Tell us what’s happening:
the instruction is to create a new string. then add that string to output_string.
string =f’\n\n{equation!s:-^24}\n\n’. The format use ‘!s’ and str().
the instruction:
Create a string containing the string representation of your equation centered in a width of 24 characters. Make the string begin and end with two newline characters, and add your new string to the current value of output_string .
the new output is:
----Linear Equation-----
-------2x +3 = 0--------
The check code returns ’ The solver function should return a different string.’
Not understanding the comment returned or for that matter what’s wrong.
is there something wrong with the output ? or do I need to create and return a thrid variable?
Your code so far
from abc import ABC, abstractmethod
import re
class Equation(ABC):
degree: int
type: str
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'"
)
if not hasattr(cls, "type"):
raise AttributeError(
f"Cannot create '{cls.__name__}' class: missing required attribute 'type'"
)
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'
return re.sub(r"(?<!\d)1(?=x)", "", equation_string.strip("+"))
@abstractmethod
def solve(self):
pass
@abstractmethod
def analyze(self):
pass
class LinearEquation(Equation):
degree = 1
type = 'Linear Equation'
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}
class QuadraticEquation(Equation):
degree = 2
type = 'Quadratic Equation'
def __init__(self, *args):
super().__init__(*args)
a, b, c = self.coefficients.values()
self.delta = b**2 - 4 * a * c
def solve(self):
if self.delta < 0:
return []
a, b, _ = self.coefficients.values()
x1 = (-b + (self.delta) ** 0.5) / (2 * a)
x2 = (-b - (self.delta) ** 0.5) / (2 * a)
if self.delta == 0:
return [x1]
return [x1, x2]
def analyze(self):
a, b, c = self.coefficients.values()
x = -b / (2 * a)
y = a * x**2 + b * x + c
if a > 0:
concavity = 'upwards'
min_max = 'min'
else:
concavity = 'downwards'
min_max = 'max'
return {'x': x, 'y': y, 'min_max': min_max, 'concavity': concavity}
def solver(equation):
if not isinstance(equation, Equation):
raise TypeError("Argument must be an Equation object")
# User Editable Region
string =f'\n\n{equation!s:-^24}\n\n'
output_string = f'\n{equation.type:-^24}'+ string
# User Editable Region
return output_string
lin_eq = LinearEquation(2, 3)
quadr_eq = QuadraticEquation(1, 2, 1)
print(solver(lin_eq))
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Challenge Information:
Learn Interfaces by Building an Equation Solver - Step 52