1 import re 2 import functools 3 4 5 def minus_operator_handler(formula): 6 '''處理一些特殊的減號(hào)運(yùn)算''' 7 minus_operators = re.split("-",formula) 8 calc_list= re.findall("[0-9]",formula) 9 if minus_operators[0] == '': #第一值肯定是負(fù)號(hào) 10 calc_list[0] = '-%s' % calc_list[0] 11 res = functools.reduce(lambda x,y:float(x) - float(y), calc_list) 12 print("\033[33;1m減號(hào)[%s]處理結(jié)果:\033[0m" % formula, res ) 13 return res 14 15 def remove_duplicates(formula): 16 formula = formula.replace("++","+") 17 formula = formula.replace("+-","-") 18 formula = formula.replace("-+","-") 19 formula = formula.replace("--","+") 20 formula = formula.replace("- -","+") 21 return formula 22 def compute_mutiply_and_dividend(formula): 23 '''算乘除,傳進(jìn)來的是字符串噢''' 24 operators = re.findall("[*/]", formula ) 25 calc_list = re.split("[*/]", formula ) 26 res = None 27 for index,i in enumerate(calc_list): 28 if res: 29 if operators[index-1] == "*": 30 res *= float(i) 31 elif operators[index-1] == "/": 32 res /= float(i) 33 else: 34 res = float(i) 35 36 print("\033[31;1m[%s]運(yùn)算結(jié)果=\033[0m" %formula, res ) 37 return res 38 def handle_minus_in_list(operator_list,calc_list): 39 '''有的時(shí)候把算術(shù)符和值分開后,會(huì)出現(xiàn)這種情況 ['-', '-', '-'] [' ', '14969037.996825399 ', ' ', '12.0/ 10.0 '] 40 這需要把第2個(gè)列表中的空格都變成負(fù)號(hào)并與其后面的值拼起來,惡心死了 41 ''' 42 for index,i in enumerate(calc_list): 43 if i == '': #它其實(shí)是代表負(fù)號(hào),改成負(fù)號(hào) 44 calc_list[index+1] = i + calc_list[index+1].strip() 45 def handle_special_occactions(plus_and_minus_operators,multiply_and_dividend): 46 '''有時(shí)會(huì)出現(xiàn)這種情況 , ['-', '-'] ['1 ', ' 2 * ', '14969036.7968254'],2*...后面這段實(shí)際是 2*-14969036.7968254,需要特別處理下,太惡心了''' 47 for index,i in enumerate(multiply_and_dividend): 48 i = i.strip() 49 if i.endswith("*") or i.endswith("/"): 50 multiply_and_dividend[index] = multiply_and_dividend[index] + plus_and_minus_operators[index] + multiply_and_dividend[index+1] 51 del multiply_and_dividend[index+1] 52 del plus_and_minus_operators[index] 53 return plus_and_minus_operators,multiply_and_dividend 54 def compute(formula): 55 '''這里計(jì)算是的不帶括號(hào)的公式''' 56 57 formula = formula.strip("()") #去除外面包的拓號(hào) 58 formula = remove_duplicates(formula) #去除外重復(fù)的+-號(hào) 59 plus_and_minus_operators = re.findall("[+-]", formula) 60 multiply_and_dividend = re.split("[+-]", formula) #取出乘除公式 61 if len(multiply_and_dividend[0].strip()) == 0:#代表這肯定是個(gè)減號(hào) 62 multiply_and_dividend[1] = plus_and_minus_operators[0] + multiply_and_dividend[1] 63 del multiply_and_dividend[0] 64 del plus_and_minus_operators[0] 65 66 plus_and_minus_operators,multiply_and_dividend=handle_special_occactions(plus_and_minus_operators,multiply_and_dividend) 67 for index,i in enumerate(multiply_and_dividend): 68 if re.search("[*/]" ,i): 69 sub_res = compute_mutiply_and_dividend(i) 70 multiply_and_dividend[index] = sub_res 71 72 #開始運(yùn)算+,- 73 print(multiply_and_dividend, plus_and_minus_operators) 74 total_res = None 75 for index,item in enumerate(multiply_and_dividend): 76 if total_res: #代表不是第一次循環(huán) 77 if plus_and_minus_operators[index-1] == '+': 78 total_res += float(item) 79 elif plus_and_minus_operators[index-1] == '-': 80 total_res -= float(item) 81 else: 82 total_res = float(item) 83 print("\033[32;1m[%s]運(yùn)算結(jié)果:\033[0m" %formula,total_res) 84 return total_res 85 86 def calc(formula): 87 '''計(jì)算程序主入口, 主要邏輯是先計(jì)算拓號(hào)里的值,算出來后再算乘除,再算加減''' 88 parenthesise_flag = True 89 calc_res = None #初始化運(yùn)算結(jié)果為None,還沒開始運(yùn)算呢,當(dāng)然為None啦 90 while parenthesise_flag: 91 m = re.search("\([^()]*\)", formula) #找到最里層的拓號(hào) 92 if m: 93 #print("先算拓號(hào)里的值:",m.group()) 94 sub_res = compute(m.group()) 95 formula = formula.replace(m.group(),str(sub_res)) 96 else: 97 print('\033[41;1m----沒拓號(hào)了...---\033[0m') 98 99 print('\n\n\033[42;1m最終結(jié)果:\033[0m',compute(formula)) 100 parenthesise_flag = False #代表公式里的拓號(hào)已經(jīng)都被剝除啦 101 102 if __name__ == '__main__': 103 104 #res = calc("1 - 2 * ( (60-30 +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )") 105 res = calc("1 - 2 * ( (60-30 +(-9-2-5-2*3-5/3-40*4/2-3/5+6*3) * (-9-2-5-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )")

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轉(zhuǎn)載于:https://www.cnblogs.com/JerryZao/p/8681379.html

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