Getting index of first occurrence in each row

Cannot test right now, but I think this should work

arr.argmax(axis=1).T

argmax on bools shortcircuits in numpy 1.9, so it should be preferred to where or nonzero for this use case.

EDIT OK, so the above solution doesn't work, but the approach with argmax is still useful:

In [23]: mult = np.product(arr.shape[:-1])

In [24]: np.column_stack(np.unravel_index(arr.shape[-1]*np.arange(mult) +
   ....:                                  arr.argmax(axis=-1).ravel(),
   ....:                                  arr.shape))
Out[24]:
array([[0, 0, 0],
       [0, 1, 0],
       [1, 0, 2],
       [1, 1, 0],
       [2, 0, 0],
       [2, 1, 0]])

It seems you want np.where() combined with the solution of this answer to find unique rows:

b = np.array(np.where(a)).T
#array([[0, 0, 0],
#       [0, 0, 1],
#       [0, 1, 0],
#       [1, 0, 2],
#       [1, 0, 4],
#       [1, 1, 0],
#       [2, 0, 0],
#       [2, 1, 0]], dtype=int64)
c = b[:,:2]
d = np.ascontiguousarray(c).view(np.dtype((np.void, c.dtype.itemsize * c.shape[1])))
_, idx = np.unique(d, return_index=True)

b[idx]
#array([[0, 0, 0],
#       [0, 1, 0],
#       [1, 0, 2],
#       [1, 1, 0],
#       [2, 0, 0],
#       [2, 1, 0]], dtype=int64)

Adding onto other answers, if you want it to give you the index of the first col that's True, and return n where n = # cols in a if a row doesn't contain True:

first_occs = np.argmax(a, axis=1)
all_zeros = ~a.any(axis=1).astype(int)
first_occs_modified = first_occs + all_zeros * a.shape[1]