标签:bst rac port 规则 定义 nbsp python pytho print
弄了6个小时,各种出错,伪静态规则出错1次,
def逻辑思维出错1次。哈哈
# 实现加减乘除及拓号优先级解析 # 用户输入 1 - 2 * ( (60-30 +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )等类似公式后,必须自己解析里面的(),+,-,*,/符号和公式,运算后得出结果,结果必须与真实的计算器所得出的结果一致 import re equation = "1 - 2 * ( (60-30 +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )" equation = re.sub(" ","",equation)#去除空格 ‘‘‘ 英文不好,先看翻译 addition --加 subtraction -- 减 multiplication --乘 division --除 ‘‘‘ #定义四个基础算法 def addition(a,b): return a+b def subtraction(a,b): return a-b def multiplication(a,b): return a*b def division(a,b): return a/b def operator_substitution(s): #将--或者+-转意为单符号 s = re.sub("--","+",s,1) s = re.sub("\+\-", "-", s, 1) return s obj = re.compile("(^\-)?(\d+\.)?\d+[\*|/]\-?\d+(\.\d+)?") obj2 = re.compile("(^\-)?(\d+\.)?\d+[\+|\-]\-?\d+(\.\d+)?") def yunsuan(s): if "*" in s or "/" in s: prt = re.search(obj,s).group() hao = re.match(obj,prt).group() if "*" in hao: pp = "*" k1 = prt.split(pp) print(k1) de = multiplication(float(k1[0]), float(k1[1])) elif "/" in hao: pp = "/" k1 = prt.split(pp) de = division(float(k1[0]), float(k1[1])) # de = int(de) s = re.sub(obj,str(de),s,1) s = operator_substitution(s) return s elif "+" in s or "-" in s: prt = re.search(obj2,s).group() hao = re.match(obj2,prt).group() if "+" in hao: pp = "+" k1 = prt.split(pp) de = addition(float(k1[0]), float(k1[1])) elif "-" in hao: pp = "-" k1 = prt.split(pp) de = subtraction(float(k1[0]),float(k1[1])) s = re.sub(obj2, str(de), s, 1) s = operator_substitution(s) return s def zhongji(s): while "*" in s or "/" in s or "+" in s or "-" in s and s.find("-") !=0: s = yunsuan(s) return s def teji(s): prt = re.search("\([^()]+\)", s).group() krt = re.search("[^(]+[^)]",prt).group() if krt.find("-") == 0: krt = re.sub("-","",krt) krt = -float(zhongji(krt)) else: krt = zhongji(krt) s = re.sub("\([^()]+\)",str(krt),s,1) s = operator_substitution(s) return s def yesok(s): while "(" in s: s = teji(s) print(s) s = zhongji(s) return s print(yesok(equation)) print(1-2*((60-30+(-40/5)*(9-2*5/3+7/3*99/4*2998+10*568/14))-(-4*3)/(16-3*2)))
精简了一下,没有优化程序
# 实现加减乘除及拓号优先级解析 # 用户输入 1 - 2 * ( (60-30 +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )等类似公式后,必须自己解析里面的(),+,-,*,/符号和公式,运算后得出结果,结果必须与真实的计算器所得出的结果一致 import re equation = "1 - 2 * ( (60-30 +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )" equation = re.sub(" ","",equation) def addition(a,b): return a+b def subtraction(a,b): return a-b def multiplication(a,b): return a*b def division(a,b): return a/b def operator_substitution(s): #将--或者+-转意为单符号 s = re.sub("--","+",s,1) s = re.sub("\+\-", "-", s, 1) return s obj = re.compile("(^\-)?(\d+\.)?\d+[\*|/]\-?\d+(\.\d+)?") obj2 = re.compile("(^\-)?(\d+\.)?\d+[\+|\-]\-?\d+(\.\d+)?") def yunsuan(s): if "*" in s or "/" in s: prt = re.search(obj,s).group() hao = re.match(obj,prt).group() if "*" in hao: pp = "*" k1 = prt.split(pp) de = multiplication(float(k1[0]), float(k1[1])) elif "/" in hao: pp = "/" k1 = prt.split(pp) de = division(float(k1[0]), float(k1[1])) s = re.sub(obj,str(de),s,1) s = operator_substitution(s) return s elif "+" in s or "-" in s: prt = re.search(obj2,s).group() hao = re.match(obj2,prt).group() if "+" in hao: pp = "+" k1 = prt.split(pp) de = addition(float(k1[0]), float(k1[1])) elif "-" in hao: pp = "-" k1 = prt.split(pp) de = subtraction(float(k1[0]),float(k1[1])) s = re.sub(obj2, str(de), s, 1) s = operator_substitution(s) return s def zhongji(s): while "*" in s or "/" in s or "+" in s or "-" in s and s.find("-") !=0: s = yunsuan(s) return s def teji(s): prt = re.search("\([^()]+\)", s).group() krt = re.search("[^(]+[^)]",prt).group() if krt.find("-") == 0: krt = re.sub("-","",krt) krt = -float(zhongji(krt)) else: krt = zhongji(krt) s = re.sub("\([^()]+\)",str(krt),s,1) s = operator_substitution(s) return s def yesok(s): while "(" in s: s = teji(s) s = zhongji(s) return s print(yesok(equation)) print(yesok("1+1-(105/999)")) print(yesok("8+9*85-(-898/58+666-8989/20)"))
标签:bst rac port 规则 定义 nbsp python pytho print
原文地址:https://www.cnblogs.com/bdua/p/12242995.html