Python语言实现科学计算器
本文实例为大家分享了Python语言实现科学计算器的具体代码,供大家参考,具体内容如下
自学Python语言一个月,还是小白,发一个科学计算器的代码,希望大家批评指正,共勉嘛。
calculator.py
from tkinter import * from functools import partial from calculate import * # 生成计算器主界面 def buju(root): menu = Menu(root) # 菜单 submenu1 = Menu(menu, tearoff=0) # 分窗,0为在原窗,1为点击分为两个窗口 menu.add_cascade(label='编辑', menu=submenu1) # 添加子选项(label参数为显示内容) submenu1.add_command(label='复制', command=lambda: bianji(entry, 'copy')) # 添加命令 submenu1.add_command(label='剪切', command=lambda: bianji(entry, 'cut')) submenu1.add_command(label='粘贴', command=lambda: bianji(entry, 'paste')) submenu2 = Menu(menu, tearoff=0) menu.add_cascade(label='查看', menu=submenu2) submenu2.add_command(label='帮助', command=lambda: chakan(entry, 'help')) submenu2.add_command(label='作者', command=lambda: chakan(entry, 'author')) root.config(menu=menu) # 重新配置,添加菜单 label = Label(root, width=29, height=1, bd=5, bg='#FFFACD', anchor='se', textvariable=label_text) # 标签,可以显示文字或图片 label.grid(row=0, columnspan=5) # 布局器,向窗口注册并显示控件; rowspan:设置单元格纵向跨越的列数 entry = Entry(root, width=23, bd=5, bg='#FFFACD', justify="right", font=('微软雅黑', 12)) # 文本框(单行) entry.grid(row=1, column=0, columnspan=5, sticky=N + W + S + E, padx=5, pady=5) # 设置控件周围x、y方向空白区域保留大小 myButton = partial(Button, root, width=5, cursor='hand2', activebackground='#90EE90') # 偏函数:带有固定参数的函数 button_sin = myButton(text='sin', command=lambda: get_input(entry, 'sin(')) # 按钮 button_arcsin = myButton(text='arcsin', command=lambda: get_input(entry, 'arcsin(')) button_exp = myButton(text='e', command=lambda: get_input(entry, 'e')) button_ln = myButton(text='ln', command=lambda: get_input(entry, 'ln(')) button_xy = myButton(text='x^y', command=lambda: get_input(entry, '^')) button_sin.grid(row=2, column=0) button_arcsin.grid(row=2, column=1) button_exp.grid(row=2, column=2) button_ln.grid(row=2, column=3) button_xy.grid(row=2, column=4) button_shanyige = myButton(text='←', command=lambda: backspace(entry)) # command指定按钮消息的回调函数 button_shanquanbu = myButton(text=' C ', command=lambda: clear(entry)) button_zuokuohao = myButton(text='(', command=lambda: get_input(entry, '(')) button_youkuohao = myButton(text=')', command=lambda: get_input(entry, ')')) button_genhao = myButton(text='√x', command=lambda: get_input(entry, '√(')) button_shanyige.grid(row=3, column=0) button_shanquanbu.grid(row=3, column=1) button_zuokuohao.grid(row=3, column=2) button_youkuohao.grid(row=3, column=3) button_genhao.grid(row=3, column=4) button_7 = myButton(text=' 7 ', command=lambda: get_input(entry, '7')) button_8 = myButton(text=' 8 ', command=lambda: get_input(entry, '8')) button_9 = myButton(text=' 9 ', command=lambda: get_input(entry, '9')) button_chu = myButton(text=' / ', command=lambda: get_input(entry, '/')) button_yu = myButton(text='%', command=lambda: get_input(entry, '%')) button_7.grid(row=4, column=0) button_8.grid(row=4, column=1) button_9.grid(row=4, column=2) button_chu.grid(row=4, column=3) button_yu.grid(row=4, column=4) button_4 = myButton(text=' 4 ', command=lambda: get_input(entry, '4')) button_5 = myButton(text=' 5 ', command=lambda: get_input(entry, '5')) button_6 = myButton(text=' 6 ', command=lambda: get_input(entry, '6')) button_cheng = myButton(text=' * ', command=lambda: get_input(entry, '*')) button_jiecheng = myButton(text='二进制', command=lambda: jinzhi(entry)) button_4.grid(row=5, column=0) button_5.grid(row=5, column=1) button_6.grid(row=5, column=2) button_cheng.grid(row=5, column=3) button_jiecheng.grid(row=5, column=4) button_1 = myButton(text=' 1 ', command=lambda: get_input(entry, '1')) button_2 = myButton(text=' 2 ', command=lambda: get_input(entry, '2')) button_3 = myButton(text=' 3 ', command=lambda: get_input(entry, '3')) button_jian = myButton(text=' - ', command=lambda: get_input(entry, '-')) button_dengyu = myButton(text=' \n = \n ', command=lambda: calculator(entry)) button_1.grid(row=6, column=0) button_2.grid(row=6, column=1) button_3.grid(row=6, column=2) button_jian.grid(row=6, column=3) button_dengyu.grid(row=6, column=4, rowspan=2) # rowspan:设置单元格横向跨越的行数 button_pai = myButton(text=' π ', command=lambda: get_input(entry, 'π')) button_0 = myButton(text=' 0 ', command=lambda: get_input(entry, '0')) button_xiaoshudian = myButton(text=' . ', command=lambda: get_input(entry, '.')) button_jia = myButton(text=' + ', command=lambda: get_input(entry, '+')) button_pai.grid(row=7, column=0) button_0.grid(row=7, column=1) button_xiaoshudian.grid(row=7, column=2) button_jia.grid(row=7, column=3) # 对文本框中的算式或答案进行复制、剪切或粘贴 def bianji(entry, argu): """ :param entry: 文本框 :param argu: 按钮对应的值 """ if argu == 'copy': entry.event_generate("<<Copy>>") elif argu == 'cut': entry.event_generate("<<Cut>>") clear(entry) elif argu == 'paste': entry.event_generate("<<Paste>>") # 查看使用帮助和作者信息 def chakan(entry, argu): root = Tk() root.resizable(0, 0) text = Text(root, width=20, height=2, bd=5, bg='#FFFACD', font=('微软雅黑', 12)) text.grid(padx=5, pady=5) if argu == 'help': root.title('帮助') text.insert(INSERT, '这个计算器多简单!\n') text.insert(INSERT, '就别跟我要帮助了!') elif argu == 'author': root.title('作者') text.insert(INSERT, 'Author:冀梁栋\n') text.insert(INSERT, 'Time:2019-07-08') # 删除最后一次输入内容 def backspace(entry): entry.delete(len(entry.get()) - 1) # 删除文本框的最后一个输入值 # 删除所有输入内容和显示内容 def clear(entry): entry.delete(0, END) # 删除文本框的所有内容 label_text.set('') # 点击计算器输入按钮后向文本框中添加内容 def get_input(entry, argu): formula = entry.get() for char in formula: if '\u4e00' <= char <= '\u9fa5': clear(entry) # 删除文本框中的汉字显示,减少手动删除操作 entry.insert(INSERT, argu) # 使用END时,键盘敲入和按键输入组合操作会出错 # 十进制整数转换为二进制整数 def jinzhi(entry): try: formula = entry.get() if re.match('\d+$', formula): number = int(formula) cunchu = [] # 放置每次除以2后的余数 result = '' while number: cunchu.append(number % 2) number //= 2 # 整数除法,返回商 while cunchu: result += str(cunchu.pop()) # 将所有余数倒置得到结果 clear(entry) entry.insert(END, result) label_text.set(''.join(formula + '=')) else: clear(entry) entry.insert(END, '请输入十进制整数') except: clear(entry) entry.insert(END, '出错') # 点击“=”后进行计算 def calculator(entry): try: formula = entry.get() # 输入内容只是数字或π或e时,仍显示该内容 if re.match('-?[\d+,π,e]\.?\d*$', formula): label_text.set(''.join(formula + '=')) return # 输入内容是算式时,显示其计算结果 result = final_calc(formula_format(formula)) clear(entry) entry.insert(END, result) # 将结果输出到文本框中 label_text.set(''.join(formula + '=')) except: clear(entry) entry.insert(END, '出错') if __name__ == '__main__': root = Tk() # 生成窗口 root.title('理正计算器') # 窗口的名字 root.resizable(0, 0) # 窗口大小可调性,分别表示x,y方向的可变性 global label_text # 定义全局变量 label_text = StringVar() buju(root) root.mainloop() # 进入消息循环(必需组件),否则生成的窗口一闪而过
calculate.py
import re from math import * # 将算式从字符串处理成列表,解决横杠是负号还是减号的问题 def formula_format(formula): """ :param formula: str """ formula = re.sub(' ', '', formula) # 去掉算式中的空格s # 以 '横杠数字' 分割, 其中正则表达式:(\-\d+\.?\d*) 括号内: # \- 表示匹配横杠开头;\d+ 表示匹配数字1次或多次;\.?表示匹配小数点0次或1次;\d*表示匹配数字0次或多次。 formula_list = [i for i in re.split('(-[\d+,π,e]\.?\d*)', formula) if i] final_formula = [] # 最终的算式列表 for item in formula_list: # 算式以横杠开头,则第一个数字为负数,横杠为负号 if len(final_formula) == 0 and re.match('-[\d+,π,e]\.?\d*$', item): final_formula.append(item) continue # 如果当前的算式列表最后一个元素是运算符['+', '-', '*', '/', '(', '%', '^'], 则横杠为减号 if len(final_formula) > 0: if re.match('[\+\-\*\/\(\%\^]$', final_formula[-1]): final_formula.append(item) continue # 按照运算符分割开 item_split = [i for i in re.split('([\+\-\*\/\(\)\%\^\√])', item) if i] final_formula += item_split return final_formula # 判断是否是运算符,如果是返回True def is_operator(e): """ :param e: str :return: bool """ opers = ['+', '-', '*', '/', '(', ')', '%', '^', '√', 'sin', 'arcsin', 'ln'] return True if e in opers else False # 在for循环中嵌套使用if和else语句 # 比较连续两个运算符来判断是压栈还是弹栈 def decision(tail_op, now_op): """ :param tail_op: 运算符栈的最后一个运算符 :param now_op: 从算式列表取出的当前运算符 :return: 1代表弹栈运算,0代表弹出运算符栈最后一个元素'(',-1表示压栈 """ # 定义4种运算符级别 rate1 = ['+', '-'] rate2 = ['*', '/', '%'] rate3 = ['^', '√', 'sin', 'arcsin', 'ln'] rate4 = ['('] rate5 = [')'] if tail_op in rate1: if now_op in rate2 or now_op in rate3 or now_op in rate4: return -1 # 说明当前运算符优先级高于运算符栈的最后一个运算符,需要压栈 else: return 1 # 说明当前运算符优先级等于运算符栈的最后一个运算符,需要弹栈运算 elif tail_op in rate2: if now_op in rate3 or now_op in rate4: return -1 else: return 1 elif tail_op in rate3: if now_op in rate4: return -1 else: return 1 elif tail_op in rate4: if now_op in rate5: return 0 # '('遇上')',需要弹出'('并丢掉')',表明该括号内的算式已计算完成并将结果压入数字栈中 else: return -1 # 只要栈顶元素为'('且当前元素不是')',都应压入栈中 # 传入两个数字,一个运算符,根据运算符不同返回相应结果 def calculate(n1, n2, operator): """ :param n1: float :param n2: float :param operator: + - * / % ^ :return: float """ result = 0 if operator == '+': result = n1 + n2 if operator == '-': result = n1 - n2 if operator == '*': result = n1 * n2 if operator == '/': result = n1 / n2 if operator == '%': result = n1 % n2 if operator == '^': result = n1 ** n2 return result # 括号内的算式求出计算结果后,计算√()、sin()或arcsin() def gaojie(op_stack, num_stack): if op_stack[-1] == '√': op = op_stack.pop() num2 = num_stack.pop() num_stack.append(sqrt(num2)) elif op_stack[-1] == 'sin': op = op_stack.pop() num2 = num_stack.pop() num_stack.append(sin(num2)) elif op_stack[-1] == 'arcsin': op = op_stack.pop() num2 = num_stack.pop() num_stack.append(asin(num2)) elif op_stack[-1] == 'ln': op = op_stack.pop() num2 = num_stack.pop() num_stack.append(log(num2)) # 负责遍历算式列表中的字符,决定压入数字栈中或压入运算符栈中或弹栈运算 def final_calc(formula_list): """ :param formula_list: 算式列表 :return: 计算结果 """ num_stack = [] # 数字栈 op_stack = [] # 运算符栈 for item in formula_list: operator = is_operator(item) # 压入数字栈 if not operator: # π和e转换成可用于计算的值 if item == 'π': num_stack.append(pi) elif item == '-π': num_stack.append(-pi) elif item == 'e': num_stack.append(e) elif item == '-e': num_stack.append(-e) else: num_stack.append(float(item)) # 字符串转换为浮点数 # 如果是运算符 else: while True: # 如果运算符栈为空,则无条件入栈 if len(op_stack) == 0: op_stack.append(item) break # 决定压栈或弹栈 tag = decision(op_stack[-1], item) # 如果是-1,则压入运算符栈并进入下一次循环 if tag == -1: op_stack.append(item) break # 如果是0,则弹出运算符栈内最后一个'('并丢掉当前')',进入下一次循环 elif tag == 0: op_stack.pop() gaojie(op_stack, num_stack) # '('前是'√'、'sin'或'arcsin'时,对括号内算式的计算结果作相应的运算 break # 如果是1,则弹出运算符栈内最后一个元素和数字栈内最后两个元素 elif tag == 1: if item in ['√', 'sin', 'arcsin']: op_stack.append(item) break op = op_stack.pop() num2 = num_stack.pop() num1 = num_stack.pop() # 将计算结果压入数字栈并接着循环,直到遇到break跳出循环 num_stack.append(calculate(num1, num2, op)) # 大循环结束后,数字栈和运算符栈中可能还有元素的情况 while len(op_stack) != 0: op = op_stack.pop() num2 = num_stack.pop() num1 = num_stack.pop() num_stack.append(calculate(num1, num2, op)) result = str(num_stack[0]) # 去掉无效的0和小数点,例:1.0转换为1 if result[len(result) - 1] == '0' and result[len(result) - 2] == '.': result = result[0:-2] return result if __name__ == '__main__': # formula = "2 * ( 3 - 5 * ( - 6 + 3 * 2 / 2 ) )" formula = "arcsin ( 0 )" formula_list = formula_format(formula) result = final_calc(formula_list) print("算式:", formula) print("计算结果:", result)
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