这是一个向量化的 Python 解决方案,使用NumPy broadcasting https://docs.scipy.org/doc/numpy/user/basics.broadcasting.html and matrix multiplication
with np.dot https://docs.scipy.org/doc/numpy/reference/generated/numpy.dot.html对于求和部分 -
x_mask = ((x >= X[:-1,None]) & (x < X[1:,None]))
y_mask = ((y >= Y[:-1,None]) & (y < Y[1:,None]))
z_g_out = np.dot(y_mask*z[None].astype(np.float32), x_mask.T)
# If needed to fill invalid places with NaNs
z_g_out[y_mask.dot(x_mask.T.astype(np.float32))==0] = np.nan
请注意,我们避免使用meshgrid
那里。因此,在使用创建的网格体时可以节省内存meshgrid
将是巨大的,并希望在此过程中获得性能改进。
标杆管理
# Original app
def org_app(x,y,z):
X = np.arange(min(x), max(x), 0.1)
Y = np.arange(min(y), max(y), 0.1)
x_g, y_g = np.meshgrid(X, Y)
nx, ny = x_g.shape
z_g = np.full(np.asarray(x_g.shape)-1, np.nan)
for ix in range(nx - 1):
for jx in range(ny - 1):
x_min = x_g[ix, jx]
x_max = x_g[ix + 1, jx + 1]
y_min = y_g[ix, jx]
y_max = y_g[ix + 1, jx + 1]
vals = z[(x >= x_min) & (x < x_max) &
(y >= y_min) & (y < y_max)]
if vals.any():
z_g[ix, jx] = sum(vals)
return z_g
# Proposed app
def app1(x,y,z):
X = np.arange(min(x), max(x), 0.1)
Y = np.arange(min(y), max(y), 0.1)
x_mask = ((x >= X[:-1,None]) & (x < X[1:,None]))
y_mask = ((y >= Y[:-1,None]) & (y < Y[1:,None]))
z_g_out = np.dot(y_mask*z[None].astype(np.float32), x_mask.T)
# If needed to fill invalid places with NaNs
z_g_out[y_mask.dot(x_mask.T.astype(np.float32))==0] = np.nan
return z_g_out
如所见,为了公平的基准测试,我使用原始方法的数组值,因为从数据帧中获取值可能会减慢速度。
时间安排和验证 -
In [143]: x = np.array([1, 1.12, 1.109, 2.1, 3, 4.104, 3.1])
...: y = np.array([-9, -0.1, -9.2, -8.7, -5, -4, -8.75])
...: z = np.array([10, 4, 1, 4, 5, 0, 1])
...:
# Verify outputs
In [150]: np.nansum(np.abs(org_app(x,y,z) - app1(x,y,z)))
Out[150]: 0.0
In [145]: %timeit org_app(x,y,z)
10 loops, best of 3: 19.9 ms per loop
In [146]: %timeit app1(x,y,z)
10000 loops, best of 3: 39.1 µs per loop
In [147]: 19900/39.1 # Speedup figure
Out[147]: 508.95140664961633