一、算法
???? 1、算法的主要思想就是將一個(gè)中綴表達(dá)式(Infix expression)轉(zhuǎn)換成便于處理的后綴表達(dá)式(Postfix expression)，然后借助于棧這個(gè)簡(jiǎn)單的數(shù)據(jù)結(jié)構(gòu)，計(jì)算出表達(dá)式的結(jié)果。
???? 2、關(guān)于如何講普通的表達(dá)式轉(zhuǎn)換成后綴表達(dá)式，以及如何處理后綴表達(dá)式并計(jì)算出結(jié)果的具體算法描述不在此敘述了，書上有詳細(xì)的說明。
二、簡(jiǎn)易計(jì)算器
使用說明
使用該計(jì)算器類的簡(jiǎn)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(cè)試案例
為了對(duì)這個(gè)計(jì)算器進(jìn)行有效地檢驗(yàn)，設(shè)計(jì)了幾組測(cè)試案例，測(cè)試結(jié)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(shí)現(xiàn)代碼
棧的實(shí)現(xiàn)
棧實(shí)際上就是一個(gè)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(shí)現(xiàn)的簡(jiǎn)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(jì)算器類的實(shí)現(xiàn)
在計(jì)算器類中，我們將表達(dá)式的合法性驗(yàn)證單獨(dú)放在一個(gè)函數(shù)中完成，但是實(shí)際上如果需要，也可以直接放在中綴表達(dá)式轉(zhuǎn)后綴表達(dá)式的函數(shù)中實(shí)現(xiàn)，這樣只需要一次遍歷表達(dá)式即可同時(shí)完成驗(yàn)證和轉(zhuǎn)換工作。但是為了保持結(jié)構(gòu)清晰，還是分開來實(shí)現(xiàn)比較好，每個(gè)函數(shù)盡可能最好一件事情才是比較實(shí)在的。
在該計(jì)算器類中，有很多種極端的情況沒有被考慮進(jìn)去，因?yàn)槟菢拥脑捳麄€(gè)實(shí)現(xiàn)的代碼會(huì)更多。不過，可以在后期為整個(gè)類繼續(xù)擴(kuò)展，添加新的功能也是可以的。目前實(shí)現(xiàn)的就是主要框架，包括基本的錯(cuò)誤檢測(cè)和運(yùn)算，重點(diǎn)時(shí)學(xué)習(xí)運(yùn)用棧這個(gè)看似簡(jiǎn)單卻強(qiáng)大的數(shù)據(jù)結(jié)構(gòu)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(jié)
以上就是利用Python實(shí)現(xiàn)簡(jiǎn)單四則運(yùn)算計(jì)算器的全部?jī)?nèi)容，希望本文的內(nèi)容對(duì)大家的學(xué)習(xí)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(shù)據(jù)結(jié)構(gòu)與算法(C語言)》上相關(guān)章節(jié)算法描述
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