将 2D 数组元素插入 3D 数组 Python 中

Question

我希望将计算的 2D 数组元素插入 3D 数组中。第一个 for 循环是需要多少个数组。例如，当i = 1时，这些值将进入3D数组的第一个数组。计算元素显示在代码的底行上

"dctArray[dctRow][dctColumn]".

本质上，我试图将此 2D 数组值等同于 3D 数组的第一个数组。所以像这样：

dctArrayBlocks[i][dctRow][dctColumn] = dctArray[dctRow][dctColumn] 

但这不适用于整个阵列。当我打印单个值时

dctArrayBlocks[i][dctRow][dctColumn]

这是正确的，但是当我尝试打印完整数组时，值非常不正确。这些输出位于底部。

谢谢，这是完整的代码：

noBlocksOn1Axis = (int(imageLength/blockSize))
        blocksList = []
        previousRow = 0
        for rowBlock in range(noBlocksOn1Axis):
            previousRow = rowBlock * blockSize  
            previousColumn = 0
            for columnBlock in range(noBlocksOn1Axis):
                previousColumn = columnBlock * blockSize
                block = arrayY[previousRow:previousRow+blockSize,previousColumn:previousColumn+blockSize]
                blocksList.append(block)
dctArray = np.zeros((8,8))
dctArrayBlocks = np.zeros(shape = (4,8,8))
for i in range(4):
            for dctRow in range(blockSize):
                for dctColumn in range (blockSize):
                    dctSum = 0
                    for row in range (blockSize):
                        for column in range (blockSize):
                            dctSum = dctSum + blocksList[i][row][column]*math.cos(((2.0*row+1)*dctRow*math.pi)/16.0)*math.cos(((2.0*column+1)*dctColumn*math.pi)/16.0)
                    if(dctRow == 1):
                        cRow = 1/math.sqrt(2)
                    else:
                        cRow = 1
                    if(dctColumn == 1):
                        cColumn= 1/math.sqrt(2)
                    else:
                        cColumn = 1
                    dctArray[dctRow][dctColumn] = 
                    1/4.0*cRow*cColumn*dctSum
                    dctArrayBlocks[i][dctRow][dctColumn] = 
                    dctArray[dctRow][dctColumn]
                    print(dctArrayBlocks[i][dctRow][dctColumn])

        print(dctArrayBlocks)

这是列表形式的预期输出

1160.75
2.15309774781
0.230969883128
0.325283736973
-0.883883476483
-0.332410006991
0.0956708580911
-0.370181380455
-29.1901374775
-0.0455109087818
0.543799659533
0.500630540423
-0.469718941578
0.21007762182
0.340311556876
-0.818563547551
-1.97625897152
1.34074300692
0.265165042945
-1.05682448828
0.450332944883
0.750180006989
0.109834957055
-0.46727710967
-2.38227629503
-0.720900476592
0.904589640549
1.02494023825
-0.564894042278
-1.11318780084
-0.318826625164
-0.627633469915
0.883883476483
0.151835185028
0.760559161902
0.114395653893
0.625
0.465174095446
-0.450332944883
0.732747465688
-1.08735713314
-0.336526203213
0.0151562551075
-0.582857714948
-0.652328957388
0.00538984764367
-0.131671763221
0.541152940662
2.15798528204
-0.175494902565
-0.640165042945
-0.631820804478
0.760559161902
0.315865402699
-0.265165042945
0.266936390222
0.373044516988
-0.443563547551
0.319493333954
-0.857754793709
0.355794721991
0.311551890598
-0.111193339081
0.0606917316737

这就是我得到的。

[[[  1.16075000e+03   2.15309775e+00   2.30969883e-01   3.25283737e-01
    -8.83883476e-01  -3.32410007e-01   9.56708581e-02  -3.70181380e-01]
  [ -2.91901375e+01  -4.55109088e-02   5.43799660e-01   5.00630540e-01
    -4.69718942e-01   2.10077622e-01   3.40311557e-01  -8.18563548e-01]
  [ -1.97625897e+00   1.34074301e+00   2.65165043e-01  -1.05682449e+00
     4.50332945e-01   7.50180007e-01   1.09834957e-01  -4.67277110e-01]
  [ -2.38227630e+00  -7.20900477e-01   9.04589641e-01   1.02494024e+00
    -5.64894042e-01  -1.11318780e+00  -3.18826625e-01  -6.27633470e-01]
  [  8.83883476e-01   1.51835185e-01   7.60559162e-01   1.14395654e-01
     6.25000000e-01   4.65174095e-01  -4.50332945e-01   7.32747466e-01]
  [ -1.08735713e+00  -3.36526203e-01   1.51562551e-02  -5.82857715e-01
    -6.52328957e-01   5.38984764e-03  -1.31671763e-01   5.41152941e-01]
  [  2.15798528e+00  -1.75494903e-01  -6.40165043e-01  -6.31820804e-01
     7.60559162e-01   3.15865403e-01  -2.65165043e-01   2.66936390e-01]
  [  3.73044517e-01  -4.43563548e-01   3.19493334e-01  -8.57754794e-01
     3.55794722e-01   3.11551891e-01  -1.11193339e-01   6.06917317e-02]]

Answer 1

好的，

我尝试重写代码以使其更具pythonic，并展示numpy矢量化的强大功能：

import numpy as np
np.random.seed(1)
imageLength = 16
arrayY= np.random.randint(1,255,(imageLength,imageLength))
blockSize = 8
noBlocksOn1Axis = imageLength//blockSize
blocksList = []
previousRow = 0
for rowBlock in range(noBlocksOn1Axis):
    previousRow = rowBlock * blockSize  
    previousColumn = 0
    for columnBlock in range(noBlocksOn1Axis):
        previousColumn = columnBlock * blockSize
        block = arrayY[previousRow:previousRow+blockSize , previousColumn:previousColumn+blockSize]
        blocksList.append(block)
dctArrayBlocks = np.zeros((4,8,8))
bl = np.array(blocksList)
dctSum = np.empty((blockSize,blockSize))
m = np.outer(np.arange(blockSize),np.ones(blockSize))
for i in range(bl.shape[0]):
    for n1 in range(blockSize):
        for n2 in range (blockSize):
            dctSum[n1,n2] = np.sum(
                bl[i] * 
                np.cos((2*m  +1) * n1*np.pi/16.) * 
                np.cos((2*m.T+1) * n2*np.pi/16.)
                )
    dctArrayBlocks[i] = dctSum/4
dctArrayBlocks [:,1,:] /= np.sqrt(2)
dctArrayBlocks [:,:,1] /= np.sqrt(2)

这应该完全按照你的代码正在做的事情，但要利用numpy和python的功能。

正如我在评论中指出的那样，区别只是格式问题，关于格式，您可以使用类似的东西

np.set_printoptions(formatter={'float':lambda x: str(x)})