Python how to extract number before x² and x?

Question

Python

Shalom everyone, I'm new to programming and the following problem arose:
There is an expression (it can be anything)
For example: 3x² + 16 - 7x
I need to get the number 3 in front of x² and add "a" to the variable, get the number -7 in front of x and add to the variable "b", well, get the number 16 and add this number to the variable "c", how can this be done? It is important that the order may not be observed, i.e. I could also write 13 -4x + x².
I will be very grateful for your help

Answer 1

You can split into separate numbers using split(), discard the signs. And process an array of 3 elements.
Then check each:
1: if the last character is x, then take everything before it(number[:-1] and reduce to a number.
2: if the penultimate character is x, then take everything before it(number[:-2]
3: if x is not, then just cast to a number.

s = input().split()
for i in s:
    if len(i) > 2 and i[-2] == 'x':
        a = int(i[:-2])
    elif len(i) == 2 and i[0] == 'x':
        a = 1
    elif i[-1] == 'x':
        b = int(i[:-1]) if len(i) > 1 else 0
    elif i != '0' and i.isdigit():
        c = int(i)

print(a, b, c)

PS I redid the code, because I messed with lambdas

Answer 2

Arseny, good afternoon!
I recommend that you look at the basic articles or help about the re module for regular expressions in python. This is what you need.