python Spyder交互输出

python Spyder交互输出,python,numpy,anaconda,spyder,anaconda3,Python,Numpy,Anaconda,Spyder,Anaconda3,最近,我写了这段代码,试图找到一些方法来计算抛物线的积分,但这对我来说不是什么问题,但重点是我写的代码没有给我任何输出,这很奇怪,我使用spyder,代码如下: def integrator(a,b,c): #a,b and c are the factors of the quadratic import numpy as np delta=((b**2)-4*a*c) answer1= (-b+delta)/2 answer2=(-b-delta)/2

最近,我写了这段代码,试图找到一些方法来计算抛物线的积分,但这对我来说不是什么问题,但重点是我写的代码没有给我任何输出,这很奇怪,我使用spyder,代码如下:

def integrator(a,b,c):
    #a,b and c are the factors of the quadratic
    import numpy as np
    delta=((b**2)-4*a*c)
    answer1= (-b+delta)/2
    answer2=(-b-delta)/2 
    #it gets the answers of the quadratic eruation and uses them as the lower and upper bounds
    if delta<=0:
        return "this func has no real answer"
    if delta==0:
        return "this function has two equal answers"
        return  answer1
  else:
      return answer1
      return answer2
  #it also calculate the answers of the quadratic equation too
  integ_range=[]
  for i in np.arange(answer1,answer2,0.01):
      integ_range.append(i)
  for i in integ_range:
      heights=(a*(i**2))+(b*i)+c
  sum1=sum(integ_range)    
  integrate=sum1*0.01
  return (integrate)

integrator(1,2,3)
def积分器(a、b、c): #a、 b和c是二次曲线的因子 将numpy作为np导入 增量=((b**2)-4*a*c) 回答1=(-b+增量)/2 回答2=(-b-δ)/2 #它得到二次求值的答案,并将其用作上下界
如果增量为积分器分配一个变量,如下所示:

i=integrator(1,2,-3)
print(i)

应该可以

您不能返回多个答案,只返回第一个答案。您可以返回一个
元组
,例如
返回answer1,answer2
。不过,多亏了它的有效性,我没有注意到代码在到达return后停止执行的事实