如何修复内部切片列表的问题

Question

我有一个分类数据集，我实现了LabelEncoder以数字形式更改它，然后我实现了StandardScaler()函数以2D形式更改它。现在我想使用 SVM 内核函数在 3D 模式下分离数据。它给了我X0_org、Y0_orig和X1_orig、Y1_orig的错误。错误是索引太多...

def randrange(n, vmin, vmax):
    return (vmax-vmin)*np.random.rand(n) + vmin
def fn_kernel(x1, x2):
    """ Implements a kernel phi(x1,y1) = [x1, y1, x1^2 + y1^2] """
    return np.array([x1, x2, x1**2.0 + x2**2.0])
""" Generate linearly nonseparable dataset (in R^2) """

# # Read the CSV file
dataset = pd.read_csv('raw_train.csv')
X = dataset.iloc[:, 1:3].values
y = dataset.iloc[:, 0].values
df_X = pd.DataFrame(X)
df_y = pd.DataFrame(y)
#  Encoding categorical data
le = LabelEncoder()
X_encode = df_X.apply(le.fit_transform)
# print(X)
y_encode = df_y.apply(le.fit_transform)
# print(y)
# splitting the datset into the training set and test set.
X_train, X_test, y_train, y_test = train_test_split(X_encode, y_encode,test_size=0.34, random_state=0)
# Changed the data in 2D
from sklearn.preprocessing import StandardScaler
sc_X = StandardScaler()
X_s = sc_X.fit_transform(X_train)
Y_s = sc_X.transform(X_test)
X = np.array(X_s)
Y = np.array(Y_s)
print(X)
print()
print(Y)

A = X[np.where(Y <= 0)]
B = X[np.where(Y >= 1)]
print(A)
print()
print(B)

X0_orig = A[:, 0]
Y0_orig = A[:, 1]
print(X0_orig)
X1_orig = B[:, 0]
Y1_orig = B[:, 1]
frac0 = len(np.where(Y == 0)[0]) / float(len(Y))
frac1 = len(np.where(Y == 1)[0]) / float(len(Y))
print("Percentage of '0' labels:", frac0)
print("Percentage of '1' labels:", frac1)
A = np.array([fn_kernel(x, y) for x, y in zip(np.ravel(X0_orig), np.ravel(Y0_orig))])
X0 = A[:, 0]
Y0 = A[:, 1]
Z0 = A[:, 2]
A = np.array([fn_kernel(x, y) for x, y in zip(np.ravel(X1_orig), np.ravel(Y1_orig))])
X1 = A[:, 0]
Y1 = A[:, 1]
Z1 = A[:, 2]

def plot_no_decision_boundary():
    fig = plt.figure(figsize=(20, 8))
    ax = fig.add_subplot(122, projection='3d')
    ax.scatter(X0, Y0, Z0, c='r', marker='o')
    ax.scatter(X1, Y1, Z1, c='b', marker='^')
    ax.set_xlabel('X Label')
    ax.set_ylabel('Y Label')
    ax.set_zlabel('Z Label')
    ax.set_title("Data in R^3 (separable)")
    # Project data to X/Y plane
    ax2d = fig.add_subplot(121)
    ax2d.scatter(X0, Y0, c='r', marker='o')
    ax2d.scatter(X1, Y1, c='b', marker='^')
    ax2d.set_xlabel('X Label')
    ax2d.set_ylabel('Y Label')
    ax2d.set_title("Data projected to R^2 (nonseparable)")
    plt.show()

def plot_decision_boundary():
    fig = plt.figure(figsize=(20, 8))
    ax = fig.add_subplot(121, projection='3d')
    ax.scatter(X0, Y0, Z0, c='r', marker='o')
    ax.scatter(X1, Y1, Z1, c='b', marker='^')
    ax.set_xlabel('X Label')
    ax.set_ylabel('Y Label')
    ax.set_zlabel('Z Label')
    ax.set_title("Data in R^3 (separable w/ hyperplane)")
    x = np.arange(-1.25, 1.25, 0.1)
    y = np.arange(-1.25, 1.25, 0.1)
    X, Y = np.meshgrid(x, y)
    Z = np.zeros(X.shape)
    Z[:, :] = 0.5
    ax.plot_surface(X, Y, Z, color='#09F911')
    # Project data to X/Y plane
    ax2d = fig.add_subplot(122)
    ax2d.scatter(X0, Y0, c='r', marker='o')
    ax2d.scatter(X1, Y1, c='b', marker='^')
    ax2d.add_patch(pl.Circle((0, 0), radius=sqrt(0.5), fill=False, linestyle='solid', linewidth=4.0, color='black'))
    ax2d.add_patch(pl.Circle((0, 0), radius=sqrt(0.5), fill=False, linestyle='dashed', linewidth=1.5, color='#09F911'))
    ax2d.set_xlabel('X Label')
    ax2d.set_ylabel('Y Label')
    ax2d.set_title("Data projected to R^2 (hyperplane projection shown)")
    plt.show()

def main():
    print("...Projecting dataset to R^3 (no decision boundary)...")
    plot_no_decision_boundary()
    print("...Projecting dataset to R^3 (with decision boundary)...")
    plot_decision_boundary()
    print("...Done.")

if __name__ == '__main__':
    main()

如果我打印A和B，它会给我这个输出。

A: [-0.03689752 -0.84216039 -0.27461752 -0.84216039 -0.17987541 -0.84216039
-1.4666764  -0.84216039  1.32139242  1.14423966  0.53501403 -0.84216039
0.24905826 -0.84216039 -1.10923168 -0.75178696 -0.68029802  1.99555397
1.10692559 -0.84216039 -0.84216039 -0.3228533   1.24990347  1.03543664]
B: [ 1.14423966  0.00915392 -0.84216039  1.14423966  0.57669679 -0.84216039]

但是如果我打印X0_orig或X1_orig，它会给出此错误：

Traceback (most recent call last):
X0_orig = A[:, 0]
IndexError: too many indices for array

Answer 1

它

解决了，问题是np.where被返回1D数组。在这里我改变了什么。我曾经np.argwhere，现在它给了我2D.

A = X_s[np.argwhere(Y_s <= 0)]
B = X_s[np.argwhere(Y_s >= 1)]
print(A.shape)

Answer 2

A和B在你的例子中是具有一个维度的数组。因此，尝试访问第一列将导致错误。仅当您可以准确指定要查找的信息时。

使用A.shape监视数组的形状，并在必要时A.reshape使其二维