Allocate m4 output arrays in `numba` code, avoid segfaults?

big_data_lines
Tyler Goodlet 2022-03-24 13:22:30 -04:00
parent 561d7e0349
commit b262532fd4
1 changed files with 28 additions and 19 deletions

View File

@ -207,7 +207,7 @@ def ohlc_to_m4_line(
# NOTE: found that a 16x px width brought greater
# detail, likely due to dpi scaling?
# px_width=px_width * 16,
32 / (1 + math.log(uppx, 2)),
64 / (1 + math.log(uppx, 2)),
1
)
)
@ -291,18 +291,16 @@ def ds_m4(
if r:
frames += 1
# these are pre-allocated and mutated by ``numba``
# code in-place.
y_out = np.zeros((frames, 4), y.dtype)
i_win = np.zeros(frames, x.dtype)
# call into ``numba``
nb = _m4(
nb, i_win, y_out = _m4(
x,
y,
i_win,
y_out,
frames,
# TODO: see func below..
# i_win,
# y_out,
# first index in x data to start at
x_start,
@ -322,19 +320,25 @@ def ds_m4(
@jit(
nopython=True,
nogil=True,
# nogil=True,
)
def _m4(
xs: np.ndarray,
ys: np.ndarray,
frames: int,
# TODO: using this approach by having the ``.zeros()`` alloc lines
# below, in put python was causing segs faults and alloc crashes..
# we might need to see how it behaves with shm arrays and consider
# allocating them once at startup?
# pre-alloc array of x indices mapping to the start
# of each window used for downsampling in y.
i_win: np.ndarray,
# i_win: np.ndarray,
# pre-alloc array of output downsampled y values
ds: np.ndarray,
# y_out: np.ndarray,
x_start: int,
step: float,
@ -343,6 +347,11 @@ def _m4(
# nbins = len(i_win)
# count = len(xs)
# these are pre-allocated and mutated by ``numba``
# code in-place.
y_out = np.zeros((frames, 4), ys.dtype)
i_win = np.zeros(frames, xs.dtype)
bincount = 0
x_left = x_start
@ -357,15 +366,15 @@ def _m4(
# (aka a row broadcast).
i_win[bincount] = x_left
# set all y-values to the first value passed in.
ds[bincount] = ys[0]
y_out[bincount] = ys[0]
for i in range(len(xs)):
x = xs[i]
y = ys[i]
if x < x_left + step: # the current window "step" is [bin, bin+1)
ds[bincount, 1] = min(y, ds[bincount, 1])
ds[bincount, 2] = max(y, ds[bincount, 2])
ds[bincount, 3] = y
y_out[bincount, 1] = min(y, y_out[bincount, 1])
y_out[bincount, 2] = max(y, y_out[bincount, 2])
y_out[bincount, 3] = y
else:
# Find the next bin
while x >= x_left + step:
@ -373,6 +382,6 @@ def _m4(
bincount += 1
i_win[bincount] = x_left
ds[bincount] = y
y_out[bincount] = y
return bincount
return bincount, i_win, y_out