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?

I don't understand the problem (and your code). If I have foo = ['+', x, y], the expression [foo[1], foo[0], foo[2]] will result in [x, '+', y]. Isn't that what you want? In case of nested expressions, you'd have to do simple recursion. Maybe you should give a clearer and more complex example of your input and expected output. — Björn Pollex
– Björn Pollex, Commented Jun 27, 2011 at 20:12
you could also try using a stack, that is a common way of doing prefix<->infix, the stack would also solve nested expressions. — Hunter McMillen
– Hunter McMillen, Commented Jun 27, 2011 at 20:13
appears to be related to previous question by same guy: stackoverflow.com/questions/6338440/small-language-in-python — Warren P
– Warren P, Commented Jun 27, 2011 at 20:15
Space_C0wb0y: i was aiming for something that can handle multiple terms, like ['+', 2, 3, 4, 5] would yield 2 + 3 + 4 + 5 — tekknolagi
– tekknolagi, Commented Jun 27, 2011 at 20:22
@Warren actually no. this is about prefix, the other was about postfix. i'm rethinking the structure of the language — tekknolagi
– tekknolagi, Commented Jun 27, 2011 at 20:22

Hyperboreus · Accepted Answer · 2011-06-27 20:41:34Z

5

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] )

edited Jun 27, 2011 at 20:41

answered Jun 27, 2011 at 20:16

Hyperboreus

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7 Comments

tekknolagi Over a year ago

i normally run it on expressions with variables, like: ['+', x, y]

Hyperboreus Over a year ago

Also x = 8; y = 9; print pre_in ( ['+', x, y] ) runs fine.

tekknolagi Over a year ago

what i meant was ['+', 'x', 'y']

tekknolagi Over a year ago

i'm sorry, i was feeding it bad input.

Hyperboreus Over a year ago

Aren't / x + y z and / + y z x equivalent?

|

Seyf · Accepted Answer · 2011-06-27 20:13:58Z

1

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.)

answered Jun 27, 2011 at 20:13

Seyf

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Comments

martega · Accepted Answer · 2012-06-22 20:11:19Z

1

Here's a fairly simple recursive solution.

def prefix_to_infix(expr):
    if type(expr) != type([]):
        # The expression is a number or variable.
        return str(expr)
    elif len(expr) == 2:
        # This is an operator expression with 1 argument.
        return str(expr[1])
    else:
        # This is an operator expression with 2 or more arguments.
        operator = expr[0]
        left_arg = prefix_to_infix([operator] + expr[1:-1])
        right_arg = prefix_to_infix(expr[-1])
        return "({0}{1}{2})".format(left_arg, operator, right_arg)

# prefix_to_infix(['+',1,2,3,4,5]) = '((((1+2)+3)+4)+5)'
# prefix_to_infix(['+',1,2,['*',3,4,5],6,7,8]) = '(((((1+2)+(3*4*5))+6)+7)+8)'

edited Jun 22, 2012 at 20:11

answered Jun 22, 2012 at 20:05

martega

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2 Comments

tekknolagi Over a year ago

But with trig functions and such?

martega Over a year ago

I was working under the assumption that the syntax for your prefix expressions were of the form ['operator', arg1, arg2, ... ,argN]. I also assumed that all operators were binary operators and left associative. It wouldn't really make sense to make trig functions in infix notation but maybe I'm just misunderstanding your question.

mjv · Accepted Answer · 2017-07-11 23:31:40Z

1

At the risk of being a bit overkill for this kind of simple parsing/conversion jobs, you may want to look at pyparsing.

edited Jul 11, 2017 at 23:31

answered Jun 27, 2011 at 20:22

mjv

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