def _find_local_maxima(ordered_RD, k=1):
"""
Find local maxima to the k neighbor of left and k neighbors of right.\
It equivalent to a moving windows of size 2*k+1. RD is an 1D numpy array.\
Since we are looking for max, a max heap would be most beneficial.\
For the sake of simplicity, we just take the negtive of RD, and use \
a min heap already implemented a while ago.
"""
local_maxima = heapdict()
moving_window = BinHeap()
window_size = 2 * k + 1
poi_idx = -k # index for point of interest start from -k from convinience.
RD_size = len(ordered_RD)
assert window_size < RD_size, 'Invalid k or RD, window size can not be smaller than RD size'
while poi_idx < RD_size:
w_s_idx = poi_idx - k # window start idx
w_e_idx = poi_idx + k # window end idx
if w_e_idx < RD_size:
moving_window.insert(
(-ordered_RD[w_e_idx], w_e_idx)
) # negtive is to accomadate use of min heap, since RD is always non-negtive.
if (moving_window.heap_size > window_size) or (w_e_idx >= RD_size):
heap_el_idx = moving_window.pos[w_s_idx - 1]
moving_window.extract_element(
heap_el_idx) # delete elements outside of the window.
if poi_idx >= 0:
current_RD = -ordered_RD[poi_idx]
if current_RD <= moving_window.heap[0][0]:
local_maxima[
poi_idx] = current_RD # a negative RD is also used for local maxima since heapdict is a min priority queue
poi_idx += 1
return local_maxima
