Learn Interfaces by Building an Equation Solver - Step 36

Tell us what’s happening:

in regards to other responses to this step, where people have struggled i came across the response of don’t remove anything before the 1 which i never did, however i am still confused on why this gives me an TypeError

``````from abc import ABC, abstractmethod
import re

class Equation(ABC):
degree: int

def __init__(self, *args):
if (self.degree + 1) != len(args):
raise TypeError(
f"'Equation' object takes {self.degree + 1} positional arguments but {len(args)} were given"
)
if any(not isinstance(arg, (int, float)) for arg in args):
raise TypeError("Coefficients must be of type 'int' or 'float'")
if args[0] == 0:
raise ValueError("Highest degree coefficient must be different from zero")
self.coefficients = {(len(args) - n - 1): arg for n, arg in enumerate(args)}

def __init_subclass__(cls):
if not hasattr(cls, "degree"):
raise AttributeError(
f"Cannot create '{cls.__name__}' class: missing required attribute 'degree'"
)

def __str__(self):
terms = []
for n, coefficient in self.coefficients.items():
if not coefficient:
continue
if n == 0:
terms.append(f'{coefficient:+}')
elif n == 1:
terms.append(f'{coefficient:+}x')
else:
terms.append(f"{coefficient:+}x**{n}")
equation_string = ' '.join(terms) + ' = 0'

# User Editable Region

return re.sub(r'(?<!\d)1','1(?=x)','', equation_string.strip('+'))

# User Editable Region

@abstractmethod
def solve(self):
pass

@abstractmethod
def analyze(self):
pass

class LinearEquation(Equation):
degree = 1

def solve(self):
a, b = self.coefficients.values()
x = -b / a
return x

def analyze(self):
slope, intercept = self.coefficients.values()
return {'slope': slope, 'intercept': intercept}

degree = 2

def __init__(self, *args):
super().__init__(*args)
a, b, c = self.coefficients.values()
self.delta = b**2 - 4 * a * c

def solve(self):
pass

def analyze(self):
pass

lin_eq = LinearEquation(2, 3)
print(lin_eq)

``````

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Challenge Information:

Learn Interfaces by Building an Equation Solver - Step 36

You’ve added an argument to `sub` which is of the wrong type

but it says in the instructions this `In the example above, the pattern a(?=t) contains a positive lookahead, which is used to match the a c ` this positive lookahead is in the format

my formatt is `'1(?=x)' ` so i don’t see how it isn’t positive

becuase how can it be of the wrong type when all it’s telling me to do is add the positive lookahead with character 1 and followed by x, i mean it seems like a obvious step

How many arguments does `sub()` take?

What type of arguments should they be?

yeah but i’ve not added anything to that i wasn’t suppose to, this instruction doesn’t tell me to remove/modify anything is that what i should do ?

If you’re getting a a Type Error you can be confident that you’ve added something of the wrong type somewhere.

Look at the example code again, and see how you are giving a different set of arguments to the `sub()` function.

3-4 arguments and they should be a pattern, a replacement a string and one or more regx flags to modify function

The instructions literally tell you to modify your regex:

``````return re.sub(r'(?<!\d)1'+'1(?=x)', '', equation_string.strip('+')