How to create a new numpy 2d array with conditions from the existing 2d array

Question

I have 2d array with 3 columns and N rows. In the third column there are only 0 or 1. I need to create 2 numpy arrays. They both contains first 2 columns of the given matrix, but first array has only rows corresponding to 0 from the third column, and second array has only rows to 1.

I've tried but it failed with dimension problems. I haven't used this kind of format before. onlyNormal_Xtest = np.vstack((onlyNormal_Xtest, Xy[Xy[N_train:, 2] == 0]))

Is it possible to do it faster than following?

onlyNormal_Xtest = np.array([])
Xy_test = Xy[N_train:, :]

    for i in range(np.size(Xy_test, 0)):
        if (Xy_test[i, 2] == 0):
            onlyNormal_Xtest = np.append(onlyNormal_Xtest, Xy_test[i, :2])

Actually it still doesn't work due to dimension problems.

np.append is also difficult to get right, especially when used iteratively. You really need to understand array dimensions. Do you even know what that np.array([]) is like? What's its shape? Or why you even used it? Did you read the whole np.append docs? — hpaulj
– hpaulj, Commented Feb 1, 2021 at 8:04
stackoverflow.com/questions/65983816/… - another recent append question — hpaulj
– hpaulj, Commented Feb 1, 2021 at 8:35

Gary Ong · Accepted Answer · 2021-02-01 08:11:45Z

0

Not sure if I understood your question but here is the code i think you were trying to do

a = np.array([[1,2,0],
             [3,4,1],
             [5,6,0],
             [7,8,1]])

Gives

a[a[:,2]==1][:,:2]
array([[3, 4],
       [7, 8]])

a[a[:,2]==0][:,:2]
array([[1, 2],
       [5, 6]])

answered Feb 1, 2021 at 8:11

Gary Ong

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