根据另一个数组中包含的索引有效地求和 4D 数组

Question

我有一个 4D 数组，本质上是一系列立方体。除了我知道位置的值的子立方体之外，这些立方体大部分都用零填充。我需要将所有这些立方体加在一起成一个立方体。我可以简单地使用 np.sum 沿轴 = 3 执行此操作，但这是蒙特卡洛过程的一部分，并且执行了很多次。我想知道，因为我知道子多维数据集在多维数据集中的位置，我可以更有效地对它们求和，因为大多数求和操作都会添加零 - 这些多维数据集的大小很大（> 100 ^ 3），所以如果我可以的话将是一个巨大的节省。我对 Python/Numpy 很陌生，并且发现很难摆脱循环心态！一般来说，我正在寻找一种方法来一起操作大型 n 维数组，但仅限于某些部分。我意识到这里想到了蒙面数组，但我已经尝试过并且认为在这种情况下它不会提供任何加速；除非我离开那里！

编辑：这是我正在尝试做的三个糟糕的版本 - 脱离上下文可能没有多大意义 - 基本上涉及计算彼此相距一定距离的多个电荷的影响，但每个电荷不会影响自身.其中之一决定了飞行中的距离，两个使用预先计算的数组，其中包含信息，但它必须“对齐”并求和 - 再次脱离上下文可能没有意义，但这是我到目前为止所拥有的

def coloumbicForces3(carrierArray, cubeEnergeticDisorderArray, array):
cubeLen=100

offsetArray=np.array([[0,0,0],[0,0,+1],[0,0,-1],[0,+1,0],[0,-1,0],[+1,0,0],[-1,0,0]])
indices=np.zeros((7,3,len(carrierArray)), dtype=np.int32)
indices1=a=indices.reshape((7*len(carrierArray),3))
tIndices=cubeLen-indices
superimposedArray=np.zeros((cubeLen,cubeLen,cubeLen,2+2), dtype=myFloat)
sumArray=np.zeros((cubeLen,cubeLen,cubeLen))
for i in range(len(carrierArray)):
    indices[:,:,i]=offsetArray+carrierArray[i,1:4]

for c, carrierC in enumerate(carrierArray[:,0]):
    for k, carrierK in enumerate(carrierArray[:,0]):
        if c==k:
            continue
        for (x,y,z) in indices[:,:,k]:
        #print c, indices[:,:,c]
            if(carrierC==1):
                superimposedArray[x,y,z,c]=cubeEnergeticDisorderArray[x,y,z,c]=-1*array[cubeLen-x,cubeLen-y, cubeLen-z]
            else:
                superimposedArray[x,y,z,c]=cubeEnergeticDisorderArray[x,y,z,c]=array[cubeLen-x,cubeLen-y, cubeLen-z]



b = np.ascontiguousarray(a).view(np.dtype((np.void, a.dtype.itemsize * a.shape[1])))
_,idx = np.unique(b, return_index=True)
aUnique=a[idx]

for (i,j,k) in aUnique:
    sumArray[i,j,k]=np.sum(superimposedArray[i,j,k])

for c, carrierC in enumerate(carrierArray[:,0]):
    for (i,j,k) in aUnique:
        cubeEnergeticDisorderArray[i,j,k,c]=cubeEnergeticDisorderArray[i,j,k,-2]+sumArray[i,j,k]

return cubeEnergeticDisorderArray


def coloumbicForces(carrierArray, cubeEnergeticDisorderArray, array):
cubeLen= len(cubeEnergeticDisorderArray[0,:,:,0])
superimposedArray=np.zeros((cubeLen,cubeLen,cubeLen,2+2), dtype=myFloat)

for k, carrier in enumerate(carrierArray[:,0]):
    superimposedArray[:,:,:,k]=cubeEnergeticDisorderArray[:,:,:,k]=array[cubeLen-carrierArray[k,1]:2*cubeLen-carrierArray[k,1],cubeLen-carrierArray[k,2]:2*cubeLen-carrierArray[k,2],cubeLen-carrierArray[k,3]:2*cubeLen-carrierArray[k,3]]

    if (carrier==1):
        a=superimposedArray[:,:,:,k]
        b=cubeEnergeticDisorderArray[:,:,:,k]
        superimposedArray[:,:,:,k]=ne.evaluate("a*-1")
        cubeEnergeticDisorderArray[:,:,:,k]=ne.evaluate("b*-1")

sumArray=ne.evaluate("sum(superimposedArray, axis=3)")

for k, carrier in enumerate(carrierArray[:,0]):
    a=cubeEnergeticDisorderArray[:,:,:,k]
    b=cubeEnergeticDisorderArray[:,:,:,-2]
    cubeEnergeticDisorderArray[:,:,:,k]=ne.evaluate("sumArray-a+b")

return cubeEnergeticDisorderArray

def coloumbicForces2(carrierArray, cubeEnergeticDisorderArray, array):
x0=carrierArray[0,1]
y0=carrierArray[0,2]
z0=carrierArray[0,3]
x1=carrierArray[1,1]
y1=carrierArray[1,2]
z1=carrierArray[1,3]

cubeEnergeticDisorderArray[x0,y0,z0,0]=cubeEnergeticDisorderArray[x0,y0,z0,-2]-(1.60217657e-19)*2995850595.79/(distance([x0,y0,z0], carrierArray[1,1:4])*1e-9)
cubeEnergeticDisorderArray[x0-1,y0,z0,0]=cubeEnergeticDisorderArray[x0-1,y0,z0,-2]-(1.60217657e-19)*2995850595.79/(distance([x0-1,y0,z0], carrierArray[1,1:4])*1e-9)
cubeEnergeticDisorderArray[x0+1,y0,z0,0]=cubeEnergeticDisorderArray[x0+1,y0,z0,-2]-(1.60217657e-19)*2995850595.79/(distance([x0+1,y0,z0], carrierArray[1,1:4])*1e-9)
cubeEnergeticDisorderArray[x0,y0-1,z0,0]=cubeEnergeticDisorderArray[x0,y0-1,z0,-2]-(1.60217657e-19)*2995850595.79/(distance([x0,y0-1,z0], carrierArray[1,1:4])*1e-9)
cubeEnergeticDisorderArray[x0,y0+1,z0,0]=cubeEnergeticDisorderArray[x0,y0+1,z0,-2]-(1.60217657e-19)*2995850595.79/(distance([x0,y0+1,z0], carrierArray[1,1:4])*1e-9)
cubeEnergeticDisorderArray[x0,y0,z0-1,0]=cubeEnergeticDisorderArray[x0,y0,z0-1,-2]-(1.60217657e-19)*2995850595.79/(distance([x0,y0,z0-1], carrierArray[1,1:4])*1e-9)
cubeEnergeticDisorderArray[x0,y0,z0+1,0]=cubeEnergeticDisorderArray[x0,y0,z0+1,-2]-(1.60217657e-19)*2995850595.79/(distance([x0,y0,z0+1], carrierArray[1,1:4])*1e-9)

cubeEnergeticDisorderArray[x1,y1,z1,1]=cubeEnergeticDisorderArray[x1,y1,z1,-2]+(1.60217657e-19)*2995850595.79/(distance([x1,y1,z1], carrierArray[0,1:4])*1e-9)
cubeEnergeticDisorderArray[x1-1,y1,z1,1]=cubeEnergeticDisorderArray[x1-1,y1,z1,-2]+(1.60217657e-19)*2995850595.79/(distance([x1-1,y1,z1], carrierArray[0,1:4])*1e-9)
cubeEnergeticDisorderArray[x1+1,y1,z1,1]=cubeEnergeticDisorderArray[x1+1,y1,z1,-2]+(1.60217657e-19)*2995850595.79/(distance([x1+1,y1,z1], carrierArray[0,1:4])*1e-9)
cubeEnergeticDisorderArray[x1,y1-1,z1,1]=cubeEnergeticDisorderArray[x1,y1-1,z1,-2]+(1.60217657e-19)*2995850595.79/(distance([x1,y1-1,z1], carrierArray[0,1:4])*1e-9)
cubeEnergeticDisorderArray[x1,y1+1,z1,1]=cubeEnergeticDisorderArray[x1,y1+1,z1,-2]+(1.60217657e-19)*2995850595.79/(distance([x1,y1+1,z1], carrierArray[0,1:4])*1e-9)
cubeEnergeticDisorderArray[x1,y1,z1-1,1]=cubeEnergeticDisorderArray[x1,y1,z1-1,-2]+(1.60217657e-19)*2995850595.79/(distance([x1,y1,z1-1], carrierArray[0,1:4])*1e-9)
cubeEnergeticDisorderArray[x1,y1,z1+1,1]=cubeEnergeticDisorderArray[x1,y1,z1+1,-2]+(1.60217657e-19)*2995850595.79/(distance([x1,y1,z1+1], carrierArray[0,1:4])*1e-9)
return cubeEnergeticDisorderArray

Answer 1

如果您知道子多维数据集是，您可以使用花哨的索引仅在您需要的地方求和，例如：

import numpy as np
from numpy.random import random

c1 = random((10, 10, 10, 10))
c2 = random((10, 10, 10, 10))
c3 = np.zeros_like(c2)

这里是我想要总结的索引：

i1 = [0, 2, 4, 6]
i2 = [0, 1, 3, 7]
i3 = [1, 5, 8, 9]
i4 = [1, 6, 7, 8]

c3[i1,i2,i3,i4] = c1[i1,i2,i3,i4] + c2[i1,i2,i3,i4]

仅在以下点求和：p1(0,0,1,1)、p2(2,1,5,6)、p3(4,3,8,7) 和 p4(6,7,9,8)。