I have the following NumPy array
[[ 0 935]
[ 0 331]
[ 0 322]
[ 1 339]
[ 1 773]
[ 2 124]
[ 2 340]
[ 3 810]
[ 5 936]
[ 5 252]]
and would like to separate it into
[[ 0 935]
[ 0 331]
[ 0 322]]
[[ 1 339]
[ 1 773]]
[[ 2 124]
[ 2 340]]
[[ 3 810]]
[[ 5 936]
[ 5 252]]
Are there any fast solutions to accomplish this?
If all equal numbers located in one place you can do:
import numpy as np
a = np.array([[ 0, 935],
[ 0, 331],
[ 0, 322],
[ 1, 339],
[ 1, 773],
[ 2, 124],
[ 2, 340],
[ 3, 810],
[ 5, 936],
[ 5, 252],])
print(np.array_split(a, np.flatnonzero(np.diff(a[:, 0])) + 1))
# [array([[ 0, 935],
# [ 0, 331],
# [ 0, 322]]), array([[ 1, 339],
# [ 1, 773]]), array([[ 2, 124],
# [ 2, 340]]), array([[ 3, 810]]), array([[ 5, 936],
# [ 5, 252]])]
otherwise you can first sort array:
import numpy as np
a = np.array([[ 0, 935],
[ 5, 936],
[ 2, 124],
[ 2, 340],
[ 5, 252],])
a.sort(axis=0)
print(*np.array_split(a, np.flatnonzero(np.diff(a[:, 0])) + 1), sep="\n")
# [[ 0 124]]
# [[ 2 252]
# [ 2 340]]
# [[ 5 935]
# [ 5 936]]
You can try this :
# Note: k is your array
import numpy as np
k = np.array([[ 0, 935],[ 0, 331],[ 0, 322],
[ 1, 339],[ 1, 773],[ 2, 124],
[ 2, 340],[ 3, 810],[ 5, 936],
[ 5, 252]])
# sort indices
indices = np.argsort(k[:, 0])
# use indices to get sorted array
arr_temp = k[indices]
# retrieve your answer
net = np.array_split(arr_temp,
np.where(np.diff(arr_temp[:,0])!=0)[0]+1)
net[0]
This solution was inspired from How to split a numpy array based on a column?
0 935or are they a list of length 2