python开发任意表达式求值全功能示例

import math
opDict={}
def addoptr(ch, outLev, inLev, func, parmNum=2):
    obj= {'name':ch, 'out':outLev, 'in':inLev, 'func':func, 'parmNum':parmNum}
    opDict[ch]= obj
def makeList(x):
    if isinstance(x[-2], list):
        x[-2].append(x[-1])
        return x[-2].copy()
    else:
        ret= []
        ret.append(x[-2])
        ret.append(x[-1])
        return ret
addoptr('#', 1, 1, None)
addoptr('(', 90, 2, None)
addoptr(')', 2, None, None)
addoptr('[', 90, 2, None)
addoptr(']', 2, 2, None)
addoptr(',', 8, 9, makeList)
addoptr('&', 13, 14, lambda x: x[-1] and x[-2])
addoptr('and', 13, 14, lambda x: x[-1] and x[-2])
addoptr('|', 11, 12, lambda x: x[-1] or x[-2])
addoptr('or', 11, 12, lambda x: x[-1] or x[-2])
addoptr('~', 16, 17, lambda x: not x[-1],1)
addoptr('not', 16, 17, lambda x: not x[-1],1)
addoptr('=', 22, 23, lambda x: x[-1]==x[-2])
addoptr('>', 22, 23, lambda x: x[-2]>x[-1])
addoptr('<', 22, 23, lambda x: x[-2]<x[-1])
addoptr('>=', 22, 23, lambda x: x[-2]>=x[-1])
addoptr('<=', 22, 23, lambda x: x[-2]<=x[-1])
addoptr('!=', 22, 23, lambda x: x[-2]!=x[-1])
addoptr('<>', 22, 23, lambda x: x[-2]!=x[-1])
addoptr('in', 22, 23, lambda x: x[-2] in x[-1])
addoptr('+', 31, 32, lambda x: x[-2]+x[-1])
addoptr('-', 31, 32, lambda x: x[-2]-x[-1])
addoptr('*', 41, 42, lambda x: x[-2]*x[-1])
addoptr('/', 41, 42, lambda x: x[-2]/x[-1])
addoptr('//', 41, 42, lambda x: x[-2]//x[-1])
addoptr('%', 41, 42, lambda x: x[-2]%x[-1])
addoptr('neg', 51, 52, lambda x: -x[-1],1)
addoptr('**', 55, 56, lambda x: x[-2]**x[-1])
addoptr('sin', 61, 62, lambda x: math.sin(x[-1]),1)
alphabet= [chr(ord('a')+x) for x in range(26)]+[chr(ord('A')+x) for x in range(26)]
# print(opChar)
# print(opSep)
# print(alphabet)
def isfloat(str1):
    try:
        number = float(str1)
    except ValueError:
        return False
    return True
class exprEngine:
    def __init__(this, isVar=None, getValue=None):
        this.opndStack=[]
        this.optrStack=[]
        this.isVar= isVar
        this.getValue= getValue
        # 这个状态，特为负号/减号这一特殊符的双含义号所设置
        this.negState=0
        # 内建函数
        if isVar:
            addoptr('isvar', 61, 62, lambda x: isVar(x[-1]),1)
        # 处理识别
        this.oplen= len(max(opDict, key=lambda x:len(x)))
        this.opChar=[]
        for i in range(this.oplen):
            tmp=[x[0:i+1] for x in opDict if len(x)>=i+1]
            this.opChar.append(tmp)
        this.opSep= [x[0] for x in opDict if x[0] not in alphabet]+[' ', '\t']
        print(this.oplen)
        print(this.opChar)
        print(this.opSep)
    def readWord(this, cond):
        cond= cond.strip()
        if cond=='':
            return '', '#'
        if cond[0] in this.opChar[0]:
            l1=this.oplen
            for i in range(this.oplen):
                if cond[:i+1] not in this.opChar[i]:
                    l1= i
                    break
            print(l1)
            if cond[:l1] in this.opChar[l1-1]:
                return cond[:l1], 'optr'
        part= ''
        for ch in cond:
            if ch in this.opSep:
                break
            part+=ch
        return part, 'opnd'
    def pushoptr(this, optr):
        # 对负号/减号的特殊处理
        if optr=='-' and this.negState==0:
            # 这种情况，实际的含义是负号
            optr= 'neg'
        op= opDict[optr].copy()
        if len(this.optrStack)==0:
            this.optrStack.append(op)
            return
        opTop= this.optrStack[-1]
        if op['out']> opTop['in']:
            this.optrStack.append(op)
        elif op['out']< opTop['in']:
            this.popoptr()
            # 这里递归
            this.pushoptr(optr)
        elif op['out']== opTop['in']:
            # 消括号对，简单弹出
            this.optrStack.pop()
        this.negState=0
    def popoptr(this):
        opTop= this.optrStack[-1]
        a= opTop['parmNum']
        if len(this.opndStack)<a:
            raise Exception('操作数不足，可能有语法错误！')
        ret= opTop['func'](this.opndStack[-a:])
        this.opndStack= this.opndStack[:-a]
        this.opndStack.append(ret)
        this.optrStack.pop()
    def pushopnd(this, opnd):
        if opnd[0]=='"':
            # 肯定是字符串
            this.opndStack.append(opnd[1:])
        elif this.isVar and this.isVar(opnd):
            this.opndStack.append(this.getValue(opnd))
        else:
            if opnd.isdigit():
                this.opndStack.append(int(opnd))
            elif isfloat(opnd):
                this.opndStack.append(float(opnd))
            else:
                this.opndStack.append(opnd)
        this.negState=1
    def popopnd(this):
        if len(this.opndStack)==1:
            return this.opndStack[0]
        else:
            print(this.opndStack)
            print(this.optrStack)
            raise Exception('可能存在语法错误。')
    def eval(this, cond):
        this.optrStack=[]
        this.opndStack=[]
        this.pushoptr('#')
        while True:
            aWord,kind= this.readWord(cond)
            print(aword, cond)
            cond= cond[len(aword):].strip()
            if kind=='#':
                this.pushoptr('#')
                break
            elif kind=='optr':
                this.pushoptr(aword)
            else:
                if aword=='':
                    raise Exception('操作数为空，肯定有哪里错了。')
                this.pushopnd(aword)
            print(this.optrStack)
            print(this.opndStack)
        return this.popopnd()
if __name__=='__main__':
    # print(opDict)
    a= exprEngine()
    # a.addInfo('水位', '低')
    # b= a.eval('3 + 5 *2 = 13 and (3+5)*2=16 & 7-2 in [3,5,7] & 12>=15 or a in [a, b,c]')
    # b= a.eval('sin(-1)<1 and 3+-5=-2')
    # print(b)
    # b= a.eval('7*-3')
    b= a.eval('3**3=27 and 19%5=4 and 21//6=3')
    print(b)