Python 中的中缀评估

Question

我正在尝试将此处的代码 http://www.geeksforgeeks.org/expression-evaluation/ 转换为 python。但是，我运行遇到了一些麻烦，想不通。

class evaluateString:

def evalString(self,expression):
    valueStack = []
    opStack = []
    i=0
    while(i<len(expression)):
        if(expression[i] == ' '):
            continue
        if(expression[i]>='0' and expression[i] <= '9'):
            charNumber = [] #for storing number
            while(i<len(expression) and expression[i]>='0' and expression[i] <= '9'):
                charNumber.append(expression[i])
                i+=1
            valueStack.append(int(''.join(charNumber)))

        elif (expression[i]=='('):
            opStack.append(expression[i])

        elif (expression[i]==')'):
            while(opStack[-1]!='('):
                valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
                opStack.pop()
        elif(expression[i]=='+'or expression[i]=='-'or expression[i]=='*'or expression[i]=='/'):
            while( (len(opStack)!=0) and ( self.opPrecedence(expression[i],opStack[-1]) ) ):
                valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
                opStack.append(expression[i])
        i = i + 1

    while(len(opStack)!=0):
        valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))

    return valueStack.pop()


def applyOperation(self,op,a,b):
    if op=='+':
        return a+b
    elif op=='-':
        return a-b
    elif op=='*':
        return a*b
    elif op=='/':
        return a/b
    else:
        return 0

def opPrecedence(self,op1,op2):
    if (op2 == '(' or op2 == ')'):
        return False
    if ((op1 == '*' or op1 == '/') and (op2 == '+' or op2 == '-')):
        return False
    else:
        return True

a = evaluateString()
print(a.evalString("(5+7)"))

我能够在 valueStack 中获得正确的数字。但是，最后两个 elseif 似乎有问题。有人能指出我正确的方向吗？

Answer 1

我已经做了一些修复，它适用于某些操作。但我还没有对所有情况进行测试。此外，操作只是整数，没有浮点数（例如检查下面的最后一个输出）。

class evaluateString:

  def evalString(self,expression):
    valueStack = []
    opStack = []
    i=0

    while(i<len(expression)):
        if(expression[i] == ' '):
            continue
        if(expression[i]>='0' and expression[i] <= '9'):
            charNumber = [] #for storing number
            j = i
            while(j<len(expression) and expression[j]>='0' and expression[j] <= '9'):
                charNumber.append(expression[j])
                j += 1

            i = (j-1)
            valueStack.append(int(''.join(charNumber)))

        elif (expression[i]=='('):
            opStack.append(expression[i])

        elif (expression[i]==')'):
            while(opStack[-1]!='('):
                valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
            opStack.pop()
        elif(expression[i]=='+'or expression[i]=='-'or expression[i]=='*'or expression[i]=='/'):
            while( (len(opStack)!=0) and ( self.opPrecedence(expression[i],opStack[-1]) ) ):
                valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))
            opStack.append(expression[i])
        i = i + 1

    while(len(opStack)!=0):
        valueStack.append(self.applyOperation(opStack.pop(),valueStack.pop(),valueStack.pop()))

    return valueStack.pop()


  def applyOperation(self,op,a,b):
    if op=='+':
        return a+b
    elif op=='-':
        return b-a
    elif op=='*':
        return a*b
    elif op=='/':
        return b/a
    else:
        return 0

  def opPrecedence(self,op1,op2):
    if (op2 == '(' or op2 == ')'):
        return False
    if ((op1 == '*' or op1 == '/') and (op2 == '+' or op2 == '-')):
        return False
    else:
        return True

a = evaluateString()
print(a.evalString("8*12"))        #prints 96
print(a.evalString("(122-434)"))   #prints -312
print(a.evalString("(232+12)/2"))  #print 122
print(a.evalString("232/12+2"))    #prints 21

Answer 2

在 python 中，eval() 将计算中缀表达式

print(eval("(5+7)/2"))

它将计算的中缀表达式值打印为 6。

Python 中的中缀评估

Infix evaluation in Python

python

infix-operator