我的PCA分析功能编程正确吗？

Question

我有功课可以对PCA进行编程，而无需使用库自动进行。我目前正在对此进行编程。但是尽管正在执行某些操作，但我不确定是否正确。因此，请您检查我的代码。

def input_to_output(self, x):
        # Computes output layer - transforms input pattern to new PCA
        # coordinate system.
        ### YOUR CODE GOES HERE ###
        count = x.shape[0]
        return x[0] # dummy return


def output_to_input(self, y):
    # Computes input layer from output - transforms output pattern back to
    # its original coordinate system.
    ### YOUR CODE GOES HERE ###
    return [self.w[0] * y, self.w[1] * y]


def denoise_data(self, data):
    # Transforms all data to new coordinate system, and then perform inverse
    # transformation back to original coordinate system. Side effect is
    # noise removal. Returns array of size [data_dimension x data_count] = [2 x 100]
    ### YOUR CODE GOES HERE ###
    count = data.shape[1]  # number of data points

    return self.output_to_input(self.input_to_output(data))


def train_iterative(self, data, num_epochs, alpha=0.1):
    # Trains the neural network iteratively (NN approach)
    count = data.shape[1] # number of data points

    def normalize(v):
        norm = np.linalg.norm(v)
        if norm == 0: 
           return v
        return v / norm

    for ep in range(num_epochs):
        for p in np.random.permutation(count):
            x = data[:, p]
            ### YOUR CODE GOES HERE ###
            y = self.input_to_output(x)
            for i, znak in enumerate(self.w):  
                self.w[i] = self.w[i] + alpha * y * (x[i] - y * self.w[i])

            self.w = normalize(self.w)

def train_analytic(self, data):
    # Trains the neural network analytically (eigenvector computation)
    ### YOUR CODE GOES HERE ###
    count = data.shape[1]  # number of data points

    Q = 1 / len(data[0]) * np.outer(data,data.T)

    lambdas, v = np.linalg.eig(Q)

    # tuple (eigenvalue, eigenvector)
    eig_pairs = [(np.abs(lambdas[i]), v[:,i]) for i in range(len(lambdas))]

    # sort tuple (eigenvalue, eigenvector) from highest to lowest according to eigenvalue
    eig_pairs.sort(key=lambda x: x[0], reverse=True)


    print(eig_pairs[0][1])
    self.w = eig_pairs[0][1]

图形如下：