我为纬度(Xpos
)和经度(Ypos
)用维度(125,800,000
)屏蔽了数组
我想计算数组中的经纬度差。这里是数组Xpos
(Ypos
类似)
masked_array( data = [ [-2.0551843643188477, -2.637551784515381, -2.720881223678589, ..., 2.2812530994415283, 2.281250476837158, 2.281254768371582 ],
[-2.3242127895355225, -2.804257869720459, -2.8825504779815674, ..., 2.2812530994415283, 2.281250476837158, 2.281254768371582 ],
[-2.073770523071289, -2.6198980808258057, -2.708889961242676, ..., 2.2812530994415283, 2.281250476837158, 2.281254768371582 ],
...,
[-3.517531633377075, -2.908338785171509, -2.9069409370422363, ..., 2.2812530994415283, 2.281250476837158, 2.281254768371582 ],
[-3.688690662384033, -3.0086288452148438, -3.010164260864258, ..., 2.2812530994415283, 2.281250476837158, 2.281254768371582 ],
[-3.520817518234253, -2.943941116333008, -2.941941738128662, ..., 2.2812530994415283, 2.281250476837158, 2.281254768371582 ]
],
mask = [ [ False, False, False, ..., False, False, False ],
[ False, False, False, ..., False, False, False],
[ False, False, False, ..., False, False, False],
...,
[ False, False, False, ..., False, False, False],
[ False, False, False, ..., False, False, False],
[ False, False, False, ..., False, False, False]
],
fill_value = 1e+20,
dtype = float32
)
这是我的代码,它可以工作,但需要很长的时间来计算
Dist= np.zeros((len(XposApr),len(XposApr[0])))
DiffLon=np.zeros((len(XposApr),len(XposApr[0])))
DiffLat=np.zeros((len(XposApr),len(XposApr[0])))
for i in range (1,len(XposApr),12):
for j in range (0,len(XposApr[0])):
DiffLon[i][j]=(XposApr[i][j]-XposApr[i-1][j])
DiffLat[i][j]=(YposApr[i][j]-YposApr[i-1][j])
我真的不知道如何制作著名的单行线,这是我尝试过的,但不起作用:
DistLon = [XposApr[i][j]-XposApr[i-1][j] for i in enumerate (XposApr) and j in enumerate (XposApr[0])]
有没有可能生成一个线性或另一个代码,从而使计算速度大大加快?
提前谢谢
语法格式不是实现速度的核心,
智能矢量或重新公式是:
直到光荣的
numba
-团队决定将对掩蔽数组操作的支持转移到前端,来自numba.jit( ... )
(Masked array support #2103)的JIT加速速度仍然不足那么让我们试试
numpy
-本机步骤:内环可能首先脱离缓慢的GIL驱动运动:
然而,如果我正确地阅读了布局,整个问题可能会在两个维度上得到解决,具体如下:
我相信无论是@Divakar还是@cᴏʟᴅsᴘᴇᴇᴅ,这里真正的
numpy
大师都会准备好告诉你更聪明、更有效的切片/矢量化技巧,让你做到这一点,所以要有耐心:o)感谢您的回答和@patrick的有用评论,我相信这也可以解决我的问题:
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