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I have a list of five elements. For example,

list1 = {a1, a2, a3, a4, a5};

I'm trying to get a list:

{{a2/a1}, {a3/a1, a3/a2}, {a4/a1, a4/a2, a4/a3}, {a5/a1, a5/a2, a5/a3, a5/a4}}

How can I do this? And can I get it without Table?

Then, it is necessary to combine it with another list

list2 = {{b1, c1}, {b2, c2}, {b3, c3}, {b4, c4}, {b5, c5}};

as

{{b1, c1, a2/a1}, {{b1, c1, a3/a1}, {b2, c2, a3/a2}}, {{b1, c1, a4/a1}, {b2, c2, a4/a2}, 
{b3, c3, a4/a3}}, {{b1, c1, a5/a1}, {b2, c2, a5/a2}, {b3, c3, a5/a3}, {b4, c4, a5/a4}}}

I'd like to do this with Mathematica functions. Any ideas?

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4 Answers 4

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step1 = ReplaceList[list1, {a__, b_, ___} :> b/{a}]
step2 = FlattenAt[(Flatten /@ Thread[{list2[[;; Length@#1]], #1}]) & /@step1, 1]
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list11=list1[[#]]/list1[[;; # - 1]] & /@ Range[2, Length[list1]]

Fro the second question may be this work:

list22 = list2[[;; # - 1]] & /@ Range[2, Length[list2]]
MapThread[Append, #] & /@ Transpose[{list22, list11}]
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  • $\begingroup$ This was my first thought as well. $\endgroup$ Jul 16, 2015 at 8:44
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The first thing that came to mind:

list1 = {a1, a2, a3, a4, a5};
list2 = {{b1, c1}, {b2, c2}, {b3, c3}, {b4, c4}, {b5, c5}};

Rest@FoldList[Append, {}, Most@list1]

Rest[list1] / %

Flatten /@ Thread @ {Take[list2, Length@#], #} & /@ %       (* not too pretty *)
{{a2/a1}, {a3/a1, a3/a2}, {a4/a1, a4/a2, a4/a3}, {a5/a1, a5/a2, a5/a3, a5/a4}}

{{{b1, c1, a2/a1}}, {{b1, c1, a3/a1}, {b2, c2, a3/a2}}, {{b1, c1, a4/a1},
 {b2, c2, a4/a2}, {b3, c3, a4/a3}}, {{b1, c1, a5/a1}, {b2, c2, a5/a2},
  {b3, c3, a5/a3}, {b4, c4, a5/a4}}}
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Ugly but another way (perhaps):

Join @@ With[{r = 
    GatherBy[
     Join @@ MapIndexed[#1/list1[[1 ;; #2[[1]]]] &, Rest@list1], 
     Denominator]},
  MapThread[Map[Function[x, Append[#1, x]], #2] &, {Most@list2, r}]]
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