python numpy小记1

1.参考：https://docs.scipy.org/doc/numpy/reference/generated/numpy.outer.html#numpy.outer

向量$b=[1,2,3]^T$，求$bb^T$为3x3的矩阵。因为numpy.dot得到的是向量的内积。

为计算$bb^T$，使用numpy.outer命令。故可以得到3x3的矩阵。

2.reshape(-1,1)的用法。变成一列，行数自动得到。得到2*1的维度。

x = np.array([1,2]).reshape(-1, 1)            

2.1关于np.array()直接‘’=‘’号赋值，对新array的改变将直接改变旧array的值。故使用copy()方法。

old = np.array([[1, 1, 1],
                [1, 1, 1]])
new = old
new[0, :2] = 0

print(old)
"""
我们得到的输出为: 
 [ [0 0 1],  
    [1 1 1]  ]

"""

使用copy()方法：

old = np.array([[1, 1, 1],
                [1, 1, 1]])

new = old.copy()
new[:, 0] = 0

print(old)
"""输出为原来的值：
 [[1 1 1],
 [1 1  1]  ]

""        

2.2 np.array() 的遍历：

test = np.random.randint(0, 10, (4,3))
"""
>>> test
array([[4, 7, 3],
       [8, 5, 7],
       [1, 4, 8],
       [1, 9, 8]])
"""
for row in test:
     print(row)
""" output:
[4 7 3]
[8 5 7]
[1 4 8]
[1 9 8]
"""
for i in range(len(test)):
     print i
""" output:
0
1
2
3
"""

for i, row in enumerate(test):
     print('row', i,'is', row )

"""output: 
('row', 0, 'is', array([4, 7, 3]))
('row', 1, 'is', array([8, 5, 7]))
('row', 2, 'is', array([1, 4, 8]))
('row', 3, 'is', array([1, 9, 8]))
"""

test2 = test**2
""" output:
>>> test2
array([[16, 49,  9],
       [64, 25, 49],
       [ 1, 16, 64],
       [ 1, 81, 64]])

"""

同时遍历这两个array，可以使用zip()

 for i, j in zip(test, test2):
    print i, "+" , j,'=', i+j
"""
[4 7 3] + [16 49  9] = [20 56 12]
[8 5 7] + [64 25 49] = [72 30 56]
[1 4 8] + [ 1 16 64] = [ 2 20 72]
[1 9 8] + [ 1 81 64] = [ 2 90 72]
"""

 3. 投影矩阵

 1.1.将$x$投影到$b$的向量为: $\frac{bb^T}{ {\lVert b \rVert}^2} x$ ，投影矩阵为 $\frac{bb^T}{ {\lVert b \rVert}^2}$

def projection_matrix_1d(b):
    """Compute the projection matrix onto the space spanned by `b`
    Args:
        b: ndarray of dimension (D,), the basis for the subspace
    
    Returns:
        P: the projection matrix
    """
    D, = b.shape
    P = np.eye(D) # EDIT THIS
    P = np.outer(b, b.T) / np.dot(b,b)
    return P

 投影向量：

def project_1d(x, b):
    """Compute the projection matrix onto the space spanned by `b`
    Args:
        x: the vector to be projected
        b: ndarray of dimension (D,), the basis for the subspace
    
    Returns:
        y: projection of x in space spanned by b
    """
    p = np.zeros(3) # EDIT THIS
    P = np.dot(projection_matrix_1d(b),x)
    return p

 

2.1 将$x$投影到多维空间，设该多维空间的基为 $[b_1, b_2...b_M]$，可视为DxM维矩阵。得到的投影矩阵为 $B (B^TB)^{-1} B^T$。

def projection_matrix_general(B):
    """Compute the projection matrix onto the space spanned by `B`
    Args:
        B: ndarray of dimension (D, M), the basis for the subspace
    
    Returns:
        P: the projection matrix
    """
    P = np.eye(B.shape[0]) # EDIT THIS
    P = np.dot( np.dot( B, np.linalg.pinv(B.T, B)), B.T )
    return P

 

2.2 得到的投影向量计算为  $B (B^TB)^{-1} B^T x$

def project_general(x, B):
    """Compute the projection matrix onto the space spanned by `B`
    Args:
        B: ndarray of dimension (D, E), the basis for the subspace
    
    Returns:
        y: projection of x in space spanned by b
    """
    p = np.zeros(x.shape) # EDIT THIS
    P = np.dot( projection_matrix_general(B), x )
    return p

 3.关于向量乘法中@ 符号的用法。 

参考来源：https://legacy.python.org/dev/peps/pep-0465/

import numpy as np
from numpy.linalg import inv, solve

# Using dot function:
S = np.dot((np.dot(H, beta) - r).T,
           np.dot(inv(np.dot(np.dot(H, V), H.T)), np.dot(H, beta) - r))

# Using dot method:
S = (H.dot(beta) - r).T.dot(inv(H.dot(V).dot(H.T))).dot(H.dot(beta) - r)

如果使用@操作符。

#With the @ operator, the direct translation of the above formula #becomes:
S = (H @ beta - r).T @ inv(H @ V @ H.T) @ (H @ beta - r)