numpy 基础——Vectors, Matrices and Arrays

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本文介绍使用Python的NumPy库进行矩阵操作的方法，包括矩阵创建、元素选取、矩阵描述、矩阵运算等基本操作，并深入探讨了矩阵转置、展平、求秩等高级应用。

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学习目标：

结合矩阵去学习 Python 中 numpy

学习内容：

矩阵
Python 的 numpy 的基础

学习时间：

2020.11.09

学习产出：

1. 矩阵和 numpy

1.1 创建一维矩阵

In [1]: import numpy as np

In [2]: vector_row = np.array([1, 2, 3])

In [3]: vector_column = np.array([[1], [2], [3]])		# 注意两个 中括号

In [4]: vector_row
Out[4]: array([1, 2, 3])

In [5]: vector_column
Out[5]: 
array([[1],
       [2],
       [3]])

1.2 创建二维矩阵

In [9]: matrix_2d = np.array([[1, 2, 4], [3, 6.5, 9], [3, 2, 4]])		# 注意两个 中括号


In [10]: matrix_2d
Out[10]: 
array([[1. , 2. , 4. ],
       [3. , 6.5, 9. ],
       [3. , 2. , 4. ]])

1.3 创建稀疏矩阵

\

\
\
\

1.4 选择元素

In [17]: vector_column
Out[17]: 
array([[1],
       [2],
       [3]])

In [18]: vector_row
Out[18]: array([1, 2, 3])

In [19]: matrix_2d
Out[19]: 
array([[1. , 2. , 4. ],
       [3. , 6.5, 9. ],
       [3. , 2. , 4. ]])
#############################################
# 元素选择 从 0 开始
In [20]: vector_column[1]
Out[20]: array([2])

In [21]: matrix_2d[0, 0]
Out[21]: 1.0

In [22]: matrix_2d[0][0]
Out[22]: 1.0

# Select all elements of a vector
In [23]: matrix_2d[:]
Out[23]: 
array([[1. , 2. , 4. ],
       [3. , 6.5, 9. ],
       [3. , 2. , 4. ]])

# Select the elements of the 2nd column
In [24]: matrix_2d[:, 1]
Out[24]: array([2. , 6.5, 2. ])

# Select the elements of the 1st row
In [25]: matrix_2d[0, :]
Out[25]: array([1., 2., 4.])

# Select the elements of the 2nd and 3rd column in the 1st row
# 选择第 1 行第 2，3 列的元素
In [26]: matrix_2d[0, 1:3]
Out[26]: array([2., 4.])

1.5 矩阵的描述

In [29]: matrix_2d = np.array([[1,2,3], [2,3.3,4], [6, 6, 6], [9, 9.9, 90]])

In [30]: matrix_2d
Out[30]: 
array([[ 1. ,  2. ,  3. ],
       [ 2. ,  3.3,  4. ],
       [ 6. ,  6. ,  6. ],
       [ 9. ,  9.9, 90. ]])

In [31]: matrix_2d.shape
Out[31]: (4, 3)				# 行 × 列

In [32]: matrix_2d.size		
Out[32]: 12					# 行 × 列

In [34]: matrix_2d.ndim
Out[34]: 2					# 二维矩阵

1.6 矩阵的操作符

# all elements add 10
In [30]: matrix_2d
Out[30]: 
array([[ 1. ,  2. ,  3. ],
       [ 2. ,  3.3,  4. ],
       [ 6. ,  6. ,  6. ],
       [ 9. ,  9.9, 90. ]])

In [35]: mat_10 = matrix_2d + 10				# add directly

In [36]: mat_10
Out[36]: 
array([[ 11. ,  12. ,  13. ],
       [ 12. ,  13.3,  14. ],
       [ 16. ,  16. ,  16. ],
       [ 19. ,  19.9, 100. ]])

# all elements divied by 2
In [37]: mat_d_2 = matrix_2d / 2

In [38]: mat_d_2
Out[38]: 
array([[ 0.5 ,  1.  ,  1.5 ],
       [ 1.  ,  1.65,  2.  ],
       [ 3.  ,  3.  ,  3.  ],
       [ 4.5 ,  4.95, 45.  ]])

1.7 查找最大值和最小值

In [39]: mat_1 = matrix_2d

In [40]: mat_1
Out[40]: 
array([[ 1. ,  2. ,  3. ],
       [ 2. ,  3.3,  4. ],
       [ 6. ,  6. ,  6. ],
       [ 9. ,  9.9, 90. ]])

In [44]: np.max(mat_1)
Out[44]: 90.0

In [45]: np.min(mat_1)
Out[45]: 1.0

# Find maximum element in each column  每列最大值
In [46]: np.max(mat_1, axis=0)
Out[46]: array([ 9. ,  9.9, 90. ])

# Find minimum element in each row 每行最大值
In [47]: np.min(mat_1, axis=1)
Out[47]: array([1., 2., 6., 9.])

1.8 计算期望，总体方差，总体标准差

$E(\boldsymbol{X}) = {\frac{1}{n}}\sum_{i=1}^{n}(x_i)$
$总体方差：D(X)=E(X−E(X))=1n∑i=1n(xi−xˉ)2总体方差：D(\boldsymbol{X}) = E(\boldsymbol{X} - E(\boldsymbol{X})) = {\frac{1}{n}}\sum_{i=1}^{n}(x_i - \bar{x})^2$
$总体标准差：σ=D(X)=1n∑i=1n(xi−xˉ)2总体标准差：\sigma = \sqrt{D(\boldsymbol{X})}= \sqrt{{\frac{1}{n}}\sum_{i=1}^{n}(x_i - \bar{x})^2}$

In [48]: matrix = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])

In [49]: matrix
Out[49]: 
array([[1, 2, 3], 
       [4, 5, 6], 
       [7, 8, 9]])

# 下面的操作会把整个矩阵看作是一维的:  X = array([[1, 2, 3, 4, 5, 6, 7, 8, 9]])
# Average
In [50]: np.mean(matrix)
Out[50]: 5.0

# Total Variance
In [51]: np.var(matrix)
Out[51]: 6.666666666666667

# Total Standard Deviation
In [52]: np.std(matrix)
Out[52]: 2.581988897471611

1.9 将矩阵 reshape



In [56]: matrix_2d
Out[56]: 
array([[ 1. ,  2. ,  3. ], 
       [ 2. ,  3.3,  4. ], 
       [ 6. ,  6. ,  6. ], 
       [ 9. ,  9.9, 90. ]])

# make matrix_2d (4 × 3) to (12 × 1)
In [58]: matrix_2d.reshape(12,1)
Out[58]: 
array([[ 1. ],
       [ 2. ],
       [ 3. ],
       [ 2. ],
       [ 3.3],
       [ 4. ],
       [ 6. ],
       [ 6. ],
       [ 6. ],
       [ 9. ],
       [ 9.9],
       [90. ]])

# Parameters -1 represent any number
In [61]: matrix_2d
Out[61]: 
array([[ 1. ,  2. ,  3. ], 
       [ 2. ,  3.3,  4. ], 
       [ 6. ,  6. ,  6. ], 
       [ 9. ,  9.9, 90. ]])

In [62]: matrix
Out[62]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

In [61]: matrix_2d
Out[61]: 
array([[ 1. ,  2. ,  3. ], 
       [ 2. ,  3.3,  4. ], 
       [ 6. ,  6. ,  6. ], 
       [ 9. ,  9.9, 90. ]])

In [62]: matrix
Out[62]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

In [63]: matrix_2d.reshape(1, -1)			# equals (1 × 12)
Out[63]: 
array([[ 1. ,  2. ,  3. ,  2. ,  3.3,  4. ,  6. ,  6. ,  6. ,  9. ,  9.9, 90. ]])

In [64]: matrix.reshape(1, -1)				# equals (1 × 9)
Out[64]: array([[1, 2, 3, 4, 5, 6, 7, 8, 9]])

1.10 矩阵的转置

In [65]: matrix_2d
Out[65]: 
array([[ 1. ,  2. ,  3. ],
       [ 2. ,  3.3,  4. ],
       [ 6. ,  6. ,  6. ],
       [ 9. ,  9.9, 90. ]])

In [66]: matrix_2d.T
Out[66]: 
array([[ 1. ,  2. ,  6. ,  9. ],
       [ 2. ,  3.3,  6. ,  9.9],
       [ 3. ,  4. ,  6. , 90. ]])

###########################################
In [67]: mat_1 = matrix_2d.reshape(1, -1)

In [68]: mat_1
Out[68]: 
array([[ 1. ,  2. ,  3. ,  2. ,  3.3,  4. ,  6. ,  6. ,  6. ,  9. ,  9.9,
        90. ]])

In [69]: mat_1.T					# 行矩阵的转置是列矩阵
Out[69]:      
array([[ 1. ],
       [ 2. ],
       [ 3. ],
       [ 2. ],
       [ 3.3],
       [ 4. ],
       [ 6. ],
       [ 6. ],
       [ 6. ],
       [ 9. ],
       [ 9.9],
       [90. ]])

1.11 将一个矩阵展平

In [71]: matrix
Out[71]: 
array([[1, 2, 3], 
       [4, 5, 6], 
       [7, 8, 9]])

In [72]: matrix.flatten()
Out[72]: array([1, 2, 3, 4, 5, 6, 7, 8, 9])


## 注意 用 reshape 会有两个中括号，遍历调用数据需要用两个 for
In [73]: matrix.reshape(1, -1)
Out[73]: array([[1, 2, 3, 4, 5, 6, 7, 8, 9]])

1.12 寻找矩阵的秩

In [76]: matrix					# 2
Out[76]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

In [77]: matrix2				# 1
Out[77]: 
array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])

In [78]: np.linalg.matrix_rank(matrix)
Out[78]: 2

In [79]: np.linalg.matrix_rank(matrix2)
Out[79]: 1

1.13 计算行列式的值

In [80]: matrix
Out[80]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

In [81]: np.linalg.det(matrix)			# 会发现不是 0 ，原因是 10^-16 -> 0，计算机浮点型计算的精度就是这样的。
Out[81]: 6.66133814775094e-16

In [84]: print("{d:10.7f}".format(d=np.linalg.det(matrix)))    # 将精度控制一下就行                                                                                                                                    0.0000000

1.14 获取对角元素

In [86]: matrix
Out[86]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

In [87]: matrix.diagonal()
Out[87]: array([1, 5, 9])

1.15 计算矩阵的本征值

$H^Ψ=EΨ。(Ψ是需要求的矩阵，E是本征值，有很多个——(E1,E2,E3,⋯ ,En)） \boldsymbol{\hat{H}} \boldsymbol{\Psi} = E \boldsymbol{\Psi} 。( \boldsymbol{\Psi} 是需要求的矩阵， E 是本征值，有很多个——(E_1, E_2, E_3, \cdots, E_n)）$

In [88]: matrix
Out[88]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

In [89]: eigenvalues, eigenvectors = np.linalg.eig(matrix)

In [90]: eigenvalues				# 第三个值为 -1e-15 -> 0，其实就是 0.
Out[90]: array([ 1.61168440e+01, -1.11684397e+00, -1.30367773e-15])

In [91]: eigenvectors
Out[91]: 
array([[-0.23197069, -0.78583024,  0.40824829],
       [-0.52532209, -0.08675134, -0.81649658],
       [-0.8186735 ,  0.61232756,  0.40824829]])

1.16 计算矩阵的迹（对角元素之和，也是本征值之和）

# 先看 1.15
In [88]: matrix
Out[88]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])

In [89]: eigenvalues, eigenvectors = np.linalg.eig(matrix)

In [90]: eigenvalues
Out[90]: array([ 1.61168440e+01, -1.11684397e+00, -1.30367773e-15])

In [91]: eigenvectors
Out[91]: 
array([[-0.23197069, -0.78583024,  0.40824829],
       [-0.52532209, -0.08675134, -0.81649658],
       [-0.8186735 ,  0.61232756,  0.40824829]])

#####  矩阵的迹
In [92]: matrix.trace()
Out[92]: 15

# 本征值之和
In [93]: np.sum(eigenvalues)
Out[93]: 14.999999999999998				# 14.999999999999998 -> 15

# 本征值之和 == 矩阵的迹，在计算机里，需要考虑精度范围

1.17 计算向量的内积

主要讲的一维向量的内积是，多维的以后用上再写

$\vec{a} \cdot \vec{b} = \sum_{i=1}^{n}a_i b_i$

In [99]: a
Out[99]: array([1, 2, 3, 4, 5, 6])

In [100]: b
Out[100]: array([ 6,  7,  8,  9, 10, 11])

In [101]: np.dot(a,b)
Out[101]: 196

1.18 矩阵相乘

(m, n) × （n，l）= (m, l)，就是线性代数中的矩阵相乘

In [105]: a				# 3 × 2
Out[105]: 
array([[1, 3],
       [4, 3],
       [5, 6]])

In [106]: b				# 2 × 3
Out[106]: 
array([[9, 8, 7],
       [1, 2, 4]])

In [107]: np.dot(a,b)	# 结果为 3 × 3
Out[107]: 
array([[12, 14, 19], 
       [39, 38, 40], 
       [51, 52, 59]])

1.19 矩阵求逆

# 前提：满秩矩阵————即矩阵对应行列式不为 0 


In [112]: n
Out[112]: 
array([[1, 2],
       [3, 4]])

In [113]: np.linalg.inv(n)
Out[113]: 
array([[-2. ,  1. ],
       [ 1.5, -0.5]])

#  求其行列式的值
In [114]: np.linalg.det(n)
Out[114]: -2.0000000000000004

## 如果行列式为 0 ，求逆会报 Singular matrix 错误
In [119]: m
Out[119]: 
array([[12, 14, 19],
       [39, 38, 40],
       [51, 52, 59]])

In [120]: np.linalg.det(m)
Out[120]: 0.0

In [121]: np.linalg.inv(m)
---------------------------------------------------------------------------
LinAlgError                               Traceback (most recent call last)
<ipython-input-121-83af191aaacb> in <module>
----> 1 np.linalg.inv(m)

<__array_function__ internals> in inv(*args, **kwargs)

C:\ProgramData\Anaconda3\lib\site-packages\numpy\linalg\linalg.py in inv(a)
    545     signature = 'D->D' if isComplexType(t) else 'd->d'
    546     extobj = get_linalg_error_extobj(_raise_linalgerror_singular)
--> 547     ainv = _umath_linalg.inv(a, signature=signature, extobj=extobj)
    548     return wrap(ainv.astype(result_t, copy=False))
    549

C:\ProgramData\Anaconda3\lib\site-packages\numpy\linalg\linalg.py in _raise_linalgerror_singular(err, flag)
     95
     96 def _raise_linalgerror_singular(err, flag):
---> 97     raise LinAlgError("Singular matrix")
     98
     99 def _raise_linalgerror_nonposdef(err, flag):

LinAlgError: Singular matrix

1.19 numpy 产生随机数据行列式\

可以用于产生随机数据进行学习
具体看官网：

https://docs.scipy.org/doc/numpy-1.15.0/reference/routines.random.html

# Set seed
In [122]: np.random.seed(0)

# Generate three random floats between 0.0 and 1.0
In [123]: np.random.random(3)
Out[123]: array([0.5488135 , 0.71518937, 0.60276338])

# 
In [124]: np.random.normal(0.0, 1.0, 3)
Out[124]: array([-2.2683282 ,  1.33354538, -0.84272405])

1.20 向量的点乘和叉乘

# 空间中三个不共线的向量，可以构成一个平行六面体
# 它的体积可以由: V = a1 点乘 (a2 叉乘 a3)
a1 = np.array([4.09278, 0, 0])
a2 = np.array([-2.04639, 3.54445365, 0])
a3 = np.array([0, 0, 20])

# 求分母，也就是体积
V = np.dot(a1, np.cross(a2, a3))
print(V)

# 结果： 290.13338019294