在保留格式的同時將多維數組轉換為元組-python

Question

我正在嘗試將具有不同格式的多維數組轉換為元組，同時保留不同的格式。 這是構建數組的代碼

nodeData=np.empty((npoints*npoints,4))
ii=0
for i in range (npoints):
    for j in range (npoints):
        nodeData[ii,0]=int(ii+1)
        nodeData[ii,1]=float(X[i,j])
        nodeData[ii,2]=float(Y[i,j])
        nodeData[ii,3]=float(Z[i,j])
        ii+=1

哪個輸出

[[  1.          10.           2.           0.        ]
 [  2.           9.23463314   1.84775906   0.        ]
 [  3.           8.58578682   1.41421354   0.        ]
 [  4.           8.15224103   0.76536686   0.        ]
 [  5.           8.           0.           0.        ]
 [  6.          10.           4.           0.        ]
 [  7.           8.14486726   3.94570562   0.        ]
 [  8.           6.43933982   3.56066012   0.        ]
 [  9.           6.05429438   1.8551327    0.        ]
 [ 10.           6.           0.           0.        ]
 [ 11.          10.           6.           0.        ]
 [ 12.           7.07624214   6.00295346   0.        ]
 [ 13.           4.29289317   5.70710671   0.        ]
 [ 14.           3.99704657   2.92375783   0.        ]
 [ 15.           4.           0.           0.        ]
 [ 16.          10.           8.           0.        ]
 [ 17.           6.02814844   8.02067549   0.        ]
 [ 18.           2.14644665   7.85355332   0.        ]
 [ 19.           1.97932457   3.97185155   0.        ]
 [ 20.           2.           0.           0.        ]
 [ 21.          10.          10.           0.        ]
 [ 22.           5.          10.           0.        ]
 [ 23.           0.          10.           0.        ]
 [ 24.           0.           5.           0.        ]
 [ 25.           0.           0.           0.        ]]

現在，當我嘗試使用nodeData = tuple(map(tuple, nodeData))將其轉換為元nodeData = tuple(map(tuple, nodeData))我獲得

((1.0, 10.0, 2.0, 0.0), (2.0, 9.234633143257458, 1.847759058732358, 0.0), (3.0, 8.5857868194580078, 1.4142135381698608, 0.0), (4.0, 8.1522410342120963, 0.76536686381751795, 0.0), (5.0, 8.0, 0.0, 0.0), (6.0, 10.0, 4.0, 0.0), (7.0, 8.144867260307727, 3.9457056195410947, 0.0), (8.0, 6.439339816570282, 3.5606601238250732, 0.0), (9.0, 6.0542943801670015, 1.8551327030961533, 0.0), (10.0, 6.0, 0.0, 0.0), (11.0, 10.0, 6.0, 0.0), (12.0, 7.0762421416553485, 6.0029534580819224, 0.0), (13.0, 4.2928931713104248, 5.7071067094802856, 0.0), (14.0, 3.997046571142258, 2.9237578279641867, 0.0), (15.0, 4.0, 0.0, 0.0), (16.0, 10.0, 8.0, 0.0), (17.0, 6.028148440014629, 8.0206754926510051, 0.0), (18.0, 2.1464466452598572, 7.8535533249378204, 0.0), (19.0, 1.9793245736169094, 3.9718515524632179, 0.0), (20.0, 2.0, 0.0, 0.0), (21.0, 10.0, 10.0, 0.0), (22.0, 5.0, 10.0, 0.0), (23.0, 0.0, 10.0, 0.0), (24.0, 0.0, 5.0, 0.0), (25.0, 0.0, 0.0, 0.0))

當我嘗試將每行的第一個值保存為整數時，它已將所有值轉換為浮點數：

((1, 10.0, 2.0, 0.0),
 (2, 9.234633143257458, 1.847759058732358, 0.0), 
(3, 8.5857868194580078, 1.4142135381698608, 0.0), 
(4, 8.1522410342120963, 0.76536686381751795, 0.0), 
(5, 8.0, 0.0, 0.0), 
(6, 10.0, 4.0, 0.0), 
(7, 8.144867260307727, 3.9457056195410947, 0.0), 
(8, 6.439339816570282, 3.5606601238250732, 0.0), 
(9, 6.0542943801670015, 1.8551327030961533, 0.0), 
(10, 6.0, 0.0, 0.0), 
(11, 10.0, 6.0, 0.0), 
(12, 7.0762421416553485, 6.0029534580819224, 0.0), 
(13, 4.2928931713104248, 5.7071067094802856, 0.0), 
(14, 3.997046571142258, 2.9237578279641867, 0.0), 
(15, 4.0, 0.0, 0.0), 
(16, 10.0, 8.0, 0.0), 
(17, 6.028148440014629, 8.0206754926510051, 0.0), 
(18, 2.1464466452598572, 7.8535533249378204, 0.0), 
(19, 1.9793245736169094, 3.9718515524632179, 0.0), 
(20, 2.0, 0.0, 0.0), 
(21, 10.0, 10.0, 0.0), 
(22, 5.0, 10.0, 0.0), 
(23, 0.0, 10.0, 0.0), 
(24, 0.0, 5.0, 0.0), 
(25, 0.0, 0.0, 0.0))

或者，無論如何都可以在不構建數組的情況下獲得最終結果，而是直接以正確的格式構建元組？ 任何幫助將不勝感激。

Answer 1

您不能逐步構建元組，因為元組是不可變的。 但是您可以構建一個列表，然后將其轉換為元組。

nodeData = []
ii = 1
for i in range (npoints):
    for j in range (npoints):
        nodeData.append((ii, float(X[i,j]), float(Y[i,j]), float(Z[i,j])))
        ii += 1
nodeData = tuple(nodeData)

Answer 2

當你第一次創建nodeData陣列，您可以檢查dtype和通知該數組類型float64 （見下文print語句）：

npoints = 25
nodeData=np.empty((npoints*npoints,4))
print(nodeData.dtype)

'float64'

這意味着當您稍后分配整數時，它會自動轉換為浮點數（請參閱下面的打印語句）：

npoints = 25
nodeData=np.empty((npoints*npoints,4))
ii=0
for i in range (npoints):
    for j in range (npoints):
        nodeData[ii,0]=int(ii+1)
        print(nodeData[ii,0].dtype)

'float64'...

正如@barmar 所建議的那樣，您必須以不會將整數值轉換為浮點數的方式構建一組元組（在 numpy 數組中設置值的方式將自動轉換為匹配數組類型）。

Answer 3

作為 Barmar 答案的替代方案，您還可以使用生成器：

def generate():
    for i in range(npoints):
        for j in range(npoints):
            yield (int(ii+1), float(X[i,j]), float(Y[i,j]), float(Z[i,j]))
            ii += 1

nodeData = tuple(generate())