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Let's say I have the following array:

a = np.random.randint(5, size=(2000, 2000, 1))
a = np.repeat(a, 3, axis=2) # Using this method to have a (m,n,3) array with the same values

and the next arrays:

val_old = np.array([[0,  0,   0],  [3,  3,  3]])
val_new = np.array([[12, 125, 13], [78, 78, 0]])

What I want to do is to replace the values from the array a with the values specified in the array val_new. So, all [0,0,0] arrays would become [12,125,13] and all [3,3,3] would become [78, 78, 0].

I can't find an efficient way to do this... I tried to adapt this solution but it's only for 1-d arrays...

Does anyone know a fast way/method to replace these values ?

Etienne Landuré
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  • Would all three values in each of the row in `val_old` be the same? Also, are all rows from `val_old` guaranteed to be in `a`? And can we use `a = np.random.randint(5, size=(2000, 2000, 1))` as the starting point, that is since the repeated version would have the values repeated anyway, so start off with the original "2D" version? – Divakar Apr 15 '19 at 18:14

1 Answers1

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Assuming you have a "map" for each integer, you can use a (2000, 2000) index on a (5,) array to broadcast to a (2000,2000, 5) array. example:

val_new = np.array([[12, 125, 13], [0,0,0], [1,3,3], [78, 78, 0]])  #0, 1, 2, 3
a = np.random.randint(4,size=(4,5))

val_new[a]  # (4,5,3) shaped array

>>array([[[  0,   0,   0],
    [ 78,  78,   0],
    [ 78,  78,   0],
    [ 12, 125,  13],
    [  0,   0,   0]],
....
   [[ 12, 125,  13],
    [ 12, 125,  13],
    [  0,   0,   0],
    [ 12, 125,  13],
    [  0,   0,   0]]])
Tarifazo
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