Why is
"Xabcde" /. "X" ~~ e__ -> e
"Xabcde"
and not
"abcde"
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1$\begingroup$ Related: mathematica.stackexchange.com/questions/8945/… $\endgroup$Sjoerd C. de Vries– Sjoerd C. de Vries2015-04-04 23:10:23 +00:00Commented Apr 4, 2015 at 23:10
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2 Answers
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Use
StringReplace["Xabcde", "X" ~~ e__ -> e].
Replace, et al are for lists/expressions...
Notice that AtomQ@"Xabcde" is True, so regular (non-string) replace operations only "see" it as a singular entity:
"Xabcde" /. "Xabcde" -> 1
(* 1 *)
From the docs for ReplaceAll: "... to transform each subpart..." - but there is no "subpart" for atoms, so regular replace operations only operate on the string as a complete entity.
If you want to do such things as part of a larger replacement program, something like this can be done:
test = {"Xabcde", {1, 2, 3}};
test /. {a_String :> StringReplace[a, "X" ~~ e__ -> e], {a_, b_, c_} :> {b, c}}
(* {"abcde", {2,3}} *)
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$\begingroup$ So that form of pattern will not work with
/.? $\endgroup$orome– orome2015-04-04 22:11:57 +00:00Commented Apr 4, 2015 at 22:11 -
$\begingroup$ @raxacoricofallapatorius: Not for your intended result on strings... $\endgroup$ciao– ciao2015-04-04 22:12:57 +00:00Commented Apr 4, 2015 at 22:12
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$\begingroup$ Is there a pattern I can use with
/.to get the result I'm looking for? $\endgroup$orome– orome2015-04-04 22:14:50 +00:00Commented Apr 4, 2015 at 22:14 -
$\begingroup$ @raxacoricofallapatorius: Perhaps you should fill in more details on what you're trying to accomplish in the OP? $\endgroup$ciao– ciao2015-04-04 22:16:48 +00:00Commented Apr 4, 2015 at 22:16
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$\begingroup$ Exactly that: Find a pattern to use with
/.to achieve string pattern matching and replacement. $\endgroup$orome– orome2015-04-04 22:18:32 +00:00Commented Apr 4, 2015 at 22:18
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"Xabcde" /. a_ :> StringCases[a, _] /. {"X", b__} :> StringJoin[b]
(*"abcde"*)
