CODEWITHSUNDEEP
pythonnumpynormalization
Daniel Zapata
Daniel Zapata

Reputation: 852

Normalization: Euclidean distance between a numpy array and one point

I have a numpy array (68x2) which correspond to 68 different points of a detected face. 

[16.0000 93.0000]
[17.0000 116.0000]
[20.0000 139.0000]
[25.0000 162.0000]
[33.0000 184.0000]
[47.0000 205.0000]
[66.0000 219.0000] ... until 68

These points have the origin at the left bottom corner of the picture. I want to normalize according to a new center. Two questions, is there a way to do this without a loop? And is this the correct way to normalize according to a new origin?

new_origin = [112,135]
new_X
for point in X[0][0]:
    new_X.append(point-new_origin)

Upvotes: 1

Views: 1999

Answers (1)

Eric Duminil
Eric Duminil

Reputation: 54293

If you just want to translate those points, all you need to do is subtract a value to the left column (X values) and another to the right column (Y values):

>>> import numpy as np
>>> a = np.arange(10).reshape(5,2)
>>> a
array([[0, 1],
       [2, 3],
       [4, 5],
       [6, 7],
       [8, 9]])
>>> a[:,0] = a[:,0] - 112
>>> a[:,1] = a[:,1] - 135
>>> a
array([[-112, -134],
       [-110, -132],
       [-108, -130],
       [-106, -128],
       [-104, -126]])

You can do directly with np.subtract:

>>> np.subtract(a, [112, 135])
array([[-112, -134],
       [-110, -132],
       [-108, -130],
       [-106, -128],
       [-104, -126]])

or just :

>>> a - [112, 135]
array([[-112, -134],
       [-110, -132],
       [-108, -130],
       [-106, -128],
       [-104, -126]])

Note that with numpy, you almost never have to manually iterate over each element.

Upvotes: 1

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