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Parsing a phrase with Sprache(Words seperated by spaces)

By : Fery Prasetyanto
Date : October 18 2020, 06:10 AM
I think the issue was by ths following , to @PanagiotisKanavos, the trick would be to use the .Then() operator. The following works:
code :
public static Parser<string> WordParser =

public static Parser<string> PhraseParser =
    from leading in Parse.LetterOrDigit.Many().Text()
    from rest in Parse.Char(' ').Then(_ => WordParser).Many()
    select leading + " " + String.Join(" ", rest);

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In Ruby, how can I remove numeric words from a phrase without splitting phrase into words and losing word delimiters?

By : eu3dfx
Date : March 29 2020, 07:55 AM
like below fixes the issue I am trying to remove numeric words from a phrase. I am writing the following: , This regex does the job:
code :
phrase.gsub!(/\b\d+\b/, "")

How to find number of words in a phrase with spaces removed by checking v. dictionary

By : Jo Be
Date : March 29 2020, 07:55 AM
Hope that helps You can use Dynamic Programming for this task:
code :
f(0) = 0
f(i) = MIN { f(i-j) + (Dictionary.contais(s.substring(i-j,i)?1:INFINITY } for each j=1,...,i

Getting exception while parsing file using Sprache "Parsing failure: Unexpected end of input reached; expected =&qu

By : a2fong
Date : March 29 2020, 07:55 AM
may help you . Two problems: first, values can contain _ or . in your example, so LetterOrDigit won't cover it. Should be:
code :
internal static Parser<string> ValueText =
  Parse.LetterOrDigit.Or(Parse.Chars('_', '.')).AtLeastOnce().Text().Token();
internal static Parser<string> Identifier =

Recursive expression parsing in Sprache

By : Mike Myers
Date : March 29 2020, 07:55 AM
this one helps. What you have so far is a basic comparison expression parser. It looks like you want to wrap that in a parser that handles logical expressions (and, or, etc.) with sub-expression support.
The code I posted at first was ripped from poorly-tested code I was still working on, which didn't handle statements with multiple terms. My understanding of the ChainOperator method was clearly incomplete.
code :
// Helpers to make access simpler
public static class Condition
    // For testing, will fail all variable references
    public static Expression<Func<object, bool>> Parse(string text)
        => ConditionParser<object>.ParseCondition(text);

    public static Expression<Func<T, bool>> Parse<T>(string text)
        => ConditionParser<T>.ParseCondition(text);

    public static Expression<Func<T, bool>> Parse<T>(string text, T instance)
        => ConditionParser<T>.ParseCondition(text);

public static class ConditionParser<T>
    static ParameterExpression Parm = Expression.Parameter(typeof(T), "_");

    public static Expression<Func<T, bool>> ParseCondition(string text)
        => Lambda.Parse(text);

    static Parser<Expression<Func<T, bool>>> Lambda =>
        OrTerm.End().Select(body => Expression.Lambda<Func<T, bool>>(body, Parm));

    // lowest priority first
    static Parser<Expression> OrTerm =>
        Parse.ChainOperator(OpOr, AndTerm, Expression.MakeBinary);

    static Parser<ExpressionType> OpOr = MakeOperator("or", ExpressionType.OrElse);

    static Parser<Expression> AndTerm =>
        Parse.ChainOperator(OpAnd, NegateTerm, Expression.MakeBinary);

    static Parser<ExpressionType> OpAnd = MakeOperator("and", ExpressionType.AndAlso);

    static Parser<Expression> NegateTerm =>

    static Parser<Expression> NegatedFactor =>
        from negate in Parse.IgnoreCase("not").Token()
        from expr in Factor
        select Expression.Not(expr);

    static Parser<Expression> Factor =>

    static Parser<Expression> SubExpression =>
        from lparen in Parse.Char('(').Token()
        from expr in OrTerm
        from rparen in Parse.Char(')').Token()
        select expr;

    static Parser<Expression> BooleanValue =>

    static Parser<Expression> BooleanLiteral =>
        .Select(value => Expression.Constant(bool.Parse(value)));

    static Parser<Expression> BooleanVariable =>
        .Select(name => VariableAccess<bool>(name));

    static Expression VariableAccess<TTarget>(string name)
        MemberInfo mi = typeof(T).GetMember(name, MemberTypes.Field | MemberTypes.Property, BindingFlags.Instance | BindingFlags.Public).FirstOrDefault();
        var targetType = typeof(TTarget);
        var type = 
            (mi is FieldInfo fi) ? fi.FieldType : 
            (mi is PropertyInfo pi) ? pi.PropertyType : 
            throw new ParseException($"Variable '{name}' not found.");

        if (type != targetType)
            throw new ParseException($"Variable '{name}' is type '{type.Name}', expected '{targetType.Name}'");

        return Expression.MakeMemberAccess(Parm, mi);

    // Helper: define an operator parser
    static Parser<ExpressionType> MakeOperator(string token, ExpressionType type)
        => Parse.IgnoreCase(token).Token().Return(type);
static class Program
    static void Main()
        // Parser with no input
        var condition1 = Condition.Parse("true and false or true");
        var fn1 = condition1.Compile();
        Console.WriteLine("\t={0}", fn1(null));

        // Parser with record input
        var record1 = new { a = true, b = false };
        var record2 = new { a = false, b = true };
        var condition2 = Condition.Parse("a and b or not a", record);
        var fn2 = condition2.Compile();
        Console.WriteLine("\t{0} => {1}", record1.ToString(), fn2(record1));
        Console.WriteLine("\t{0} => {1}", record2.ToString(), fn2(record2));
static Parser<Expression> BooleanValue => 

How to read words seperated by spaces as a single value

By : user3144574
Date : March 29 2020, 07:55 AM
wish of those help Without seeing actual code, it's difficult to know exactly what's gone wrong here.
readlines will split a file on a newline delimiter. For complete cross-platform compatibility, open files in the "universal" compatibility mode (PEP-278 https://www.python.org/dev/peps/pep-0278/), which avoids questions about whether your lines end in '\n', '\r\n', or some other variation (depends on whether you're on a DOS or Unix-like system):
code :
with open('input.txt', 'rU') as t:
    lines = t.readlines()
results = list()
for line in lines:
    key,value = line.strip().split('\t')
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