如何使用Numpy求解化学方程得到整数答案

Question

我被要求写一段Python代码，用来解决一个化学方程式，使用的是阶梯矩阵的方法，课程是代数。整体上没问题，但我希望最后的结果是整数，我不知道该怎么做。在我的solve_equation函数里，我把自由变量设为1，这样我就可以根据这个值计算其他变量。我不知道还能做些什么，因为我对Python不太熟悉，而且刚开始学习编程。这是我目前的函数（它接收简化后的阶梯矩阵作为输入，并返回系数）：

def solve_equation(matrix):
    m, n = matrix.shape
    solution = {}
    
    # Iterate over each row to find the pivot variable and calculate its value
    for i in range(m):
        pivot_column = np.argmax(matrix[i, :-1])
        if matrix[i, pivot_column] == 0:
            continue  # Skip trivial equations
        variable_index = pivot_column
        variable_value = matrix[i, -1] / matrix[i, pivot_column]
        if(variable_value < 0):
            variable_value *= -1
        solution[f'X{variable_index + 1}'] = variable_value

    # Handle free variables
    for j in range(n - 1):
        if f'X{j + 1}' not in solution:
            # Assign a value of 1 to free variables
            solution[f'X{j + 1}'] = 1
            free_var_index = j
            # Iterate through the matrix to adjust values of basic variables
            for i in range(m):
                if i != free_var_index:
                    coefficient = matrix[i, free_var_index]
                    if coefficient < 0 :
                        coefficient *= -1
                    solution[f'X{i + 1}'] += coefficient * solution[f'X{j + 1}']

    return solution

我觉得我应该根据简化阶梯矩阵中的系数来改变自由变量的输入。

Answer 1

我找到的答案（在ChatGPT和另一个问题的帮助下）：

def solve_equation(matrix):

    m, n = matrix.shape
    solution = {}
    coeffs = {}

    # Iterate over each row to find the pivot variable and calculate its value
    for i in range(m):
        pivot_column = np.argmax(matrix[i, :-1])
        if matrix[i, pivot_column] == 0:
            continue  # Skip trivial equations
        variable_index = pivot_column
        variable_value = matrix[i, -1] / matrix[i, pivot_column]
        solution[f'X{variable_index + 1}'] = variable_value

    # finding the free variables
    for j in range(n - 1):
        if f'X{j + 1}' not in solution:
            # here I am assiging 1 to the free variable so that I can calculate others based on it
            solution[f'X{j + 1}'] = 1
            free_var_index = j
            # Iterate through the matrix to adjust values of basic variables
            for i in range(m):
                if i != free_var_index:
                    coefficient = matrix[i, free_var_index]
                    solution[f'X{i + 1}'] += coefficient * solution[f'X{j + 1}']

    # I wanted to have integer answers in the end so here we convert numbers to integer by claculating the LCM
    for var, value in solution.items():
        if var.startswith('X'):
            coeffs[var] = value

    fractions = [Fraction(val).limit_denominator(MAX_DENOM) for val in coeffs.values()]
    ratios = np.array([(f.numerator, f.denominator) for f in fractions])
    factor = np.lcm.reduce(ratios[:,1])


    for var, value in solution.items():
        if var.startswith('X'):
            solution[var] *= factor

    return solution