利用Python进行数学分析

1.数据分析之Numpy教程

2018-10-05  本文已影响0人  VickeyLiu

1.快速构造一个矩阵

a = np.zeros(shape=(3, 5), dtype=int)
print(a)
[[0 0 0 0 0]
 [0 0 0 0 0]
 [0 0 0 0 0]]
a = np.ones(shape=(2, 4), dtype=int)
print(a)
[[1 1 1 1]
 [1 1 1 1]]
a = np.full((3, 5), 666)
print(a)
[[666 666 666 666 666]
 [666 666 666 666 666]
 [666 666 666 666 666]]
a = np.random.randint(0, 10, (3, 5))
print(a)
[[7 5 0 9 3]
 [7 3 9 8 7]
 [9 1 6 9 7]]
a = np.random.normal(0, 1, (3, 5))
print(a)
[[ 0.43121389 -1.45724221  0.84369408 -1.62208387  0.42111614]
 [-2.70974994  0.46920864  1.48373216 -1.72006643 -0.29006381]
 [-0.70056842  0.22743593  0.6276454  -0.78630736 -1.17294585]]
b = np.mat(np.eye(3, 3, dtype=int))
print(b)
# 生成一个3*3的对角矩阵
[[1 0 0]
 [0 1 0]
 [0 0 1]]
b = np.mat(np.diag([1, 2, 3]))
print(b)
[[1 0 0]
 [0 2 0]
 [0 0 3]]

2.矩阵的常用操作

a = np.mat([[1, 1, 1], [2, 3, 6], [4, 5, 7]])
[[1 1 1]
 [2 3 6]
 [4 5 7]]
print(a.ndim)  # 查询几维数组 2
print(a.shape)  # 查询几行几列 (3, 3)
print(a.size)  # 查询元素个数 9
print(a[1, 1])  # 输出1行1列(0开始算起)元素 3
print(a.max())  # 计算矩阵中最大元素 7
print(max(a[:, 1]))  # 计算第一列中的最大值,得到的是矩阵 [[5]]
print(a[1, :].max())  # 计算第二行的最大值,得到一个数 6
print(np.max(a, 0))  # 计算所有列的最大值
[[4 5 7]]
print(np.max(a, 1))  # 计算所有行的最大值
[[1]
 [6]
 [7]]
print(np.argmax(a, 0))  # 求所有列的最大值的索引
[[2 2 2]]
print(np.argmax(a[1, :]))  # 计算第二行的最大值在该行的索引 2
print(a[:2, :2])  # 从原矩阵中分割出(0-2)*(0-2)的子矩阵
[[1 1]
 [2 3]]
d1 = np.mat(np.ones((2, 2)))
d2 = np.mat(np.eye(2))
print(d1, d2)
[[1. 1.]
 [1. 1.]]
 [[1. 0.]
 [0. 1.]]
d3 = np.vstack((d1, d2))  # 两个矩阵按列合并
print(d3)
[[1. 1.]
 [1. 1.]
 [1. 0.]
 [0. 1.]]
d4 = np.hstack((d1, d2))
print(d4)  # 按行合并
[[1. 1. 1. 0.]
 [1. 1. 0. 1.]]

3.矩阵的运算

b1 = np.mat([1, 2, 3])
b2 = np.mat([[3], [2], [1]])
print(b1, b2)
b = b1 * b2
print(b)
[[1 2 3]]
 [[3]
 [2]
 [1]]
[[10]]
b1 = np.mat([1, 2])
b2 = np.mat([3, 4])
print(b1, b2)
b = np.multiply(b1, b2)
print(b)
[[1 2]] [[3 4]]
[[3 8]]
b1 = np.mat([1, 1])
b = b1 * 2
print(b)
# 矩阵与数的点乘
[[2 2]]
c1 = np.mat(np.eye(2, 2)*0.5)  # 构建一个2行2列的对角矩阵,元素为0.5
print(c1)
c = c1.I
print(c)
[[0.5 0. ]
[0.  0.5]]
[[2. 0.]
[0. 2.]]
c1 = np.array([[1, 2], [3, 4]])
c = np.linalg.inv(c1)
print(c)
[[-2.   1. ]
 [ 1.5 -0.5]]
c1 = np.mat([[1, 2], [3, 4]])
c = c1.T
print(c)
[[1 3]
 [2 4]]
c1 = np.array([[1, 2], [3, 4]])
c = c1.transpose()
print(c)
[[1 3]
 [2 4]]
c = np.array([[1, 2], [3, 4]])
print(np.linalg.det(c))
-2.0000000000000004
c = np.array([[1, 2], [3, 4]])
print(np.linalg.eig(c))
# 所得的元组中,第一个为特征值元组,第二个为相对应的特征向量
(array([-0.37228132,  5.37228132]), array([[-0.82456484, -0.41597356],
       [ 0.56576746, -0.90937671]]))
c = np.array([[1, 2], [3, 4]])
d = np.array([[5], [10]])
print(np.linalg.solve(c, d))
"""
求解线性方程组
1X + 2Y = 5
3X + 4Y = 10
"""
[[0. ]
 [2.5]]
c = np.mat([[1, 1], [2, 3], [4, 5]])
c1 = c.sum(axis=0)
print(c1)
# 求列和,得到1*2的矩阵
c2 = c.sum(axis=1)
print(c2)
# 求行和,得到3*1的矩阵
[[7 9]]
[[2]
 [5]
 [9]]
上一篇下一篇

猜你喜欢

热点阅读