Prefix notation to infix notation in Python

Question

I am writing a small calculator (with prefix notation) and I'm curious how I'd convert prefix notation to infix notation. I currently have a function, but it's being weird, and I'm not sure how to fix it. By being weird, I mean that if given ['+', x, y] it will return (() + x + () + y) which is confusing me. Here's the code.

def pre_in(read):
    #print read
    tempOp = read[0]
    body = read[1:]
    expr = []
    for i in range(len(body)-1):
        if not isinstance(body[i], list) and body[i] != " ":
            expr.append(str(body[i]))
            expr.append(tempOp)
        else:
            expr.append(str(pre_in(body[i])))
            expr.append(tempOp)
    try:
        if not isinstance(body[-1], list):
            expr.append(str(body[-1]))
        else:
            expr.append(str(pre_in(body[-1])))
    except:
        pass
    if expr != None: return "("+' '.join(expr)+")"

What am I doing wrong?

Answer 1

Actually your code works fine.

print pre_in ( ['+', 8, 9] )

yields

(8 + 9)

EDIT: As the others have stated, maybe you want to use a stack. Here a simple sandbox implementation with some examples (it produces many parenthesis but those don't hurt):

class Calculator:
    def __init__ (self):
        self.stack = []

    def push (self, p):
        if p in ['+', '-', '*', '/']:
            op1 = self.stack.pop ()
            op2 = self.stack.pop ()
            self.stack.append ('(%s %s %s)' % (op1, p, op2) )
        elif p == '!':
            op = self.stack.pop ()
            self.stack.append ('%s!' % (op) )
        elif p in ['sin', 'cos', 'tan']:
            op = self.stack.pop ()
            self.stack.append ('%s(%s)' % (p, op) )
        else:
            self.stack.append (p)

    def convert (self, l):
        l.reverse ()
        for e in l:
            self.push (e)
        return self.stack.pop ()

c = Calculator ()

print c.convert ( ['+', 8, 9] )
print c.convert ( ['!', 42] )
print c.convert ( ['sin', 'pi'] )
print c.convert ( ['+', 'sin', '/', 'x', 2, 'cos', '/', 'x', 3] )

Answer 2

If your aim is not to develop the algorithm on your own, go to this page. There are links to two pages which explain the infix->postfix and postfix->infix algorithm. (And also, if you want to know how the algorithms are implemented in javascript, you can take a look at the source code of the page.)