本篇内容介绍了“python如何开发任意表达式的求值全功能”的有关知识，在实际案例的操作过程中，不少人都会遇到这样的困境，接下来就让小编带领大家学习一下如何处理这些情况吧！希望大家仔细阅读，能够学有所成！
完整的源代码
import mathopDict={}def addoptr(ch, outLev, inLev, func, parmNum=2): obj= {'name':ch, 'out':outLev, 'in':inLev, 'func':func, 'parmNum':parmNum} opDict[ch]= objdef 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 retaddoptr('#', 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 Trueclass 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)“python如何开发任意表达式的求值全功能”的内容就介绍到这里了，感谢大家的阅读。如果想了解更多行业相关的知识可以关注编程网网站，小编将为大家输出更多高质量的实用文章！
