我怎样才能把答案打印成分数，而不是那样？

Question

**我试图让函数返回二次方程的结果，但我不知道如何将解打印为分数。请帮帮我！**

def cube_root(x):
  return x**(1/3)
def  Quadratic(a, b, c):
  delta = (b**2)-4*a*c
  if delta == 0:
    x = (-b)/2*a
    return f"This Quadratic equation has 1 solution: {x}"
  else:
    if delta  < 0 :
      return "This Quadratic equation has no solutions: "
    else:
      x1 = ((-b)-cube_root(delta))/2*a
      x2 = ((-b)+cube_root(delta))/2*a 
      return f"This Quadratic equation has 2 solutions: {x1} & {x2}"


print(Quadratic(12, 0, -1))

Answer 1

如果您不需要额外的软件包，以下内容可能会有所帮助：

from fractions import Fraction

def is_square(x):
    if x < 0: return False
    s = int(x**0.5)
    return s*s == x

def sqrt_frac_str(frac):
    if frac < 0:
        return f'i {sqrt_frac_str(-frac)}'
    num_isq = is_square(frac.numerator)
    den_isq = is_square(frac.denominator)
    if num_isq and den_isq:
        return f'{int(frac.numerator**0.5)}/{int(frac.denominator**0.5)}'
    elif num_isq:
        return f'{int(frac.numerator**0.5)}/sqrt({frac.denominator})'
    elif den_isq:
        return f'sqrt({frac.numerator})/{int(frac.denominator**0.5)}'
    else:
        return f'sqrt({frac})'

def  quadratic_frac(a, b, c):
    delta = Fraction(b**2 - 4 * a * c)
    rootcenter = Fraction(-b, 2 * a)
    rootdeltasq = delta / Fraction(2 * a)**2
    return rootcenter, rootdeltasq

def quadsol_str(rootcenter, rootdeltasq):
    return f'{rootcenter} +/- {sqrt_frac_str(rootdeltasq)}'

测试：

rc, rd = quadratic_frac(2, 1, -1)
rc, rd
# (Fraction(-1, 4), Fraction(9, 16))

quadsol_str(*quadratic_frac(2, 1, -1))
# '-1/4 +/- 3/4'

quadsol_str(*quadratic_frac(2, 0, -1))
# '0 +/- 1/sqrt(2)'

quadsol_str(*quadratic_frac(2, 0, 1))
# '0 +/- i 1/sqrt(2)'

quadsol_str(*quadratic_frac(3, 2, -1))
# '-1/3 +/- 2/3'

quadsol_str(*quadratic_frac(5, 3, -7))
# '-3/10 +/- sqrt(149)/10'

Answer 2

您可以从sympy包中使用simplify（不在标准库中-您必须安装它）：

from sympy import simplify, sqrt

def quadratic(a, b, c):
    a = simplify(a)  # convert inputs into objects used by simplify
    b = simplify(b)
    c = simplify(c)

    delta = (b**2)-4*a*c
    if delta == 0:
        x = (-b)/2*a
        return f"This Quadratic equation has 1 solution: {x}"
    elif delta < 0 :
        return "This Quadratic equation has no real solutions: "
    else:
        x1 = ((-b)-sqrt(delta))/2*a  # using sqrt from sympy
        x2 = ((-b)+sqrt(delta))/2*a
        return f"This Quadratic equation has 2 solutions: {x1} & {x2}"

print(quadratic(12, 0, -1))

这使得：

This Quadratic equation has 2 solutions: -24*sqrt(3) & 24*sqrt(3)

不同的例子：

print(quadratic(12, 2, -1))

给出：

This Quadratic equation has 2 solutions: -12*sqrt(13) - 12 & -12 + 12*sqrt(13)

实际上sympy也可以为您处理复数，这样您就可以摆脱没有实际解决方案的测试（即删除elif，这样delta < 0由else:块处理）

如果您这样做，然后给出一个示例：

print(quadratic(12, 2, 1))

你会得到：

This Quadratic equation has 2 solutions: -12 - 12*sqrt(11)*I & -12 + 12*sqrt(11)*I