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I have the list: 

l = 
  {{"d", "w", "x"}, {"a", "d"}, {"e", "h", "u", "d", "c", "q"}, 
   {"e", "c", "e", "a", "d", "e"}, {"s", "x", "r", "t", "v", "f"}, {"s"}, 
   {"c", "d", "e"}, {"s", "t", "y", "m"}, {"t", "d", "t", "u", "i", "q"}};

I want to select those elements that they are of length 6, so I evaluate 

sizedl = Select[l, Length@# == 6 &]

and get

{{"e", "h", "u", "d", "c", "q"}, {"e", "c", "e", "a", "d", "e"}, 
 {"s","x", "r", "t", "v", "f"}, {"t", "d", "t", "u", "i", "q"}}

I next want to find the most common characters, pair them and capitalise them. I have the following function to do that.

replace = 
  With[
    {commonest = 
       Commonest[Flatten[Map[Subsets[#, {2}] &, #], 1]][[1]]}, 
    # //. {OrderlessPatternSequence[## & @@ commonest, p___]} :> 
             {StringJoin[ToUpperCase[commonest]], p}] &;

Alpha = NestList[replace, sizedl, 1]

The second part of Alpha gives:

{{"ED", "c", "h", "q", "u"}, {"ED", "a", "c", "e", "e"}, 
 {"s", "x", "r", "t", "v", "f"}, {"t", "d", "t", "u", "i", "q"}}

Now, in this new list I have two elements of 5 and two elements of size 6. I want to get rid of the ones that are size 5 and repeat the same procedure. 

sizedl1 = Select[Alpha[[2]], Length @ # == 6 &];
Alpha = NestList[replace, sizedl1, 1];

I wonder how can I do this procedure for t times without repeating the code as above.

Clarification: The output for t=3 should look like this: 

l = {{"d", "w", "x"}, {"a", "d"}, {"e", "h", "u", "d", "c", 
"q"}, {"e", "c", "e", "a", "d", "e"}, {"s", "x", "r", "t", "v", 
"f"}, {"s"}, {"c", "d", "e"}, {"s", "t", "y", "m"}, {"t", "d", 
"t", "u", "i", "q"}};

k = 2;
ClearAll[replace]
replace = 
  With[{commonest = 
      Commonest[
        Flatten[Map[Subsets[#, {k}] &, #], 
         1]][[1]]}, # //. {OrderlessPatternSequence[## & @@ commonest,
         p___]} :> {StringJoin[ToUpperCase[commonest]], p}] &;

m=1;
sized1 = Select[l, Length@# == 6 &];
a1 = NestList[replace, sizedl, m][[2]]

sized2 = Select[a1, Length@# == 6 &];
a2 = NestList[replace, sized2, m][[2]]

sized3 = Select[a2, Length@# == 6 &];
a3 = NestList[replace, sized3, m][[2]]

OUTPUT:

{{"ED", "c", "h", "q", "u"}, {"ED", "a", "c", "e", "e"}, {"s", "x", 
  "r", "t", "v", "f"}, {"t", "d", "t", "u", "i", "q"}}

{{"s", "x", "r", "t", "v", "f"}, {"TU", "d", "i", "q", "t"}}

{{"SX", "f", "r", "t", "v"}}
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  • $\begingroup$ what is "replace" in your code? $\endgroup$ – J42161217 Oct 15 '18 at 13:21
  • $\begingroup$ Thank for the notice, I corrected it. replace it the function that does the replacement. $\endgroup$ – Wiliam Oct 15 '18 at 13:23
  • $\begingroup$ what about p and k in the replace function? $\endgroup$ – J42161217 Oct 15 '18 at 13:38
  • $\begingroup$ k is a number which indicates pairs and is equal to 2. Fixed it. and p is variable that indicates what shall be capitalised. $\endgroup$ – Wiliam Oct 15 '18 at 13:40
  • $\begingroup$ for the sake of example lets us take t=2. But in general I am dealing with a list that has 56000 elements and I want to apply the method for arbitrary t. $\endgroup$ – Wiliam Oct 15 '18 at 13:51
2
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replaceselect = Composition[replace, Select[Length @ # == 6 &]];
Rest @ NestWhileList[replaceselect , l, Length@# > 1 &]

{{{"ED", "c", "h", "q", "u"}, {"ED", "a", "c",    "e", "e"}, {"s", "x", "r", "t", "v", "f"}, {"t", "d", "t", "u", "i", "q"}},
{{"s", "x", "r", "t", "v", "f"}, {"TU", "d", "i", "q",    "t"}},
{{"SX", "f", "r", "t", "v"}}}

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  • $\begingroup$ Such short and neat answer deserves a wow. I learnt a lot from you so far @kglr $\endgroup$ – Wiliam Oct 15 '18 at 15:34
  • 1
    $\begingroup$ Thank you for the accept and kind words @William. $\endgroup$ – kglr Oct 15 '18 at 15:36

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