What is an appropriate command that does the opposite of the following?
StringSplit["a b c d e f g"," "]
$\begingroup$
$\endgroup$
5
5 Answers
$\begingroup$
$\endgroup$
8
A combination of StringJoin and Riffle:
res = StringSplit["a b c d e f g"," "];
StringJoin@Riffle[res," "]
-
$\begingroup$ you beat me by 10 seconds! I'll delete my answer... $\endgroup$– bill sMar 27, 2015 at 17:59
-
-
$\begingroup$ @WRI, why not just StringJoin[res," "] $\endgroup$ Mar 27, 2015 at 20:28
-
1$\begingroup$ @William
StringSplit[" a b c "," "]is just{"a", "b", "c"}so information is lost. There is no way you could know if there were a leading or ending space in the original string. This is the closest to an inverse that you can come. In Javascript that information is preserved bysplit. $\endgroup$– C. E.Mar 27, 2015 at 23:44 -
1$\begingroup$ @William ...if you want it to work like your Javascript example you can use
StringSplit[" a b c ", " ", All]. $\endgroup$– C. E.Mar 27, 2015 at 23:48
$\begingroup$
$\endgroup$
1
In version 10.1 you can use StringRiffle:
res = StringSplit["a b c d e f g", " "];
StringRiffle[res]
Use no second argument for spaces, or something else for something else. A nice advantage of StringRiffle is that res elements can be non string elements, and it will be automatically converted. It's something I miss in StringJoin.
PS: this answer is based on Docs, I don't have V10.1.
Take care with ToString@Row in V9 or older. See this post, about some problems.
-
$\begingroup$ This fails for example it doesn't match the following js
" a b ".split(" ").join(" ")" $\endgroup$– WilliamMar 27, 2015 at 22:47
$\begingroup$
$\endgroup$
Sometimes you can you this alternative:
res = StringSplit["a b c d e f g", " "];
ToString @ Row[res, " "]
$\begingroup$
$\endgroup$
I know, I know, Kuba's and Pickett's solutions are preferable. For those who fancy Patterns here is an alternative.
StringReplace[res, t : __ :> t <> " "] // StringJoin // #~StringDrop~ -1 &
Doesn't look elegant, anyhow.....it works...hi,hi,hi.!
$\begingroup$
$\endgroup$
res = StringSplit["a b c d e f g", " "];
Fold[#1 <> " " <> #2 &, res]
(*"a b c d e f g"*)
answered Mar 27, 2015 at 21:04
res = StringSplit[s = " a b c d e f g ", " "]; StringJoin@Riffle[res, " "]; % == swill be False... fails with trailing spaces also. Don't think one can correctly reconstruct all stings from a string split... $\endgroup$s, is it... bottom line, there is no direct inverse for the title form for all strings. $\endgroup$Allas the optional third argument to the split allows correct reconstruction of all strings, but then you will have the "excess" (any leading/trailing/adjacent whitespace) as members of the split. $\endgroup$