numpy

2021-11-26

切片

a = np.arange(9)
a = a.reshape((3, 3))
print(a)
print(a[0, 1:2])
print(a[::2, ::2])
'''
[[0 1 2]
 [3 4 5]
 [6 7 8]]
 
[0 1]

[[0 2]
 [6 8]]

'''

运算

'''
a.dot(b)
.dot()是矩阵乘法
需要满足矩阵乘法的条件
a*b是直接将两个矩阵各个元素相乘
'''
a = np.arange(25)
a = a.reshape((5, 5))

b = np.array([10, 62, 1, 14, 2, 56, 79, 2, 1, 45,
              4, 92, 5, 55, 63, 43, 35, 6, 53, 24,
              56, 3, 56, 44, 78])
b = b.reshape((5,5))
print(a*b)
print(a.dot(b))
'''
[[   0   62    2   42    8]
 [ 280  474   14    8  405]
 [  40 1012   60  715  882]
 [ 645  560  102  954  456]
 [1120   63 1232 1012 1872]]
 
[[ 417  380  254  446  555]
 [1262 1735  604 1281 1615]
 [2107 3090  954 2116 2675]
 [2952 4445 1304 2951 3735]
 [3797 5800 1654 3786 4795]]

'''

索引

利用数组索引

a = np.arange(0, 100, 10)
indices = [1, 5, -1]
b = a[indices]
print(a) # >>>[ 0 10 20 30 40 50 60 70 80 90]
print(b) # >>>[10 50 90]

利用布尔值索引

1
2
3

a=np.array([1,2,3,4,5,6,7]) 
isp = a > 3
print(a[isp])# >>>[4,5,6,7]

where索引

返回了一个元组还有dtype类型

a = np.arange(0, 100, 10)
b = np.where(a < 50) 
c = np.where(a >= 50)[0]
print(b) # >>>(array([0, 1, 2, 3, 4]，dtype=int64),)
print(c) # >>>[5 6 7 8 9]

创建

1	a = np.arrange(20).reshape(4,5)

1	a = np.zeros((4,4)) # ones`

1	np.full((2,2), 3)

1	np.eye(3,3)

1	np.linspace(0, 10, num=5) # [ 0. 2.5 5. 7.5 10. ]

矩阵操作

a = np.array([[1,2,3]]) # 注意此处矩阵应是至少二维？
'''
[[1]
 [2]
 [3]]
 '''

1
2
3

A = np.array([[2,1,-2],[3,0,1],[1,1,-1]])
b = np.transpose(np.array([[-3,5,-2]]))
x = np.linalg.solve(A,b) # 求解矩阵