Python - Concatenation in an array of redondant values

Question

I would like to "concatenate" results of detections of positions of specific values in my array "Coord3".

I have a double criteria on my array for 2 specific dimension of it. I get some redondant positions and i would like to gather it in order to apply a mask of this values.

As example with a moderate array :

import numpy as np

Coord3 = np.array([[[[ 0.,  0.],
        [ 0.,  1.],
        [ 0.,  2.]],

       [[ 1.,  0.],
        [ 1.,  1.],
        [ 1.,  2.]],

       [[ 2.,  0.],
        [ 2.,  1.],
        [ 2.,  2.]]],


        [[[ 1.,  0.],
        [ 1.,  1.],
        [ 1.,  2.]],

       [[ 2.,  0.],
        [ 2.,  1.],
        [ 2.,  2.]],

       [[ 4.,  0.],
        [ 3.,  1.],
        [ 4.,  2.]]],


        [[[ 2.,  0.],
        [ 2.,  1.],
        [ 2.,  2.]],

       [[ 3.,  0.],
        [ 3.,  1.],
        [ 3.,  2.]],

       [[ 4.,  0.],
        [ 4.,  1.],
        [ 4.,  4.]]]])


#I apply my double criteria in 2 shapes of my array Coord3

plaY=[]

for i in range(Coord3.shape[0]):
    holding_list = zip(*np.where(Coord3[i,:,:,0] > 3))
    plaY.append(holding_list)

plaY_array = np.asarray(plaY)

#plaY_array
#Out[1088]: array([[], [(2, 0), (2, 2)], [(2, 0), (2, 1), (2, 2)]], dtype=object)



plaX=[]

for i in range(Coord3.shape[0]):
    holding_list = zip(*np.where(Coord3[i,:,:,1] > 2))
    plaX.append(holding_list)

plaX_array = np.asarray(plaX)


#plaX_array
#Out[1097]: array([[], [], [(2, 2)]], dtype=object)

So (2,0) and (2,2) are redondant and i would like to supress it and to gather all in an unique array... as :

np.array([[2, 0], [2, 2],....])

--- EDIT LATER ------------------------------------------

I could concatenate for each specific time

plaY_array = plaY_array[:,np.newaxis]

plaX_array = plaX_array[:,np.newaxis]


test = plaX_array + plaY_array

#I get that :

#test
#array([[[[]]],

       #[[[(2, 0), (2, 2)]]],

       #[[[(2, 0), (2, 1), (2, 2), (2, 2)]]]], dtype=object)

So now i have to suppress just the (2,2) redondant in the "third slice", it could be interesting to know the coordinates for each specific time (so i let the (2,2) in "2nd slice"

unutbu · Accepted Answer · 2014-11-12 13:48:47Z

1

You could turn the list of coordinates into a set to remove duplicates:

In [21]: set(zip(*(np.where(Coord3[:,:,:,0] > 3)[1:])))
Out[21]: {(2, 0), (2, 1), (2, 2)}

plaY_array = np.array(list(set(zip(*(np.where(Coord3[:,:,:,0] > 3)[1:])))))
plaX_array = np.array(list(set(zip(*(np.where(Coord3[:,:,:,1] > 2)[1:])))))    

print(plaY_array)
# [[2 0]
#  [2 1]
#  [2 2]]    

print(plaX_array)
# [[2 2]]

Also note that you can eliminate the for-loop

for i in range(Coord3.shape[0]):

by calling np.where(Coord3[:,:,:,0] > 3) instead of np.where(Coord3[i,:,:,0] > 3) for each i:

In [16]: np.where(Coord3[:,:,:,0] > 3)
Out[16]: (array([1, 1, 2, 2, 2]), array([2, 2, 2, 2, 2]), array([0, 2, 0, 1, 2]))

The i values are in the first array, but since you don't care about those, you can just drop the first array.

answered Nov 12, 2014 at 13:48

unutbu

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6 Comments

user3601754 Over a year ago

Thanks for the answer! ;) as i have just said, it could be interesting to consider the different frames. I consider 3 times and i would to stock the results as an array of shape (3,2) with 3 for times and 2 for the positions. Sorry to bring later conditions...

user3601754 Over a year ago

My difficulty is due to the special shape of the "test array"(3, 1, 1). I dont know how to get acess to coordinates separately...

unutbu Over a year ago

I'm confused. What are the extra conditions and what is the desired final result?

user3601754 Over a year ago

Sorry i m not clear! :s as you can see in my second part of the edit, i got a "test array" which represents the couple of coordinates satisfactory the 2 conditions >3 and >2, but, in this array, i got : [[[(2, 0), (2, 1), (2, 2), (2, 2)]]]], dtype=object) in third line, (2,2) is redondant i would like to suppress it

user3601754 Over a year ago

and I would like to reshape the results in an array whose shape is (3,2) in order to get access easily to my results (because with the shape (3, 1, 1), i dont manage to access to values).But perhaps for this part, i should ask in another question...

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