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 as np
x=np.ones((10,10))
x[1:9,1:9]=0
print(x)

 

 

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
x = np.arange(1, 5)
print(np.diag(x))

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 intermediate

import numpy as np
X = np.random.random((5, 5))
Xmax, Xmin = X.max(), X.min()
X = (X - Xmin)/(Xmax - Xmin)
print(X)

 

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Origin www.cnblogs.com/huangdatou/p/11592776.html