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。
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
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\
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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
期望：E(X)=1n∑i=1n(xi)期望： E(\boldsymbol{X}) = {\frac{1}{n}}\sum_{i=1}^{n}(x_i)期望：E(X)=n1i=1∑n(xi)
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 总体方差：D(X)=E(X−E(X))=n1i=1∑n(xi−xˉ)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} 总体标准差：σ=D(X)=n1i=1∑n(xi−xˉ)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)）H^Ψ=EΨ。(Ψ是需要求的矩阵，E是本征值，有很多个——(E1,E2,E3,⋯,En)）
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
result=a⃗⋅b⃗=∑i=1naibi result = \vec{a} \cdot \vec{b} = \sum_{i=1}^{n}a_i b_iresult=a⋅b=i=1∑naibi
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
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