一、算法
???? 1、算法的主要思想就是將一個中綴表達式(Infix expression)轉換成便于處理的后綴表達式(Postfix expression)，然后借助于棧這個簡單的數據結構，計算出表達式的結果。
???? 2、關于如何講普通的表達式轉換成后綴表達式，以及如何處理后綴表達式并計算出結果的具體算法描述不在此敘述了，書上有詳細的說明。
二、簡易計算器
使用說明
使用該計算器類的簡單示例如下：
# usage
c = Calculator()
print('result: {:f}'.formart(c.get_result('1.11+2.22-3.33*4.44/5.55')))
# output:
result: 0.666000
測試案例
為了對這個計算器進行有效地檢驗，設計了幾組測試案例，測試結果如下：
Test No.1: (1.11) = 1.110000
Test No.2: 1.11+2.22-3.33*4.44/5.55 = 0.666000
Test No.3: 1.11+(2.22-3.33)*4.44/5.55 = 0.222000
Test No.4: 1.11+(2.22-3.33)*(4.44+5.55)/6.66 = -0.555000
Test No.5: 1.11*((2.22-3.33)*(4.44+5.55))/(6.66+7.77) = -0.852992
Test No.6: (1.11+2.22)*(3.33+4.44)/5.55*6.66 = 31.048920
Test No.7: (1.11-2.22)/(3.33+4.44)/5.55*(6.66+7.77)/(8.88) = -0.041828
Test No.8: Error: (1.11+2.22)*(3.33+4.44: missing ")", please check your expression
Test No.9: Error: (1.11+2.22)*3.33/0+(34-45): divisor cannot be zero
Test No.10: Error: 12+89^7: invalid character: ^
實現代碼
棧的實現
棧實際上就是一個被限制操作的表，所有的操作只能在棧的頂端（入棧、出棧等），以下是使用Python代碼實現的簡單的棧：
class Stack(object):
"""
The structure of a Stack.
The user don't have to know the definition.
"""
def __init__(self):
self.__container = list()
def __is_empty(self):
"""
Test if the stack is empty or not
:return: True or False
"""
return len(self.__container) == 0
def push(self, element):
"""
Add a new element to the stack
:param element: the element you want to add
:return: None
"""
self.__container.append(element)
def top(self):
"""
Get the top element of the stack
:return: top element
"""
if self.__is_empty():
return None
return self.__container[-1]
def pop(self):
"""
Remove the top element of the stack
:return: None or the top element of the stack
"""
return None if self.__is_empty() else self.__container.pop()
def clear(self):
"""
We'll make an empty stack
:return: self
"""
self.__container.clear()
return self
計算器類的實現
在計算器類中，我們將表達式的合法性驗證單獨放在一個函數中完成，但是實際上如果需要，也可以直接放在中綴表達式轉后綴表達式的函數中實現，這樣只需要一次遍歷表達式即可同時完成驗證和轉換工作。但是為了保持結構清晰，還是分開來實現比較好，每個函數盡可能最好一件事情才是比較實在的。
在該計算器類中，有很多種極端的情況沒有被考慮進去，因為那樣的話整個實現的代碼會更多。不過，可以在后期為整個類繼續擴展，添加新的功能也是可以的。目前實現的就是主要框架，包括基本的錯誤檢測和運算，重點時學習運用棧這個看似簡單卻強大的數據結構解決問題。
class Calculator(object):
"""
A simple calculator, just for fun
"""
def __init__(self):
self.__exp = ''
def __validate(self):
"""
We have to make sure the expression is legal.
1. We only accept the `()` to specify the priority of a sub-expression. Notes: `[ {` and `] }` will be
replaced by `(` and `)` respectively.
2. Valid characters should be `+`, `-`, `*`, `/`, `(`, `)` and numbers(int, float)
- Invalid expression examples, but we can only handle the 4th case. The implementation will
be much more sophisticated if we want to handle all the possible cases.:
1. `a+b-+c`
2. `a+b+-`
3. `a+(b+c`
4. `a+(+b-)`
5. etc
:return: True or False
"""
if not isinstance(self.__exp, str):
print('Error: {}: expression should be a string'.format(self.__exp))
return False
# Save the non-space expression
val_exp = ''
s = Stack()
for x in self.__exp:
# We should ignore the space characters
if x == ' ':
continue
if self.__is_bracket(x) or self.__is_digit(x) or self.__is_operators(x) \
or x == '.':
if x == '(':
s.push(x)
elif x == ')':
s.pop()
val_exp += x
else:
print('Error: {}: invalid character: {}'.format(self.__exp, x))
return False
if s.top():
print('Error: {}: missing ")", please check your expression'.format(self.__exp))
return False
self.__exp = val_exp
return True
def __convert2postfix_exp(self):
"""
Convert the infix expression to a postfix expression
:return: the converted expression
"""
# highest priority: ()
# middle: * /
# lowest: + -
converted_exp = ''
stk = Stack()
for x in self.__exp:
if self.__is_digit(x) or x == '.':
converted_exp += x
elif self.__is_operators(x):
converted_exp += ' '
tp = stk.top()
if tp:
if tp == '(':
stk.push(x)
continue
x_pri = self.__get_priority(x)
tp_pri = self.__get_priority(tp)
if x_pri > tp_pri:
stk.push(x)
elif x_pri == tp_pri:
converted_exp += stk.pop() + ' '
stk.push(x)
else:
while stk.top():
if self.__get_priority(stk.top()) != x_pri:
converted_exp += stk.pop() + ' '
else:
break
stk.push(x)
else:
stk.push(x)
elif self.__is_bracket(x):
converted_exp += ' '
if x == '(':
stk.push(x)
else:
while stk.top() and stk.top() != '(':
converted_exp += stk.pop() + ' '
stk.pop()
# pop all the operators
while stk.top():
converted_exp += ' ' + stk.pop() + ' '
return converted_exp
def __get_result(self, operand_2, operand_1, operator):
if operator == '+':
return operand_1 + operand_2
elif operator == '-':
return operand_1 - operand_2
elif operator == '*':
return operand_1 * operand_2
elif operator == '/':
if operand_2 != 0:
return operand_1 / operand_2
else:
print('Error: {}: divisor cannot be zero'.format(self.__exp))
return None
def __calc_postfix_exp(self, exp):
"""
Get the result from a converted postfix expression
e.g. 6 5 2 3 + 8 * + 3 + *
:return: result
"""
assert isinstance(exp, str)
stk = Stack()
exp_split = exp.strip().split()
for x in exp_split:
if self.__is_operators(x):
# pop two top numbers in the stack
r = self.__get_result(stk.pop(), stk.pop(), x)
if r is None:
return None
else:
stk.push(r)
else:
# push the converted number to the stack
stk.push(float(x))
return stk.pop()
def __calc(self):
"""
Try to get the result of the expression
:return: None or result
"""
# Validate
if self.__validate():
# Convert, then run the algorithm to get the result
return self.__calc_postfix_exp(self.__convert2postfix_exp())
else:
return None
def get_result(self, expression):
"""
Get the result of an expression
Suppose we have got a valid expression
:return: None or result
"""
self.__exp = expression.strip()
return self.__calc()
"""
Utilities
"""
@staticmethod
def __is_operators(x):
return x in ['+', '-', '*', '/']
@staticmethod
def __is_bracket(x):
return x in ['(', ')']
@staticmethod
def __is_digit(x):
return x.isdigit()
@staticmethod
def __get_priority(op):
if op in ['+', '-']:
return 0
elif op in ['*', '/']:
return 1
總結
以上就是利用Python實現簡單四則運算計算器的全部內容，希望本文的內容對大家的學習或者工作能有所幫助，如果有疑問大家可以留言交流。
參考
《數據結構與算法(C語言)》上相關章節算法描述
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