markdown Python numpy矩阵数组操作基本语法

Posted 2021-05-05

# numpy 向量操作

## 求向量模长
``````
import numpy as np
x = np.array([1,2,3,4,5])
np.linalg.norm(x)  # 默认求二范数
np.linalg.norm(x,ord=1) # 求一范数
``````
![](https://ws1.sinaimg.cn/large/cdd040eely1g1ivoq3tmhj20h005swep.jpg)

## 求两个向量叉乘（cross product）  
``````
numpy.cross(a,b,axisa=-1, axisb=-1, axisc=-1, axis=None)
``````
``````
x = [1, 2, 3]
y = [4, 5, 6]
np.cross(x, y)
# Output:array([-3,  6, -3])
``````  

# numpy矩阵操作  
## 矩阵乘法  
``````
# 矩阵：
A = np.array([[1, 2, 3], [4, 5, 6]]) # 2 x 3矩阵
B = np.array([[1, 2], [3, 4], [5, 6]]) # 3 x 2矩阵
np.dot(A,B)
# Output: [[22 28]
          [49 64]]
# 向量：
a = np.array([1, 2, 3])
b = np.array([4, 5, 6])
np.dot(a,b)
# Output: 32
``````
## 两矩阵对应元素相乘（数乘）  
``````
A = np.array([[1, 2, 3], [4, 5, 6]])
B = np.array([[7, 8, 9], [4, 7, 1]])
A * B
# 等价于np.multiply(A,B)
# Output：[[ 7 16 27]
          [16 35  6]]
``````
## 两矩阵的拼接  
``````
a = array(［0, 1, 2],
       [3, 4, 5],
       [6, 7, 8］)
b = array(［ 0, 2, 4],
       [ 6, 8, 10],
       [12, 14, 16］)
       
# 按垂直方向堆叠
c = np.vstack((a,b))
# Output:
c = array(［ 0, 1, 2],
       [ 3, 4, 5],
       [ 6, 7, 8],
       [ 0, 2, 4],
       [ 6, 8, 10],
       [12, 14, 16］)

# 按水平方向拼接  
d = np.hstack((a,b))
# Output：
d = array(［ 0, 1, 2, 0, 2, 4],
       [ 3, 4, 5, 6, 8, 10],
       [ 6, 7, 8, 12, 14, 16］)

# 按深度方向堆叠
e = np.dstack((a,b))
# Output：
e = array(［[ 0, 0],
        [ 1, 2],
        [ 2, 4］,

       ［ 3, 6],
        [ 4, 8],
        [ 5, 10］,

       ［ 6, 12],
        [ 7, 14],
        [ 8, 16］])

# 多维矩阵按任意axis堆叠
f = np.concatenate((a,b),axis=1) # axis = 1相当于水平堆叠
``````

## 所有矩阵元素取exp  
``````
np.exp(A)
``````
# 矩阵的保存和读取  
``````
    filename1 = 'train_leftleg_pos.dat'
    import _pickle as pickle
    output1 = open(filename1, 'wb')
    pickle.dump(train, output1)
    
    file1 = open(filename1, 'rb')
    train = pickle.load(file1)
    file1.close()
``````