Question

I am currently working on the python problemsets on a website called singpath. The question is:

Prefix Evaluation
Create a function that evaluates the arithmetic expression in the form of prefix notation without spaces or syntax errors. The expression is given as a string, all the numbers in the expression are integer 0~9, and the operators are +(addition), -(subtraction), *(multiplication), /(division), %(modulo), which operate just the same as those in Python.
Prefix notation, also known as Polish notation, is a form of notation for logic, arithmetic, and algebra. it places operators to the left of their operands. If the arity of the operators is fixed, the result is a syntax lacking parentheses or other brackets that can still be parsed without ambiguity.

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This seems simple enough but the string is condensed with no spaces in the input to splice out the data. How could I separate the data from the string without importing modules? Furthermore how could I use the results of the data to solve the given equation? Also please keep in minf that Singpath solutions must be in ONE function that cannot use methods that couldn’t be found in the standard python library. This also includes functions declared within the solution :S

Examples:

>>> eval_prefix("+34")
7
>>> eval_prefix("*−567")
-7
>>> eval_prefix("-*33+2+11")
5
>>> eval_prefix("-+5*+1243")
14
>>> eval_prefix("*+35-72")
40
>>> eval_prefix("%3/52")
1

See my point no spaces D:

Answer 1

OK, not as snazzy as alex jordan’s lamba/reduce solution, but it doesn’t choke on garbage input. It’s sort of a recursive descent parser meets bubble sort abomination (I’m thinking it could be a little more efficient when it finds a solvable portion than just jumping back to the start. 😉

import operator
def eval_prefix(expr):
    d = {'+': operator.add,
         '-': operator.sub,
         '*': operator.mul,
         '/': operator.div, # for 3.x change this to operator.truediv
         '%': operator.mod}
    for n in range(10):
        d[str(n)] = n
    e = list(d.get(e, None) for e in expr)
    i = 0
    while i + 3 <= len(e):
        o, l, r = e[i:i+3]
        if type(o) == type(operator.add) and type(l) == type(r) == type(0):
            e[i:i+3] = [o(l, r)]
            i = 0
        else:
            i += 1
    if len(e) != 1:
        print 'Error in expression:', expr
        return 0
    else:
        return e[0]

def test(s, v):
    r = eval_prefix(s)
    print s, '==', v, r, r == v

test("+34", 7)
test("*-567", -7)
test("-*33+2+11", 5)
test("-+5*+1243", 14)
test("*+35-72", 40)
test("%3/52", 1)
test("****", 0)
test("-5bob", 10)