Replace NumPy array elements with other element

Question

Suppose I have a randomly generated 3d array srouceArray:

ex: np.random.rand(3, 3, 3)

array([[[0.61961383, 0.26927599, 0.03847151],
        [0.03497162, 0.77748313, 0.15807293],
        [0.15108821, 0.36729448, 0.19007034]],

      [[0.67734758, 0.88312758, 0.97610746],
       [0.5643174 , 0.20660141, 0.58836553],
       [0.59084109, 0.77019768, 0.35961768]],

      [[0.19352397, 0.47284641, 0.97912889],
       [0.48519117, 0.37189048, 0.37113941],
       [0.94934848, 0.92755083, 0.52662299]]])

I would like to randomly replace all 3rd dimennsion elements to zeroes.

Expected array:

array([[[0, 0, 0],
        [0.03497162, 0.77748313, 0.15807293],
        [0.15108821, 0.36729448, 0.19007034]],

      [[0.67734758, 0.88312758, 0.97610746],
       [0 , 0, 0],
       [0.59084109, 0.77019768, 0.35961768]],

      [[0, 0, 0],
       [0, 0, 0],
       [0.94934848, 0.92755083, 0.52662299]]])

I was thinking about generating "mask"? using random

np.random.choice([True, False], sourceArray.shape, p=[...])

and somehow transforming it to 3d array where False=[0, 0, 0] and True=[1, 1, 1] and multiplying with source...

But I do not know how to achieve that transformation. And I bet there is a simpler way I do not know about.

The pattern in your example of expected array is not clear to me. Could you provide an example in 3x3x3 array? — pmarcol
– pmarcol, Commented May 29, 2019 at 10:39
@pmarcol updated examples of the current array and expected array. [0, 0, 0] in expected array is random. — Martynas Jurkus
– Martynas Jurkus, Commented May 29, 2019 at 11:11

Lante Dellarovere · Accepted Answer · 2019-05-29 11:19:35Z

1

If I correctly understood data structure, can use this (this will change original array):

import numpy as np

l = np.random.rand(5, 4, 3)
m = np.random.choice([True, False], size=(l.shape[0], l.shape[1]))
l[m] = [0, 0, 0]
l
array([[[0.62551611, 0.26268253, 0.51863006],
        [0.        , 0.        , 0.        ],
        [0.45038189, 0.97229114, 0.63736078],
        [0.        , 0.        , 0.        ]],

       [[0.54282399, 0.14585025, 0.80753245],
        [0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        ],
        [0.18190234, 0.19806439, 0.3052623 ]],

       [[0.        , 0.        , 0.        ],
        [0.46409806, 0.39734112, 0.21864433],
        [0.        , 0.        , 0.        ],
        [0.65046231, 0.78573179, 0.76362864]],

       [[0.05296007, 0.50762852, 0.18839052],
        [0.52568072, 0.8271628 , 0.24588153],
        [0.92039708, 0.8653368 , 0.96737845],
        [0.        , 0.        , 0.        ]],

       [[0.        , 0.        , 0.        ],
        [0.37039626, 0.64673356, 0.01186108],
        [0.        , 0.        , 0.        ],
        [0.        , 0.        , 0.        ]]])

edited May 29, 2019 at 11:19

answered May 29, 2019 at 11:12

Lante Dellarovere

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4 Comments

Martynas Jurkus Over a year ago

This was exactly what I wanted. Could you refer me where I could read more about what is happening here?

Lante Dellarovere Over a year ago

@MartynasJurkus I just fixed and continued your solution based on mask application. Basically, we reassign values at the indexes provided by the mask. Maybe you can start from this docs.scipy.org/doc/numpy-1.13.0/user/basics.indexing.html and in particular "mask index arrays" part

Lante Dellarovere Over a year ago

I also like @Masoud approach, which exploits multiplication properties

Martynas Jurkus Over a year ago

This approach np.random.choice allows to leverage probabilities via p=[]

Masoud · Accepted Answer · 2019-05-29 11:38:08Z

1

Mathematically, its possible to generate another random 0-1 array, multiply to original array:

import numpy as np

ar = np.random.rand(3,3,3)
ar2 = np.random.randint(2, size = (3,3,1))
ar3 = ar*ar2

answered May 29, 2019 at 11:38

Masoud

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pmarcol · Accepted Answer · 2019-05-29 11:30:17Z

0

You could do it like this:

a = np.ones((3, 3, 3)) # your original array
b = a.reshape((-1,3)) # array of just rows from 3rd dim
temp = np.random.random(b.shape[0]) # get random value from 0 to 1 for each row from b
prob = 0.4 # up to you - probability of making a row all zeros
mask = temp<prob
b[mask]=0
result = b.reshape(a.shape) # getting back to original shape

Example output:

[[[0. 0. 0.]
  [1. 1. 1.]
  [1. 1. 1.]]

 [[1. 1. 1.]
  [1. 1. 1.]
  [0. 0. 0.]]

 [[0. 0. 0.]
  [1. 1. 1.]
  [0. 0. 0.]]]

answered May 29, 2019 at 11:30

pmarcol

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