

The expected final output

 k2={
 {{(a + b),(-1 - e)},{ k },{(x + y),(2 + e)}},
 {{(a + b),(-1 - e)},{-z},{(x + y),(2 + e)}}
    }

or

  k2={
  {{{a , b},{-1, - e}},{ k },{{x , y},{2 , e}}},
  {{{a , b},{-1, - e}},{-z},{{x , y},{2 , e}}}
     }

Such that finally I get each of the two terms ( which are separated by +/-)

  (1.)   (a + b)^(-1 - e) k (x + y)^(2 + e) 
  (2.) - (a + b)^(-1 - e) (x + y)^(2 + e) z

Also from (1.) and (2.) I will get each terms which are multiplied i.e.

  (1.)  (a + b)^(-1 - e),  k,  (x + y)^(2 + e)
  (2.)  (a + b)^(-1 - e), -z, (x + y)^(2 + e)

I find the difficult part is to handle this -ve sign.

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    $\begingroup$ does k1 /. {Power[a_, b_] :> Power[a /. Plus | Times -> List, b] , Plus | Times -> List} give what you need? $\endgroup$
    – kglr
    Commented Sep 1, 2020 at 23:02
  • $\begingroup$ partially,(I can work on this) since my final goal is to first separate out each terms first by +- and then for each of them I want to find out the terms which are multiplied. Like in this example first k2 = k2[[1]] + k2[[2]]. Then for each k2[[i]] I want to take out each term which is multipled. Like from k2[[1]] I want (a + b)^(-1 - e), (x + y)^(2 + e), k. $\endgroup$
    – BabaYaga
    Commented Sep 1, 2020 at 23:08
  • $\begingroup$ can you post the desired output for k2? and for -e /. rule1 and -1+e /. rule2? $\endgroup$
    – kglr
    Commented Sep 1, 2020 at 23:13
  • $\begingroup$ @kglr, pls see update. I can infact take the above comment of yours and make it work to get the desired output by just simple final substitution: k2[[1]][[1]] /. {{x_^y_, a_^b_} -> (x+a)^y}. $\endgroup$
    – BabaYaga
    Commented Sep 1, 2020 at 23:32
  • $\begingroup$ The problem comes in handling - ve sign in a general case. In your scond list k2[[2]], it contains -1 as one of the list element, which I like to keep with any of the terms. I can probably do a workaround. Basically the top level lists will have same Length. $\endgroup$
    – BabaYaga
    Commented Sep 1, 2020 at 23:39

1 Answer 1

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$\begingroup$ 
ReplaceAll[Power -> List] @ Replace[SortBy[Length] /@ (List @@@ List @@ k1), 
   {a_, b_, c___} :> If[a === -1, {{-b}, c}, {{a}, b, c}], 2] 

{{{k}, {a + b, -1 - e}, {x + y, 2 + e}}, 
 {{-z}, {a + b, -1 - e}, {x + y, 2 + e}}}

Use ReplaceAll[Power | Plus -> List] to get

{{{k}, {{a, b}, {-1, -e}}, {{x, y}, {2, e}}},
 {{-z}, {{a, b}, {-1, -e}}, {{x, y}, {2, e}}}}
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