limour/b.limour.top

title: Python后缀表达式在解方程中的应用 tags:
Python
后缀表达式
栈
解方程
逆波兰表达式 id: '496' categories:
Python练习
date: 2020-12-23 16:11:33

import re
from fractions import Fraction

class Stack(list):
def isEmpty(self):
return self == []
def peek(self):
if self == []: return None
else: return self[-1]
def size(self):
return len(self)
push = list.append
def pop(self):
if self == []: return None
else: return super().pop()

class Polynomial(list):
def __init__(self, value):
for item in value: self.append(Fraction(item))
def add(self, value):
if len(self) < len(value): self += [Fraction(0)]*(len(value)-len(self))
for i in range(len(value)): self[i] += value[i]
def sub(self, value):
if len(self) < len(value): self += [Fraction(0)]*(len(value)-len(self))
for i in range(len(value)): self[i] -= value[i]
def mul(self, value):
tmp = self.copy()
size = len(self)
self.clear()
self += [Fraction(0)]*(size+len(value)-1)
for i,item in enumerate(value):
    for j in range(size):
        self[i+j] += tmp[j]*item
def divn(self, n):
if type(n) is Polynomial:
    for i in range(len(self)): self[i] /= n[0]
else:
    _n = Fraction(n)
    for i in range(len(self)): self[i] /= _n
def __str__(self):
if self == []:
    return '0'
elif len(self) == 1:
    return str(self[0])
elif len(self) == 2:
    return f'({self[1]})x + {self[0]}'
else: pass

def get_Formula(equation):
return equation.replace(' ','').split('=')

_ep = re.compile(r'([\+\-\*/()][^\+\-\*/()]+)')
_op = {
'+': lambda x,y:x.add(y),
'-': lambda x,y:x.sub(y),
'*': lambda x,y:x.mul(y),
'/': lambda x,y:x.divn(y)
}
def _middle2behind(Fma, res, s, e):
sta = Stack()
_s = s
while s < e:
if Fma[s] in '+-':
    if sta.isEmpty():
        sta.push(Fma[s])
        s += 1
    elif sta.peek() in '*/':
        while sta: res.push(sta.pop())
        sta.push(Fma[s])
        s += 1
    elif sta.peek() in '+-':
        res.push(sta.pop())
        sta.push(Fma[s])
        s += 1
elif Fma[s] in '*/':
    if sta.isEmpty():
        sta.push(Fma[s])
        s += 1
    elif sta.peek() in '+-':
        sta.push(Fma[s])
        s += 1
    elif sta.peek() in '*/':
        res.push(sta.pop())
        sta.push(Fma[s])
        s += 1
elif Fma[s] == '(':
    s += 1
    d = _middle2behind(Fma, res, s, e)
    s += d
elif Fma[s] == ')':
    s += 1
    break
else:
    res.push(Fma[s])
    s += 1
while sta: res.push(sta.pop())
return s-_s
def middle2behind(Formula):
if Formula.startswith('-'): Formula = '0' + Formula
expr = _ep.findall(Formula.replace('(-','(0-'))
res = Stack()
_middle2behind(expr, res, 0, len(expr))
return res
def str2polynomial(_str):
if _str.endswith('x'):
if _str == 'x': return Polynomial((0, 1))
return Polynomial((0, _str.rstrip('x')))
else: return Polynomial((_str,))
def not_eval(Formula):
expr = middle2behind(Formula)
#print(expr)
sta = Stack()
for item in expr:
if item in _op:
    y = sta.pop()
    x = sta.pop()
    _op[item](x, y)
    sta.push(x)
else: sta.push(str2polynomial(item))
#print(sta)
return sta.pop()

_x = re.compile(r'[a-zA-Z]+')
def solve_eq(equation):
print(f'equation is \t\t{equation}')
xname = set(_x.findall(equation))
if len(xname) != 1: 
#print(xname)
print(f'别逗，{equation}是一元一次方程吗?')
return
xname = xname.pop()
Fma = get_Formula(_x.sub('x', equation))
if len(Fma) != 2:
print('{equation}不是标准方程！')
return
Fma = [not_eval(expr) for expr in Fma]
print(f'Simplification is \t{Fma[0]} = {Fma[1]}')
tmp = Fma[0]
tmp.sub(Fma[1])
print(f'Transposition is \t{tmp} = 0')
if len(tmp) < 2 or tmp[1] == 0:
print(f'{xname} 无解')
else:
x = -tmp[0]/tmp[1]
print(f'{xname} = x = {x}')
python后缀表达式在解方程中的应用.md 4.4 KB Permalänk Historik Rå

date: 2020-12-23 16:11:33

python后缀表达式在解方程中的应用.md 4.4 KB

Permalänk Historik Rå
