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I have large StringExpression that is used in StringCases. Some of the parts being pulled out are integer strings. These pull out fine. However, when I go to convert them from strings to integers with FromDigits (it can take a string of integers) in StringCases it errors but still converts the string to an integer. Minimal example:

ClearAll[BigStringExpr];
Off[RuleDelayed::rhs];
BigStringExpr[anInteger_Symbol] := 
 StartOfString ~~ 
  anInteger : (Except["0", DigitCharacter] ~~ DigitCharacter ...) ~~ 
  EndOfString
On[RuleDelayed::rhs];

StringCases["1234", BigStringExpr[y] -> y]
(* {"1234"} *)

StringCases["1234", BigStringExpr[y] -> FromDigits@y]
FromDigits::nlst: The expression y is not a list of digits or a string of valid digits. >>
(* {1234} *)

Any ideas why this is happening? I know that I can convert after StringCases but I would prefer to keep it in there since it is a simple call and also helps with code readability.

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    $\begingroup$ Use RuleDelayed (:>) rather than Rule (->) in your StringCases; this will force evaluation of your FromDigits function at the opportune time, i.e. when $y$ has been given a value. $\endgroup$
    – MarcoB
    Aug 19, 2015 at 14:19
  • $\begingroup$ @MarcoB That did it. I have to keep that whole delayed thing in mind as I always forget its significance. Executes after replacement. {x, x} /. x -> RandomReal[] verses {x, x} /. x :> RandomReal[]. Thanks. $\endgroup$
    – Edmund
    Aug 19, 2015 at 14:25

1 Answer 1

Reset to default
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ClearAll[BigStringExpr];
BigStringExpr = (StartOfString ~~ 
    anInteger : (Except["0", DigitCharacter] ~~ DigitCharacter ...) ~~
      EndOfString :> ToExpression@anInteger)
StringCases["1234", BigStringExpr]
Head[%[[1]]]

{1234}

Integer

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  • $\begingroup$ Thanks, but I need to define the string expression as I have because it is in a package. The method I've used allows me to reference the named parts of the expression from the package without using the full context path. See question 91851. $\endgroup$
    – Edmund
    Aug 19, 2015 at 14:29
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    $\begingroup$ @Edmund It is not the question of how you define the string expression that is the problem. It is the use of Rule. george2079 is using RuleDelayed (i.e., :>). This is the key point. This works StringCases["1234", BigStringExpr[y] :> FromDigits@y] $\endgroup$
    – Jack LaVigne
    Aug 19, 2015 at 21:11

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