Can I write a function that carries out symbolic calculations in Python 2.7?

I'm currently transitioning from Java to Python and have taken on the task of trying to create a calculator that can carry out symbolic operations on infix-notated mathematical expressions (without using custom modules like Sympy). Currently, it's built to accept strings that are space delimited and can only carry out the (, ), +, -, *, and / operators. Unfortunately, I can't figure out the basic algorithm for simplifying symbolic expressions.

For example, given the string '2 * ( ( 9 / 6 ) + 6 * x )', my program should carry out the following steps:

1. 2 * ( 1.5 + 6 * x )
2. 3 + 12 * x

But I can't get the program to ignore the x when distributing the 2. In addition, how can I handle 'x * 6 / x' so it returns '6' after simplification?

EDIT: To clarify, by "symbolic" I meant that it will leave letters like "A" and "f" in the output while carrying out the remaining calculations.

EDIT 2: I (mostly) finished the code. I'm posting it here if anyone stumbles on this post in the future, or if any of you were curious.

    def reduceExpr(useArray):

        # Use Python's native eval() to compute if no letters are detected.
        if (not hasLetters(useArray)):
            return [calculate(useArray)] # Different from eval() because it returns string version of result

        # Base case. Returns useArray if the list size is 1 (i.e., it contains one string). 
        if (len(useArray) == 1):
            return useArray

        # Base case. Returns the space-joined elements of useArray as a list with one string.
        if (len(useArray) == 3):
            return [' '.join(useArray)]

        # Checks to see if parentheses are present in the expression & sets.
        # Counts number of parentheses & keeps track of first ( found. 
        parentheses = 0
        leftIdx = -1

        # This try/except block is essentially an if/else block. Since useArray.index('(') triggers a KeyError
        # if it can't find '(' in useArray, the next line is not carried out, and parentheses is not incremented.
        try:
            leftIdx = useArray.index('(')
            parentheses += 1
        except Exception:
            pass

        # If a KeyError was returned, leftIdx = -1 and rightIdx = parentheses = 0.
        rightIdx = leftIdx + 1

        while (parentheses > 0):
            if (useArray[rightIdx] == '('):
                parentheses += 1
            elif (useArray[rightIdx] == ')'):
                parentheses -= 1
            rightIdx += 1

        # Provided parentheses pair isn't empty, runs contents through again; else, removes the parentheses
        if (leftIdx > -1 and rightIdx - leftIdx > 2):
            return reduceExpr(useArray[:leftIdx] + [' '.join(['(',reduceExpr(useArray[leftIdx+1:rightIdx-1])[0],')'])] + useArray[rightIdx:])
        elif (leftIdx > -1):
            return reduceExpr(useArray[:leftIdx] + useArray[rightIdx:])

        # If operator is + or -开发者_开发问答, hold the first two elements and process the rest of the list first
        if isAddSub(useArray[1]):
            return reduceExpr(useArray[:2] + reduceExpr(useArray[2:]))
        # Else, if operator is * or /, process the first 3 elements first, then the rest of the list
        elif isMultDiv(useArray[1]):
            return reduceExpr(reduceExpr(useArray[:3]) + useArray[3:])
        # Just placed this so the compiler wouldn't complain that the function had no return (since this was called by yet another function).
        return None

You need much more processing before you go into operations on symbols. The form you want to get to is a tree of operations with values in the leaf nodes. First you need to do a lexer run on the string to get elements - although if you always have space-separated elements it might be enough to just split the string. Then you need to parse that array of tokens using some grammar you require.

If you need theoretical information about grammars and parsing text, start here: http://en.wikipedia.org/wiki/Parsing If you need something more practical, go to https://github.com/pyparsing/pyparsing (you don't have to use the pyparsing module itself, but their documentation has a lot of interesting info) or http://www.nltk.org/book

From 2 * ( ( 9 / 6 ) + 6 * x ), you need to get to a tree like this:

      *
2           +
         /     *
        9 6   6 x

Then you can visit each node and decide if you want to simplify it. Constant operations will be the simplest ones to eliminate - just compute the result and exchange the "/" node with 1.5 because all children are constants.

There are many strategies to continue, but essentially you need to find a way to go through the tree and modify it until there's nothing left to change.

If you want to print the result then, just walk the tree again and produce an expression which describes it.

If you are parsing expressions in Python, you might consider Python syntax for the expressions and parse them using the ast module (AST = abstract syntax tree).

The advantages of using Python syntax: you don't have to make a separate language for the purpose, the parser is built in, and so is the evaluator. Disadvantages: there's quite a lot of extra complexity in the parse tree that you don't need (you can avoid some of it by using the built-in NodeVisitor and NodeTransformer classes to do your work).

>>> import ast
>>> a = ast.parse('x**2 + x', mode='eval')
>>> ast.dump(a)
"Expression(body=BinOp(left=BinOp(left=Name(id='x', ctx=Load()), op=Pow(),
right=Num(n=2)), op=Add(), right=Name(id='x', ctx=Load())))"

Here's an example class that walks a Python parse tree and does recursive constant folding (for binary operations), to show you the kind of thing you can do fairly easily.

from ast import *

class FoldConstants(NodeTransformer):
    def visit_BinOp(self, node):
        self.generic_visit(node)
        if isinstance(node.left, Num) and isinstance(node.right, Num):
            expr = copy_location(Expression(node), node)
            value = eval(compile(expr, '<string>', 'eval'))
            return copy_location(Num(value), node)
        else:
            return node

>>> ast.dump(FoldConstants().visit(ast.parse('3**2 - 5 + x', mode='eval')))
"Expression(body=BinOp(left=Num(n=4), op=Add(), right=Name(id='x', ctx=Load())))"