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It's easy to search if you break it down: Regex Meaning Mathematica command ------------------------------------------------- \w word character WordCharacter {2,3} repeat 2 to 3 times Repeated[..., {2, 3}] Combine it and use as: StringMatchQ[{"a", "ab", "abc", "abcd"}, Repeated[WordCharacter, {2, 3}]] (* {False, True, True, ...

What about this? StringSplit[string, i : NumberString :> i] Ok, everyone's giving answers that actually work with the $, so here's an edit, as @kguler and @MrWizard suggested StringSplit[string, i : ("" | "$" ~~ NumberString) :> i] // StringTrim

Short Version: Use ".*\".*" to match an embedded quote, ".*\\\\.*" to match an embedded backslash. This question deals with two distinct syntaxes -- Mathematica string syntax and regular expression syntax. Both syntaxes use \ as an escape character, so we'll need separate the two levels to see what is happening. First, let's deal with the Mathematica ...

Would this do the trick? selectWords[chars_, min_, max_] := Module[{charsset = Union[chars], charstally = Tally[chars], baselist, baselistchars, baselistpicks}, baselist = ToLowerCase[DictionaryLookup[x__ /; min <= StringLength[x] <= max]]]; baselistchars = Select[Characters /@ baselist, Complement[#, charsset] === {} &]; ...

Note that Rojo's solution splits the expression containing the dollar sign as well: StringSplit["there are 1234 words and numbers 5678 in here $999", i : NumberString :> i] {"there are ", "1234", " words and numbers ", "5678", " in here $", "999"} If you don't want that splitting to happen, here's one way, using a regex: StringSplit["there are 1234 ...

My attempt: First we define the existing row, using dots to represent empty squares, and our hand of 7 letters. row="...t.t...r..e.."; letters="aodalip"; Next define a function to count how many times each of our letters appears in a given string. Also run this function on our letters, to count how many of each we have. ...

Further variations: using StringReplace: List @@ StringTrim /@ StringReplace[string, a : Except[{"$", DigitCharacter}] .. | NumberString | ("$" ~~ NumberString) :> {a}] or, using the same replacement rule in StringCases: StringTrim /@ StringCases[string, a : Except[{"$", DigitCharacter}] .. | NumberString | ("$" ~~ NumberString) :> {a}] ...

I prefer the solutions given by Rojo and J.M. to the following one. But if you want to see a working version of your original approach with StringCases and RegularExpression, here is one possibility StringCases[string, RegularExpression["([A-Za-z]|\\s)+|(\\$|\\d)+"]] It returns {"there are ", "1234", " words and numbers ", "5678", " in here ","$999"} ...

The following may not fully solve your problem, but seems to logically belong here and is too long for a comment. I do use tabs in my code formatter. This question prompted me to write a palette for the formatter, which was long overdue (it surely can be improved). The palette should work with both "Input"-style and "Program"-style cells. The palette ...

I originally wrote this to help in guessing a word of known length from a bunch of letters, The word guessing game has 10 letters given so I tried to optimize this for speed. I reworked it here for scrabble: string = "hkxefri"; words = DictionaryLookup[{Apply[Alternatives, Characters[string]]} ..]; joined = StringJoin /@ ...

I find great utility in setting up grids using Ctrl, which inserts a place holder at the cursor. The input is interpreted as a List and you can create additional rows using CtrlReturn. Here's a walk-through of your Grid example above: Start with Grid[ and enter the first element in the list. Then press Ctrl,, which produces: Next, enter the second ...

I am not sure if this qualifies for an answer but it is too long for a comment therefore I put it here. Feel free to delete. In the examples the string matches the regular expression, meaning the result is not empty. Looking at the output we get StringCases["foo\"bar", RegularExpression[".*\".*"]] // InputForm (* {"foo\"bar"} *) StringCases["foo\"bar", ...