relational/parser.py

# coding=UTF-8
# Relational
# Copyright (C) 2008  Salvo "LtWorf" Tomaselli
# 
# Relation is free software: you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
# 
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
# GNU General Public License for more details.
# 
# You should have received a copy of the GNU General Public License
# along with this program.  If not, see <http://www.gnu.org/licenses/>.
# 
# author Salvo "LtWorf" Tomaselli <tiposchi@tiscali.it>
def parse(expr):
    '''This function parses a relational algebra expression, converting it into python, 
    executable by eval function to get the result of the expression.
    It has 2 class of operators:
    without parameters
    *, -, ᑌ, ᑎ, ᐅᐊ, ᐅLEFTᐊ, ᐅRIGHTᐊ, ᐅFULLᐊ
    with parameters:
    σ, π, ρ
    
    Syntax for operators without parameters is:
    relation operator relation
    
    Syntax for operators with parameters is:
    operator parameters (relation)
    
    Since a*b is a relation itself, you can parse π a,b (a*b).
    And since π a,b (A) is a relation, you can parse π a,b (A) ᑌ B.
    
    You can use parenthesis to change priority: a ᐅᐊ (q ᑌ d).
    
    IMPORTANT: The encoding used by this module is UTF-8
    
    EXAMPLES
    σage > 25 and rank == weight(A)
    Q ᐅᐊ π a,b(A) ᐅᐊ B
    ρid➡i,name➡n(A) - π a,b(π a,b(A)) ᑎ σage > 25 or rank = weight(A)
    π a,b(π a,b(A))
    ρid➡i,name➡n(π a,b(A))
    A ᐅᐊ B
    '''
    symbols=("σ","π","ρ")
    
    starts=[]#List with starts and ends
    parenthesis=0
    lexpr=list(expr)
    #Parses the string finding all 1st level parenthesis
    for i in range(len(lexpr)):
        if lexpr[i]=="(":
            if parenthesis==0:
                starts.append(i+1)
            parenthesis+=1
        elif lexpr[i]==")":
            parenthesis-=1
            if parenthesis==0:
                starts.append(i)
            
    
    if len(starts)==0: #No parenthesis: no operators with parameters
        return parse_op(expr)
    
    while len(starts)>0:
        #Converting the last complex operator into python
        end=starts.pop()
        start=starts.pop()
        
        internal=parse(expr[start:end])
        
        endp=start-1
        symbol=""
        for i in range(endp,-1,-1):
            if expr[i:i+2] in symbols:
                symbol=expr[i:i+2]
                start=i+2
                break
            if expr[i:i+1] ==")":
                break #No symbol before
            
        parameters=expr[start:endp]
        
        res="" #String for result
        if symbol=="π":#Projection
            params=""
            count=0
            for i in parameters.split(","):
                if count!=0:
                    params+=","
                else:
                    count=1
                params+="\"%s\"" % (i.strip())

            res="%s.projection(%s)" % (internal,params)
            expr= ("%s%s%s") % (expr[0:start-2],res,expr[end+1:])
        elif symbol== "σ": #Selection
            res="%s.selection(\"%s\")" % (internal,parameters)
            expr= ("%s%s%s") % (expr[0:start-2],res,expr[end+1:])
        elif symbol=="ρ": #Rename
            params=parameters.replace(",","\",\"").replace("➡","\":\"").replace(" ","")
            res="%s.rename({\"%s\"})" % (internal,params)
            expr= ("%s%s%s") % (expr[0:start-2],res,expr[end+1:])
        else:
            res="(%s)" % (internal)
            expr= ("%s%s%s") % (expr[0:start-1],res,expr[end+1:])
        #Last complex operator is replaced with it's python code
        #Next cycle will do the same to the new last unparsed complex operator
        #At the end, parse_op will convert operators without parameters
    return parse_op(expr)

def parse_op(expr):
    '''This function parses a relational algebra expression including only operators 
    without parameters, converting it into python.
    Executable by eval function to get the result of the expression.'''
    
    result=""
    symbols={}
    
    symbols["*"]=".product(%s)"
    symbols["-"]=".difference(%s)"
    symbols["ᑌ"]=".union(%s)"
    symbols["ᑎ"]=".intersection(%s)"
    symbols["ᐅLEFTᐊ"]=".outer_left(%s)"
    symbols["ᐅRIGHTᐊ"]=".outer_right(%s)"
    symbols["ᐅFULLᐊ"]=".outer(%s)"
    symbols["ᐅᐊ"]=".join(%s)"
    
    
    #We want to avoid to parse expressions within quotes.
    #We split the string into an array, and we parse only the ones with even index
    quotes=expr.split('"');
    
    
    
    for i in range (0,len(quotes),2):
        for j in symbols:
            quotes[i]=quotes[i].replace(j,"_____%s_____"% (j))
    
    #The parts outside the quotes was parsed, put the string together again 
    if (len(quotes)>1):  
        expr= '"'.join(quotes)
    else:
        expr= quotes[0]
    
    tokens=expr.split("_____")
    
    i=0;
    tk_l=len(tokens)
    while i<tk_l:
        if tokens[i] not in symbols:
            result+=tokens[i].strip()
        else:
            result+=symbols[tokens[i]] % (tokens[i+1].strip())
            i+=1
        i+=1
    return result
            
if __name__=="__main__":
    while True:
        e=raw_input("Expression: ")
        print parse(e)