I am still a beginner with Mathematica and I have a simple problem with splitting this list:

List1 = {"Id=\"1\"", "PostTypeId=\"1\"", "AcceptedAnswerId=\"2\"", \
"CreationDate=\"2013-11-05T20:22:07.323\"", "OwnerUserId=\"8\"", \
"LastEditorUserId=\"8\"", "Tags=\"<word-choice><articles>\
\"", "Id=\"2\"", "PostTypeId=\"2\"", "ParentId=\"1\"", "CreationDate=\
\"2013-11-05T20:36:02.473\"", "OwnerUserId=\"17\"", \
"LastEditorUserId=\"17\"", "Id=\"3\"", "PostTypeId=\"2\"", "ParentId=\
\"1\"", "CreationDate=\"2013-11-05T20:37:28.327\"", \
"OwnerUserId=\"6\"", "Id=\"4\"", "PostTypeId=\"2\"", \
"ParentId=\"1\"", "CreationDate=\"2013-11-05T20:38:07.760\"", \
"OwnerUserId=\"18\"", "LastEditorUserId=\"8\"", "Id=\"5\"", \
"PostTypeId=\"1\"", "CreationDate=\"2013-11-05T20:46:59.993\"", \
"OwnerUserId=\"18\"", "LastEditorUserId=\"707\"", \
"Tags=\"<etymology><proverbs>\"", "Id=\"6\"", \
"PostTypeId=\"1\"", "AcceptedAnswerId=\"8\"", \
"CreationDate=\"2013-11-05T21:01:56.857\"", "OwnerUserId=\"18\"", \
"LastEditorUserId=\"707\"", \
"Tags=\"<grammar><grammatical-number>\"", "Id=\"8\"", \
"PostTypeId=\"2\"", "ParentId=\"6\"", \
"CreationDate=\"2013-11-05T21:09:18.067\"", "OwnerUserId=\"6\"", \
"LastEditorUserId=\"1243\"", "Id=\"9\"", "PostTypeId=\"2\"", \
"ParentId=\"6\"", "CreationDate=\"2013-11-05T21:15:24.207\"", \
"OwnerUserId=\"22\"", "LastEditorUserId=\"57\"", "Id=\"10\""}

What I would like to do is divide the list in such a way that I have a lists with only: List1 = {{Id1, CreationDate1, Tags1},{Id2, CreationDate2, Tags2},{Id3, CreationDate3, Tags3}...}.

The approach I tried is to split on the >\". Unfortunately this doesn't work. Can someone help me in finding a quick way to do it? (as plus, it would nice if we could get rid of &alt; and > in somehow)

I hereby post the code I used

StringSplit[StringJoin[{List1}], ">\""]

But it doesn't help a lot, since then I have to get back then to the original list with another StringSplit. I am sure there is a smarter way I am not able to see here.

Thanks a lot in advance!

  • $\begingroup$ Try Partition[List1,3] $\endgroup$ – mikado May 1 '18 at 20:16
  • $\begingroup$ It doesn't work because not all the records have "Tags". Note: A record is from id to tag (when is present) $\endgroup$ – Carmine Ragone May 1 '18 at 20:25
  • $\begingroup$ StringSplit only works on strings, not lists. $\endgroup$ – ktm May 1 '18 at 20:40
  • $\begingroup$ @user6014 I am not sure you're right. Take a look to the documentation center. It states that "StringSplit automatically threads over lists of strings:" and uses as example: StringSplit[{"a:b:c:d", "listable:element"}, ":"] which works perfectly. $\endgroup$ – Carmine Ragone May 1 '18 at 21:00
  • 1
    $\begingroup$ Nevermind, my mistake. I apologize. $\endgroup$ – ktm May 1 '18 at 21:08

Maybe this does help?

list = Most[List1];
pos = Flatten[Position[StringTake[#, 1 ;; 3] & /@ list, "Id="]];
AppendTo[pos, Length[list] + 1];
data =
   Association[Rule @@@ StringSplit[#, "="]] &,
data = MapAt[DateObject, data, {All, "CreationDate"}];
newdata = data[[All, {"Id", "CreationDate", "Tags"}]];
newdata = MapAt[
  If[! MissingQ[#], StringReplace[#, {"&lt;" -> "<", "&gt;" -> ">"}],# ] &, 
  {All, "Tags"}
| improve this answer | |
  • $\begingroup$ This is really nice! The only thing I am sure about is the function To Associations, because on my Mathematica does not exist. Could you please take a look? However, thank you very much for the help! :) $\endgroup$ – Carmine Ragone May 1 '18 at 21:06
  • $\begingroup$ Ah right, one has to load Needs["GeneralUtilities`"]. But I removed ToAssociations altogether. Have look. $\endgroup$ – Henrik Schumacher May 1 '18 at 21:13
  • $\begingroup$ It's perfect, love it! Thanks a lot! $\endgroup$ – Carmine Ragone May 1 '18 at 22:04
  • $\begingroup$ You're welcome! $\endgroup$ – Henrik Schumacher May 1 '18 at 22:04

uses ReplaceRepeated in a recursive manner to generate the lists

x_ /; StringStartsQ[x, ("Id" | "CreationDate" | "Tags")]], 
"&gt;" | "&lt;" -> ""] //. {p___, x_ /; StringStartsQ[x, "Id"], y__,z_ /;
StringStartsQ[z, "Id"], t___} :> {p, {x, y}, z, t}

(* {{"Id=\"1\"", "CreationDate=\"2013-11-05T20:22:07.323\"", 
"Tags=\"word-choicearticles\""}, {"Id=\"2\"", 
"CreationDate=\"2013-11-05T20:36:02.473\""}, {"Id=\"3\"", 
"CreationDate=\"2013-11-05T20:37:28.327\""}, {"Id=\"4\"", 
"CreationDate=\"2013-11-05T20:38:07.760\""}, {"Id=\"5\"", 
"Tags=\"etymologyproverbs\""}, {"Id=\"6\"", 
"Tags=\"grammargrammatical-number\""}, {"Id=\"8\"", 
"CreationDate=\"2013-11-05T21:09:18.067\""}, {"Id=\"9\"", 
"CreationDate=\"2013-11-05T21:15:24.207\""}, "Id=\"10\""} *)
| improve this answer | |

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