翻译

Question

我有我生成使用

A1, M1, A2, M2, A3, M3, E = meshgrid(Grid.aGrid, Grid.mGrid, Grid.aGrid, Grid.mGrid, Grid.aGrid, Grid.mGrid, Grid.eGrid, indexing='ij') 

，其中Grid.aGrid使用linspace(aMin, aMax, nA)产生形状(A1, M1, A2, M2, A3, M3, E)的网格，并且类似地对于其它的网格。

考虑一些Z = f(A1, ...)，其中f()将标记一些网格点为无关紧要。为简单起见，假设它是

Z = A1 + A2 + A3 
Z[Z < 0] = NaN 

考虑Z[0, 1, 2, 3, 4, 5, 6]。它包含对应于实际值(aGrid[0], mGrid[1], aGrid[2], mGrid[3], aGrid[4], mGrid[5], eGrid[6])的值。这正是我尝试实现未标记的f()上Z所有点：

我想创建一个字典

foo = {z1, z2, z3, ... zn} 

其中z1等都是那种

z1 = (aGrid[0], mGrid[1], aGrid[2], mGrid[3], aGrid[4], mGrid[5], eGrid[6]) 

，这是Z内部对应z1位置的网格值。

我想出了：

aGrid = arange(0, 10) 
mGrid = arange(100, 110) 
eGrid = arange(1000, 1200) 

A,M,E = meshgrid(aGrid, mGrid, eGrid, indexing='ij') 

# contains the grid index 
Z = (A + M + E).astype(float) 
Z[A < 3] = nan 
# will contain the actual values, as tuples 
Z2 = {} 

for i, idx in enumerate(ndindex(Z.shape)): 
    a = aGrid[idx[0]] 
    m = mGrid[idx[1]] 
    e = eGrid[idx[2]] 

    if isnan(Z[idx]): 
     Z2[i] = NaN 
    else: 
     Z2[i] = (a, m, e) 

效率是关键。有没有更快更清洁的方式可以实现这一目标？任何使用字典的替代方法？

我特别不喜欢我必须写下aGrid[idx[0]]等。是否有可能让算法更一般？沿的

for i, idx in enumerate(ndindex(Z.shape)): 
    # some magic happens here. What exactly? 
    someMagicList = magic(aGrid, mGrid, eGrid) 
    Z2[i] = someMagicList[idx] 

Answer 1

broadcast_arrays()使用的线，并且将结果Z2一些事情是具有形状(20000, 3)阵列。

import numpy as np 

aGrid = np.arange(0, 10, dtype=float) 
mGrid = np.arange(100, 110, dtype=float) 
eGrid = np.arange(1000, 1200, dtype=float) 

A,M,E = np.meshgrid(aGrid, mGrid, eGrid, indexing='ij') 

# contains the grid index 
Z = (A + M + E).astype(float) 
Z[A < 3] = np.nan 

grids = [A, M, E] 
grid_bc = np.broadcast_arrays(*grids) 
Z2 = np.column_stack([g.ravel() for g in grid_bc]) 
Z2[np.isnan(Z.ravel())] = np.nan 

print Z2[5900], Z2[6000]