如果這個問題看起來很長而且很基本，我真誠地道歉。
鑒于：
import numpy as np
import time
c, q = int(3e5), int(5e5)
a = np.full( (c,q,3), None )
# fillout with some non None arrays: 3D (x,y,z) positions
a[0,0, :] = np.array([-4,0.1,0])
a[0,1, :] = np.array([9.2,3.1,0])
a[0,5, :] = np.array([3,-4.3,0])
a[0,6, :] = np.array([-1,12.8,0])
a[2,1, :] = np.array([4.5,-9,0])
a[2,3, :] = np.array([-0.1,6.1,0])
a[2,8, :] = np.array([-7,1,0])
a[3,0, :] = np.array([-1,0.7,0])
a[3,6, :] = np.array([-15,26,0])
a[5,0, :] = np.array([0.1,-1.1,0])
a[7,5, :] = np.array([0,0,0])
a[8,2, :] = np.array([5,6,0])
a[9,10, :] = np.array([-1.1,1,0])
a[10,3, :] = np.array([-32,15,0])
a[11,7, :] = np.array([0,9.3,0])
a[12,2, :] = np.array([0.9,6.2,0])
a[14,9, :] = np.array([8.6,5.6,0])
a[15,5, :] = np.array([0.5,8.5,0])
目標：
我想None從a. 目前，我的以下代碼非常耗時且效率很低，因為我使用的是基本的for loop：
bt = time.time()
for ci in range(c):
if any(ci == value for value in [2, 5]):
print(f">> Generating {ci} ranks ...")
poseNplus = []
aNplus = a[ci:]
for ci_i in range(aNplus.shape[0]):
aNplus_Q = aNplus[ci_i]
for qi in range(aNplus_Q.shape[0]):
if all(aNplus_Q[qi] != None):
poseNplus.append( aNplus_Q[qi] )
print(len(poseNplus), poseNplus)
et = time.time()
print(f"Took {(et-bt):.3f} s")
這很花時間：
Took 580.888 s
按照@Marc Felix 的回答，我可以按如下方式提取所有非None三元組：首先更改a = np.full( (c,q,3), np.nan )，然后：
bt = time.time()
nan_values = np.any(np.isnan(a), axis=-1)
result = a[nan_values==False].reshape((-1, 3))
et = time.time()
print(f"Took {(et-bt):.3f} s")
print(result.shape)
print(result)
回傳：
Took 0.318 s
(18, 3)
[[ -4. 0.1 0. ]
[ 9.2 3.1 0. ]
[ 3. -4.3 0. ]
[ -1. 12.8 0. ]
[ 4.5 -9. 0. ] <<<--- rank2 - END: from here till end
[ -0.1 6.1 0. ]
[ -7. 1. 0. ]
[ -1. 0.7 0. ]
[-15. 26. 0. ]
[ 0.1 -1.1 0. ] <<<--- rank5 - END: from here till end
[ 0. 0. 0. ]
[ 5. 6. 0. ]
[ -1.1 1. 0. ]
[-32. 15. 0. ]
[ 0. 9.3 0. ]
[ 0.9 6.2 0. ]
[ 8.6 5.6 0. ]
[ 0.5 8.5 0. ]]
但我想要的結果應該是這樣的：
>> Generating 2 ranks ...
[[ 4.5 -9. 0. ]
[ -0.1 6.1 0. ]
[ -7. 1. 0. ]
[ -1. 0.7 0. ]
[-15. 26. 0. ]
[ 0.1 -1.1 0. ]
[ 0. 0. 0. ]
[ 5. 6. 0. ]
[ -1.1 1. 0. ]
[-32. 15. 0. ]
[ 0. 9.3 0. ]
[ 0.9 6.2 0. ]
[ 8.6 5.6 0. ]
[ 0.5 8.5 0. ]]
------------------------------------------------------------
>> Generating 5 ranks ...
[[ 0.1 -1.1 0. ]
[ 0. 0. 0. ]
[ 5. 6. 0. ]
[ -1.1 1. 0. ]
[-32. 15. 0. ]
[ 0. 9.3 0. ]
[ 0.9 6.2 0. ]
[ 8.6 5.6 0. ]
[ 0.5 8.5 0. ]]
------------------------------------------------------------
題：
有沒有其他時間有效的方法來做到這一點？
我知道這篇文章，但它導致：
b = a[a != None]
print(b)
[-4.0 0.1 0.0 9.2 3.1 0.0 3.0 -4.3 0.0 -1.0 12.8 0.0 4.5 -9.0 0.0 -0.1 6.1
0.0 -7 1 0 -1.0 0.7 0.0 -15 26 0 0.1 -1.1 0.0 0 0 0 5 6 0 -1.1 1.0 0.0
-32 15 0 0.0 9.3 0.0 0.9 6.2 0.0 8.6 5.6 0.0 0.5 8.5 0.0]
uj5u.com熱心網友回復：
您可以使用 np.isnan() 檢測 nan 值。這將如下所示：
nan_values = np.any(np.isnan(a), axis=-1)
那么以下應該給你正確的結果：
result = a[nan_values==False].reshape((-1, 3))
uj5u.com熱心網友回復：
a使用np.full提問者的更新修改@Marc Felix 答案和修改：
nan_values = np.any(np.isnan(a[2:]), axis=-1)
result = a[2:][nan_values==False].reshape((-1, 3))
print(f">> Generating {2} ranks ...\n", result, '\n ------------------------------------------------------------')
nan_values = np.any(np.isnan(a[5:]), axis=-1)
result = a[5:][nan_values==False].reshape((-1, 3))
print(f">> Generating {5} ranks ...\n", result, '\n ------------------------------------------------------------')
將得到預期的結果。
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標籤：Python 麻木 numpy-ndarray 无类型
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