一个表达式怎么提取出系数和选项。(语言-python)

Question

Ax**3+Bcos(x+y)+Cexp(-x**2)-bx+cy
怎么提取出。(系数与项之间有⭐，这里没显示出来)
A x3
B cos(x+y)
C exp(-x2)
-b x
c y
A x3
B cos(x+y)
C exp(-x2)
-b x
c y

Answer 1

下有代码，可直接复制使用。如有帮助，敬请采纳，你的采纳是我前进的动力，O(∩_∩)O谢谢！！！！！！！！
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测试公式
A*x**3+B*cos((x+y)+(x1*x-y**y1))+C*exp(-x**2)-b*x+c*y
A*x**3+B*cos((x+y)+(x1*x-y**y1))+C*x*exp(-x**2)-b*x+c*y*x
A*x**3+B*cos((x+y)+(x1*x-y**y1))+C*x*exp(-x**2)-b*x+c*y*x+A/B*x+B/A*y
不可完成:系数相同的表达式，有这方面的要求再增加

import re

representation1 = "A*x**3+B*cos(x+y)+C*exp(-x**2)-b*x+c*y"
representation1 = representation1.replace(' ','')
if representation1[0] not in '-+':
    representation1 = '+' + representation1
representation2 = representation1.replace('**', '^')

# 去掉最短括号,防干扰
# representation3 = re.sub('\(.*?\)', '', representation2)
brackets_num_left = 0
brackets_num_right = 0
representation3 = representation1
replace = ''
for x in representation1:
    if '(' in replace:
        replace += x
    if x == '(':
        if replace == '':
            replace = x
        brackets_num_left += 1
    if x == ')':
        brackets_num_right += 1
    if replace != '' and brackets_num_left == brackets_num_right:
        representation3 = representation3.replace(replace, '')
        replace = ''
# 获取系数
系数 = re.findall('[+\-].*?[A-Za-z]', representation3)
# 通过系数获取系数项
result = {}
result_str = ''
for x_i in range(len(系数)):
    try:
        res = re.findall(f'\{系数[x_i]}(.*?)\{系数[x_i + 1]}', representation1)
        result[系数[x_i]] = res[0]
    except:
        result[系数[x_i]] = representation1.split(系数[-1])[-1]
    result_str += 系数[x_i]+','+result[系数[x_i]]+','

print(result_str[:-1])