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I made the following list:

ELCo = Alphabet["English"];    
Characters[ToLowerCase[WordList[Language -> "English"]]];
Select[%, SubsetQ[ELCo, ToLowerCase[#1]] &];
LPr = Map[Sort, Map[DeleteDuplicates, %]];

This list LPr takes the English words and separates them into their characters while getting rid of duplication and capital letters. Now I make another list, basically the first five letters of the English language:

letters = Alphabet[][[1 ;; 5]];

Now I want to look at LPr and select all those words that are made of letters "a", "b", "c", "d", or "e" or the combination of them. I wonder how one would to this, I tried:

tmp = Table[SubsetQ[letters, LPr[[i]]], {i, Length[LPr]}];

and then,

Pick[LPr, tmp]

but this does not give me all those required words.

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4
  • $\begingroup$ LPr does not contain any reference to original words anymore so what exactly is a desired output? Words or your LPr entires? $\endgroup$
    – Kuba
    Nov 7, 2018 at 10:04
  • $\begingroup$ LPr enteries that contains "a", "b", "c", "d", or "e" or the combination of them. $\endgroup$
    – Wiliam
    Nov 7, 2018 at 10:06
  • $\begingroup$ I tried different approach and I get exactly what your Pick returns. So I guess you should explain why do you think your result is not correct. $\endgroup$
    – Kuba
    Nov 7, 2018 at 10:10
  • $\begingroup$ If you run LPr you see for instance the second entry is {"a","h"} this element has "a" but was not picked. As I mentioned I need the return to contain contains "a" or "b" or "c" or "d" or "e" or the combination of them. $\endgroup$
    – Wiliam
    Nov 7, 2018 at 10:15

1 Answer 1

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Is this what you need?

Select[ ContainsAny @ CharacterRange["a", "e"] ] @ LPr

{<<36k>>}

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