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I have the following list: 

l={"1 atattaggtt tttacctacc caggaaaagc caaccaacct
61 ctctaaacga actttaaaat ctgtgtagct gtcgctcggc
121 gcagtataaa caataataaa"}

which I want to turn into: 

{{{1,{"atattaggtt","tttacctacc","caggaaaagc","caaccaacct"}},
    {61,{"ctctaaacga","actttaaaat","ctgtgtagct","gtcgctcggc"}},
    {121,{"gcagtataaa","caataataaa"}}}

I wonder how one does this?

Update: I know that StringReplace[l, WhitespaceCharacter -> ","] gives me: 

{"1,atattaggtt,tttacctacc,caggaaaagc,caaccaacct,61,ctctaaacga,\
actttaaaat,ctgtgtagct,gtcgctcggc,121,gcagtataaa,caataataaa"}

Yet I don't know how to sublist this as above.

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3
  • 5
    $\begingroup$ try StringSplit[lis, ","] $\endgroup$
    – Nasser
    Commented Apr 2, 2020 at 19:21
  • 2
    $\begingroup$ @Nasser it was not a simple StringSplit[] $\endgroup$
    – Wiliam
    Commented Apr 2, 2020 at 19:50
  • $\begingroup$ I was replying to what you said Yet I don't know how to sublist this as above. not to the main question. $\endgroup$
    – Nasser
    Commented Apr 2, 2020 at 19:55

9 Answers 9

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8
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You may use StringCases with StringExpression.

StringCases[
  n : NumberString ~~ s : (WhitespaceCharacter ~~ LetterCharacter ..) .. :> 
    {FromDigits@n, StringSplit@s}
]@l

{{{1, {"atattaggtt", "tttacctacc", "caggaaaagc", "caaccaacct"}}, 
  {61, {"ctctaaacga", "actttaaaat", "ctgtgtagct", "gtcgctcggc"}}, 
  {121, {"gcagtataaa", "caataataaa"}}}}

Hope this helps.

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9
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{First[#], Rest[#]} & /@ ImportString[#, "Table"] & /@ l

(*    {{{1, {"atattaggtt", "tttacctacc", "caggaaaagc", "caaccaacct"}},
        {61, {"ctctaaacga", "actttaaaat", "ctgtgtagct", "gtcgctcggc"}},
        {121, {"gcagtataaa", "caataataaa"}}}}                              *)
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$\begingroup$

Another way using regular expressions.

StringSplit[l, RegularExpression["(?=[A-Z]|(?<!\\d)\\d)"]] // 
  Map[StringSplit /* ({First@#, Rest@#} &), #, {-1}] &
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StringCases[l,
  RegularExpression["([1-9]+)([a-t\s]+)"]:> {"$1", StringSplit["$2"]}
]

{{{1, {atattaggtt, tttacctacc, caggaaaagc, caaccaacct}}, {61, {ctctaaacga, actttaaaat, ctgtgtagct, gtcgctcggc}}, {121, {gcagtataaa, caataataaa}}}}

Original Answer 

StringCases[l,RegularExpression["([1-9]+)([a-z\s]+)"]:> {"$1", "$2"}]
  // MapAt[StringSplit, #, {All,All,2}]&
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3
$\begingroup$ 
Map[{FromDigits @ #, {##2}} & @@@ # & @* StringSplit] @ StringSplit[l, "\n"]

{{{1, {"atattaggtt", "tttacctacc", "caggaaaagc", "caaccaacct"}}, 
  {61, {"ctctaaacga", "actttaaaat", "ctgtgtagct", "gtcgctcggc"}},
  {121, {"gcagtataaa", "caataataaa"}}}}

Also

f = Normal[{#, DeleteCases[{##2}, ""]} & @@@ SemanticImportString[#]] &;
f /@ l

{{{1, {"atattaggtt", "tttacctacc", "caggaaaagc", "caaccaacct"}}, 
  {61, {"ctctaaacga", "actttaaaat", "ctgtgtagct", "gtcgctcggc"}}, 
  {121, {"gcagtataaa", "caataataaa"}}}}

and

f2 = {FromDigits @ #, {##2}} & @@@ 
      Split[StringSplit @ #, Not @* StringMatchQ[NumberString] @* First] &;
f2 /@ l

 {{{1, {"atattaggtt", "tttacctacc", "caggaaaagc", "caaccaacct"}},
   {61, {"ctctaaacga", "actttaaaat", "ctgtgtagct", "gtcgctcggc"}},
   {121, {"gcagtataaa", "caataataaa"}}}}
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$\begingroup$ 
list = {
 "1 atattaggtt tttacctacc caggaaaagc caaccaacct
 61 ctctaaacga actttaaaat ctgtgtagct gtcgctcggc
 121 gcagtataaa caataataaa"};

Using SequenceSplit (new in 11.3)

{SequenceSplit[
  ToExpression @ First @ StringSplit @ list,
  {a_, b__Symbol} :> {a, ToString /@ {b}}]}

{{{1, {"atattaggtt", "tttacctacc", "caggaaaagc", "caaccaacct"}}, {61, {"ctctaaacga", "actttaaaat", "ctgtgtagct", "gtcgctcggc"}}, {121, {"gcagtataaa", "caataataaa"}}}}

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$\begingroup$ 
list = {
 "1 atattaggtt tttacctacc caggaaaagc caaccaacct
 61 ctctaaacga actttaaaat ctgtgtagct gtcgctcggc
 121 gcagtataaa caataataaa"};

Following @eldo's idea, an equivalent form using SequenceCases:

{SequenceCases[
 ToExpression@First@StringSplit@list,
 {a_, b__Symbol} :> {a, ToString /@ {b}}]}

{{{1, {"atattaggtt", "tttacctacc", "caggaaaagc", "caaccaacct"}}, {61, {"ctctaaacga", "actttaaaat", "ctgtgtagct", "gtcgctcggc"}}, {121, {"gcagtataaa", "caataataaa"}}}}

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2
$\begingroup$

So after playing a bit, I found the solution: 

x0 = StringSplit[l, Whitespace] // Flatten // ToExpression;
x1 = Split[x0, ! IntegerQ[#2] &];
x2 = Map[Drop[#, 1] &, x1];
x3 = Table[Map[ToString[#] &, x2[[i]]], {i, Length[x2]}];
x4 = Map[Take[#, 1] &, x1] // Flatten;
final = Transpose[{x4, x3}]

To which I get: 

{{1, {"atattaggtt", "tttacctacc", "caggaaaagc", 
   "caaccaacct"}}, {61, {"ctctaaacga", "actttaaaat", "ctgtgtagct", 
   "gtcgctcggc"}}, {121, {"gcagtataaa", "caataataaa"}}}

As desired. Maybe you guys can make this more efficient.

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1
  • 1
    $\begingroup$ Your 'as desired' differs somewhat to that given in your question $\endgroup$
    – user1066
    Commented Apr 2, 2020 at 22:13
2
$\begingroup$ 
l = {"1 atattaggtt tttacctacc caggaaaagc caaccaacct
61 ctctaaacga actttaaaat ctgtgtagct gtcgctcggc
121 gcagtataaa caataataaa"};

res = {First@l // StringSplit // 
    Split[#, ! StringMatchQ[#2, NumberString] &] & // 
   MapAt[ToExpression, {All, 1}]}

{{{1, "atattaggtt", "tttacctacc", "caggaaaagc", "caaccaacct"}, {61,
"ctctaaacga", "actttaaaat", "ctgtgtagct", "gtcgctcggc"}, {121,
"gcagtataaa", "caataataaa"}}}

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