The first job --NumPy practice

1. Create a value of 1 and the inner boundary of the array is 0, the following legend:
[Note:] solving this problem can first of all values are set to 1, this is a big square; Next, the setting of all boundaries except small square 0.
The title uses a slice of numpy principle. X multidimensional arrays follow the same principle [start: step: stop] is.
[1. 1. 1. 1. 1. 1. 1. 1. 1. 1.]
[1. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
[1. 0.0 0. 0. 0. 0. 0. 0. 1.]
[1. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
[1. 0. 0. 0. 0. 0 0. 0. 0. 1.]
[1. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
[1. 0. 0. 0. 0. 0. 0. 0. 0 1.]
[1. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
[1. 0. 0. 0. 0. 0. 0. 0. 0. 1.]
[1 1. 1. 1. 1. 1. 1. 1. 1. 1.]

Import numpy NP AS 
A = np.ones ((10,10),) # Create 10 rows and 10 columns, all the elements of an array 
a [1: 9:, [ 1,2,3,4,5,6, 7,8]] = 0 # to all boundaries except small square to 0 
Print (A)

 

2. Create a value on the main diagonal array 1,2,3,4 5x5 matrix, the following legend:

[1000]
[0200]
[0030]
[0004 ]

import numpy as np 
a = np.array([[1,0,0,0],[0,2,0,0],[0,0,3,0],[0,0,0,4]])
print(a)

 

3. The operation of the normalization array
generating a random 5 * 5 matrix, to find the maximum and minimum values, and the maximum and minimum values 1 and 0, respectively, said other values between 0 and 1 in the middle.

import numpy as np
a= np.random.rand(5,5)
print(a)
print(np.max(a))
print(np.min(a))
a[a == a.max()] = 1
a[a == a.min()] = 0
print(a)