Question

我试图通过梯形方法整合多项式函数（我可以在以后更改为更精确的方法）。我的代码并不完美，我想知道为什么它不起作用。我遇到的一个问题是while循环没有结束。到目前为止，我的代码如下。

    def Integrate_Trapezoidal(x_LoBound,x_HiBound,N):
        """
        INPUT   :
                    x_LoBound   --      lower bound of integral
                    x_HiBound   --      upper bound of integral
                    N           --      number of slices (N --> inf ==> integral)
        OUTPUT  :
                                --      approximate value of integral
        """
        ## CREATE ALPHABET
        alphabet = [chr(i) for i in range(ord('a'),ord('z')+1)]
        ## alphabet = ['a','b','c',...,'z'] ##
        ## WOULD LOVE TO TRY FLOATING INPUTS VIA ARRAY COMPREHENSION
        a = float(input("What is the coefficient of the lowest order term:  "))
        CoeffList = []
        CoeffNumList = []
        LengthCoeffList = [] ## [1,2,3,...,max] where max = coefficient of highest-order term
        for letter in alphabet:
            AddOne = int(1)
            AddOne += int(1)
            for i in range(int(1),int(AddOne)):
                letter = alphabet[int(i)]
                while letter in alphabet:
                    CoeffList.append(letter)
                    LengthCoeffList.append(len(CoeffList))
                    # alphabet[i]
                    # i = i + 1
                    letter = float(input("What is the coefficient of the next-order term:  ")) ## GO FROM a = ___ TO b = ___ TO c = ___ ...
                    CoeffNumList.append(letter)
                    if float(input("What is the coefficient of the next-order term:  ")) == '0':
                        print("Type 'Y for YES and 'N' for NO")
                        YESorNO = str(input("Is that the last term of the polynomial:  "))
                        endterm = YESorNO[-1] ## look at last character of string
                        if endterm == 'N' or endterm == 'n' or endterm == 'no' or endterm == 'NO' or endterm == 'No':
                            pass
                        elif endterm == 'Y' or endterm == 'y' or endterm == 'YES' or endterm == 'yes' or endterm == 'Yes':
                            break
                            def f(x):
                                """
                                INPUT   :
                                                   x    --  variable of function
                                                        EX: x = x_LoBound OR x = x_HiBound
                                OUTPUT  :
                                            function    --  f(x) = a x^0 + b x^1 + ...
                                                        EX: f(x_LoBound) OR f(x_HiBound)
                                """
                                for expval in LengthCoeffList and CoeffNum in CoeffNumList:
                                    # function = 0
                                    function += CoeffNum * x**expval
                                    return function
                    letter = alphabet[int(i+1)] ## GO FROM a TO b TO c ...
        ## TRAPEZOIDAL RULE
        # def f(x):
        #     return x**4 - 2*x + 1
        ht = (x_HiBound - x_LoBound) / N
        ss = 0.5 * f(x_LoBound) + 0.5 * f(x_HiBound)
        for num in range(1,N):
            ss += f(x_LoBound + num*ht)
        return ht*ss
    checkanswer = Integrate_Trapezoidal(0,2,10)
    print(checkanswer)

Answer 1

我已经去查看你的代码并找到了我觉得有用的东西，检查了我下载的几个大学讲义。正如你在评论中所说的，有很多额外的列表是不必要的，所以我在那里减少了很多代码。

特别是，如果假设每个系数从最低到最高顺序依次添加，并且为任何不存在的系数添加0，那么您只需要知道列表中元素的编号x的力量。

我还移动了f()的定义来创建帮助函数solve_point()，它的工作方式和我想的一样。特别是，内置sum和enumerate，enumerate遍历coeff_list并返回计数以提供权力（0向上）。

get_coefficients()来自您原来的Integrate_Trapezoidal()，但更专注于一件事 - 这就是为什么它会在最后返回CoeffList进行最终处理。

def solve_point(x, coeff_list):
    return sum(coeff * x**e for e, coeff in enumerate(coeff_list))


def get_coefficients():
    CoeffList = []
    while True:
        # GO FROM a = ___ TO b = ___ TO c = ___ ...
        coeff = float(input("What is the coefficient of the next-order term:  "))
        CoeffList.append(coeff)
        if coeff == 0:
            YESorNO = raw_input("Is that the last term of the polynomial: [Y/N]  ")
            if YESorNO.upper() == 'Y':
                return CoeffList[:-1]

lo, hi, n = 0, 2, 6
coeff_list = get_coefficients()

ht = (hi - lo) / float(n)
ss = 0.5 * solve_point(lo, coeff_list) + 0.5 * solve_point(hi, coeff_list)
for num in range(1,n):
    ss += solve_point(lo + num*ht, coeff_list)
checkanswer = ht*ss


print(checkanswer)

我认为是对的 - 我做了几次检查。希望它可能有助于你的重写！如果您有任何不起作用的示例，那么最好知道，或者您可以看到任何错误......

如何迭代字母列表以指定浮点值以创建多项式？

1 个答案: