KP2.ipynb

Kernel: Python 3 (old Anaconda 3)

#num1

import numpy as np

X = np.linspace(0,20,6);print(X)
Y = np.copy(X);print(Y,'\n')
print(np.round(np.sin(Y**2),3),'\n')
X = X.reshape(2,3)
Y = Y.reshape(3,2)
print(X,'\n',Y,'\n')
print(X.dot(Y),'\n')
Y = Y.reshape(2,3)
print(np.hstack((X,Y)),'\n',np.vstack((X,Y)))

[  0.   4.   8.  12.  16.  20.]
[  0.   4.   8.  12.  16.  20.] 

[ 0.    -0.288  0.92  -0.491 -0.999 -0.851] 

[[  0.   4.   8.]
 [ 12.  16.  20.]] 
 [[  0.   4.]
 [  8.  12.]
 [ 16.  20.]] 

[[ 160.  208.]
 [ 448.  640.]] 

[[  0.   4.   8.   0.   4.   8.]
 [ 12.  16.  20.  12.  16.  20.]] 
 [[  0.   4.   8.]
 [ 12.  16.  20.]
 [  0.   4.   8.]
 [ 12.  16.  20.]]

#num2

import numpy.linalg as li

A = np.array([np.random.randint(-5,5) for x in np.arange(9)])
A = A.reshape(3,3)
B = np.array([np.random.randint(-5,5) for x in np.arange(3)])
print(A,'\n',B,'\n')
print(' det = ', li.det(A),'\n','rank = ', np.rank(A),'\n','spur = ', np.trace(A),'\n','norm = ', li.norm(B),'\n')
print(li.solve(A,B))
a = li.inv(A)
print(a.dot(B),'\n')
x = a.dot(B)
print(x.dot(B),'\n', np.cross(x,B),'\n', np.outer(x,B),'\n', '\n', li.eig(A))

[[ 4 -5  3]
 [-5  2  3]
 [-5  1 -3]] 
 [ 3 -4 -1] 

 det =  129.0 
 rank =  2 
 spur =  3 
 norm =  5.09901951359 

[ 0.3255814  -0.58139535 -0.40310078]
[ 0.3255814  -0.58139535 -0.40310078] 

3.70542635659 
 [-1.03100775 -0.88372093  0.44186047] 
 [[ 0.97674419 -1.30232558 -0.3255814 ]
 [-1.74418605  2.3255814   0.58139535]
 [-1.20930233  1.6124031   0.40310078]] 
 
 (array([ 7.94155764+0.j        , -2.47077882+3.18416658j,
       -2.47077882-3.18416658j]), array([[-0.62293876+0.j        ,  0.51135609-0.02741701j,
         0.51135609+0.02741701j],
       [ 0.70027022+0.j        ,  0.61012430+0.j        ,  0.61012430-0.j        ],
       [ 0.34866736+0.j        , -0.05698345+0.60188412j,
        -0.05698345-0.60188412j]]))

/ext/anaconda3/lib/python3.5/site-packages/ipykernel/__main__.py:5: VisibleDeprecationWarning: `rank` is deprecated; use the `ndim` attribute or function instead. To find the rank of a matrix see `numpy.linalg.matrix_rank`.

#num3

con = lambda i, j: 1/np.sqrt((i + 1)**2 + (j + 1)**2)

import time 
t1 = time.clock()  
A = np.zeros((1000,1000))
for i in np.arange(1000):
    for j in np.arange(1000):
        A[i,j] = con(i,j)
t2 = time.clock() 
print(t2-t1)

2.3240479999999977

t1 = time.clock()  
A = np.reshape([con(i,j) for i in np.arange(1000) for j in np.arange(1000)],(1000,1000)) 
t2 = time.clock() 
print(t2-t1)

2.0921179999999993

t1 = time.clock() 
A = np.fromfunction(con,(1000,1000))
t2 = time.clock() 
print(t2-t1)

0.029004000000000474

########################

import numpy as np
import numpy.linalg as li

N = [np.random.randint(-10,10) for i in range(30)];print(N)

[-10, -5, 5, -9, -10, 0, 4, 2, 5, 5, 5, 6, -2, -4, 7, 6, 9, -7, 6, 2, -8, -8, 2, 9, 9, -2, 6, 8, -7, -6]

n = 0
M = []
for i in range(len(N)):
    if N[i] > 0:
        M += [N[i]]
        n += 1

print(M,'\n',n,'\n', sum(M)/n)

[5, 4, 2, 5, 5, 5, 6, 7, 6, 9, 6, 2, 2, 9, 9, 6, 8] 
 17 
 5.647058823529412

Np = [x for x in N if x > 0]
print(sum(Np)/len(Np))

5.647058823529412