numpy.array中的运算
给定一个向量，让向量中每一个数乘以2
a=(0,1,2)
a*1=(0,2,4)
n=10
L=[i for i in range(n)]
2*L
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
%%timeit
A=[]
for e in L:
A.append(2*e)
2.82 µs ± 74.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%%timeit
A=[2*e for e in L] #比for循环快
7.42 µs ± 251 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
import numpy as np
L= np.arange(10E6)
L
array([0.000000e+00, 1.000000e+00, 2.000000e+00, ..., 9.999997e+06,
9.999998e+06, 9.999999e+06])
%%timeit
A = np.array(2*e for e in L)
5.34 µs ± 153 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%%timeit
A=2*L
47.4 ms ± 914 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
A
array([ 0, 2, 4, 6, 8, 10, 12, 14, 16, 18])
n=10
L=np.arange(n)
2*L
array([ 0, 2, 4, 6, 8, 10, 12, 14, 16, 18])
Universal Functions
X=np.arange(1,16).reshape([3,5])
X
array([[ 1, 2, 3, 4, 5],
[ 6, 7, 8, 9, 10],
[11, 12, 13, 14, 15]])
X+1
array([[ 2, 3, 4, 5, 6],
[ 7, 8, 9, 10, 11],
[12, 13, 14, 15, 16]])
X-1
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
X*2
array([[ 2, 4, 6, 8, 10],
[12, 14, 16, 18, 20],
[22, 24, 26, 28, 30]])
X/2
array([[0.5, 1. , 1.5, 2. , 2.5],
[3. , 3.5, 4. , 4.5, 5. ],
[5.5, 6. , 6.5, 7. , 7.5]])
X//2
array([[0, 1, 1, 2, 2],
[3, 3, 4, 4, 5],
[5, 6, 6, 7, 7]], dtype=int32)
X**2
array([[ 1, 4, 9, 16, 25],
[ 36, 49, 64, 81, 100],
[121, 144, 169, 196, 225]], dtype=int32)
X%2
array([[1, 0, 1, 0, 1],
[0, 1, 0, 1, 0],
[1, 0, 1, 0, 1]], dtype=int32)
1/X
array([[1. , 0.5 , 0.33333333, 0.25 , 0.2 ],
[0.16666667, 0.14285714, 0.125 , 0.11111111, 0.1 ],
[0.09090909, 0.08333333, 0.07692308, 0.07142857, 0.06666667]])
np.abs(X)
array([[ 1, 2, 3, 4, 5],
[ 6, 7, 8, 9, 10],
[11, 12, 13, 14, 15]])
np.sin(X)
array([[ 0.84147098, 0.90929743, 0.14112001, -0.7568025 , -0.95892427],
[-0.2794155 , 0.6569866 , 0.98935825, 0.41211849, -0.54402111],
[-0.99999021, -0.53657292, 0.42016704, 0.99060736, 0.65028784]])
np.cos(X)
array([[ 0.54030231, -0.41614684, -0.9899925 , -0.65364362, 0.28366219],
[ 0.96017029, 0.75390225, -0.14550003, -0.91113026, -0.83907153],
[ 0.0044257 , 0.84385396, 0.90744678, 0.13673722, -0.75968791]])
np.tan(X)
array([[ 1.55740772e+00, -2.18503986e+00, -1.42546543e-01,
1.15782128e+00, -3.38051501e+00],
[-2.91006191e-01, 8.71447983e-01, -6.79971146e+00,
-4.52315659e-01, 6.48360827e-01],
[-2.25950846e+02, -6.35859929e-01, 4.63021133e-01,
7.24460662e+00, -8.55993401e-01]])
np.exp(X)
array([[2.71828183e+00, 7.38905610e+00, 2.00855369e+01, 5.45981500e+01,
1.48413159e+02],
[4.03428793e+02, 1.09663316e+03, 2.98095799e+03, 8.10308393e+03,
2.20264658e+04],
[5.98741417e+04, 1.62754791e+05, 4.42413392e+05, 1.20260428e+06,
3.26901737e+06]])
np.power(3,X)
array([[ 3, 9, 27, 81, 243],
[ 729, 2187, 6561, 19683, 59049],
[ 177147, 531441, 1594323, 4782969, 14348907]], dtype=int32)
3**X
array([[ 3, 9, 27, 81, 243],
[ 729, 2187, 6561, 19683, 59049],
[ 177147, 531441, 1594323, 4782969, 14348907]], dtype=int32)
np.log(X)
array([[0. , 0.69314718, 1.09861229, 1.38629436, 1.60943791],
[1.79175947, 1.94591015, 2.07944154, 2.19722458, 2.30258509],
[2.39789527, 2.48490665, 2.56494936, 2.63905733, 2.7080502 ]])
np.log2(X)
array([[0. , 1. , 1.5849625 , 2. , 2.32192809],
[2.5849625 , 2.80735492, 3. , 3.169925 , 3.32192809],
[3.45943162, 3.5849625 , 3.70043972, 3.80735492, 3.9068906 ]])
np.log10(X)
array([[0. , 0.30103 , 0.47712125, 0.60205999, 0.69897 ],
[0.77815125, 0.84509804, 0.90308999, 0.95424251, 1. ],
[1.04139269, 1.07918125, 1.11394335, 1.14612804, 1.17609126]])
矩阵运算
A=np.arange(4).reshape(2,2)
A
array([[0, 1],
[2, 3]])
B=np.full((2,2),10)
B
array([[10, 10],
[10, 10]])
A+B
array([[10, 11],
[12, 13]])
A-B
array([[-10, -9],
[ -8, -7]])
A*B
array([[ 0, 10],
[20, 30]])
A/B
array([[0. , 0.1],
[0.2, 0.3]])
A.T #第一行变第一列，低行变第二列
array([[0, 2],
[1, 3]])
C = np.full((3,3),666)
A+C
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-48-e9ee25e268b8> in <module>
----> 1 A+C
ValueError: operands could not be broadcast together with shapes (2,2) (3,3)
A.dot(C) #乘法
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-49-36f3f9c6ed4d> in <module>
----> 1 A.dot(C)
ValueError: shapes (2,2) and (3,3) not aligned: 2 (dim 1) != 3 (dim 0)
向量和矩阵的运算
v=np.array([1,2])
A
array([[0, 1],
[2, 3]])
v+A
array([[1, 3],
[3, 5]])
A.shape[0]
2
np.vstack([v]*A.shape[0])#?
array([[1, 2],
[1, 2]])
np.vstack([v]*A.shape[0])+A
array([[1, 3],
[3, 5]])
np.tile(v,(2,1))# 行2磁，列1次
array([[1, 2],
[1, 2]])
np.tile(v,(2,1))+A
array([[1, 3],
[3, 5]])
v
array([1, 2])
A
array([[0, 1],
[2, 3]])
v*A
array([[0, 2],
[2, 6]])
v.dot(A) #满足矩阵乘法的定义
array([4, 7])
A.dot(v) #自动识别成行向量或列向量
array([2, 8])
矩阵的逆
A
array([[0, 1],
[2, 3]])
np.linalg.inv(A) #逆矩阵
array([[-1.5, 0.5],
[ 1. , 0. ]])
invA = np.linalg.inv(A) #逆矩阵,只有方阵才有逆矩阵
A.dot(invA)
array([[1., 0.],
[0., 1.]])
X=np.arange(16).reshape((2,8))
X
array([[ 0, 1, 2, 3, 4, 5, 6, 7],
[ 8, 9, 10, 11, 12, 13, 14, 15]])
np.linalg.inv(X) #逆矩阵
---------------------------------------------------------------------------
LinAlgError Traceback (most recent call last)
<ipython-input-75-5122d6b32d90> in <module>
----> 1 np.linalg.inv(X) #逆矩阵
<__array_function__ internals> in inv(*args, **kwargs)
~\anaconda3\lib\site-packages\numpy\linalg\linalg.py in inv(a)
538 a, wrap = _makearray(a)
539 _assert_stacked_2d(a)
--> 540 _assert_stacked_square(a)
541 t, result_t = _commonType(a)
542
~\anaconda3\lib\site-packages\numpy\linalg\linalg.py in _assert_stacked_square(*arrays)
201 m, n = a.shape[-2:]
202 if m != n:
--> 203 raise LinAlgError('Last 2 dimensions of the array must be square')
204
205 def _assert_finite(*arrays):
LinAlgError: Last 2 dimensions of the array must be square
pinvX=np.linalg.pinv(X) #伪逆矩阵
pinvX.shape
(8, 2)
X.dot(pinvX) #结果为单位矩阵
array([[ 1.00000000e+00, -2.49800181e-16],
[ 6.66133815e-16, 1.00000000e+00]])
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