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Infix expression evaluation using two stacks.

Python, 80 lines
 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80``` ```# Infix Expression Evaluation # The expression must contain only positive numbers, # aritmetic operators and parentheses. # FB - 20151107 def isOp(c): if c != "": return (c in "+-*/") else: return False def pri(c): # operator priority if c in "+-": return 0 if c in "*/": return 1 def isNum(c): if c != "": return (c in "0123456789.") else: return False def calc(op, num1, num2): if op == "+": return str(float(num1) + float(num2)) if op == "-": return str(float(num1) - float(num2)) if op == "*": return str(float(num1) * float(num2)) if op == "/": return str(float(num1) / float(num2)) def Infix(expr): expr = list(expr) stackChr = list() # character stack stackNum = list() # number stack num = "" while len(expr) > 0: c = expr.pop(0) if len(expr) > 0: d = expr[0] else: d = "" if isNum(c): num += c if not isNum(d): stackNum.append(num) num = "" elif isOp(c): while True: if len(stackChr) > 0: top = stackChr[-1] else: top = "" if isOp(top): if not pri(c) > pri(top): num2 = stackNum.pop() op = stackChr.pop() num1 = stackNum.pop() stackNum.append(calc(op, num1, num2)) else: stackChr.append(c) break else: stackChr.append(c) break elif c == "(": stackChr.append(c) elif c == ")": while len(stackChr) > 0: c = stackChr.pop() if c == "(": break elif isOp(c): num2 = stackNum.pop() num1 = stackNum.pop() stackNum.append(calc(c, num1, num2)) while len(stackChr) > 0: c = stackChr.pop() if c == "(": break elif isOp(c): num2 = stackNum.pop() num1 = stackNum.pop() stackNum.append(calc(c, num1, num2)) return stackNum.pop() # TEST expr = "1.7+2.8*3.6/(0-9.4)+5/(0.36-7.7)/9.12*11" print "EXPR: " + expr print "EVAL: " + str(eval(expr)) print "INFX: " + Infix(expr) ```
 Created by FB36 on Sun, 8 Nov 2015 (MIT)

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