3次元のnumpy配列で2次元配列を引数とする関数を呼び出す 'for'ループをベクトル化する方法

Question

したがって、k-neighboorsのXYZ座標（k =ポイントクラウドから10）点：

k_neighboors 
Out[53]: 
array([[[ 2.51508147e-01, 5.60274944e-02, 1.98303187e+00], 
     [ 2.48552352e-01, 5.95569573e-02, 1.98319519e+00], 
     [ 2.56611764e-01, 5.36767729e-02, 1.98236740e+00], 
     ..., 
     [ 2.54520357e-01, 6.23480231e-02, 1.98255634e+00], 
     [ 2.57603496e-01, 5.19787706e-02, 1.98221457e+00], 
     [ 2.43914440e-01, 5.68424985e-02, 1.98352253e+00]], 

     [[ 9.72352773e-02, 2.06699912e-02, 1.99344850e+00], 
     [ 9.91205871e-02, 2.36056261e-02, 1.99329960e+00], 
     [ 9.59625840e-02, 1.71508361e-02, 1.99356234e+00], 
     ..., 
     [ 1.03216261e-01, 2.19752081e-02, 1.99304521e+00], 
     [ 9.65025574e-02, 1.44127617e-02, 1.99355054e+00], 
     [ 9.59930867e-02, 2.72080526e-02, 1.99344873e+00]], 

     [[ 1.76408485e-01, 2.81930678e-02, 1.98819435e+00], 
     [ 1.78670138e-01, 2.81904750e-02, 1.98804617e+00], 
     [ 1.80372953e-01, 3.05109434e-02, 1.98791444e+00], 
     ..., 
     [ 1.81960404e-01, 2.47725621e-02, 1.98785996e+00], 
     [ 1.74499243e-01, 3.50728296e-02, 1.98826015e+00], 
     [ 1.83470801e-01, 2.70808022e-02, 1.98774099e+00]], 

     ..., 
     [[ 1.78178743e-01, -4.60980982e-02, -1.98792374e+00], 
     [ 1.77953839e-01, -4.73701134e-02, -1.98792756e+00], 
     [ 1.77889392e-01, -4.75468598e-02, -1.98793030e+00], 
     ..., 
     [ 1.79924294e-01, -5.08776568e-02, -1.98772371e+00], 
     [ 1.76720902e-01, -5.11409082e-02, -1.98791265e+00], 
     [ 1.83644593e-01, -4.64747548e-02, -1.98756230e+00]], 

     [[ 2.00245917e-01, -2.33091787e-03, -1.98685515e+00], 
     [ 2.02384919e-01, -5.60011715e-04, -1.98673022e+00], 
     [ 1.97325528e-01, -1.03301927e-03, -1.98705769e+00], 
     ..., 
     [ 1.95464164e-01, -6.23105839e-03, -1.98713481e+00], 
     [ 1.98985338e-01, -8.39920342e-03, -1.98688531e+00], 
     [ 1.95959195e-01, 2.68006674e-03, -1.98713303e+00]], 

     [[ 1.28851235e-01, -3.24527062e-02, -1.99127460e+00], 
     [ 1.26415789e-01, -3.27731185e-02, -1.99143147e+00], 
     [ 1.25985757e-01, -3.24910432e-02, -1.99146211e+00], 
     ..., 
     [ 1.28296465e-01, -3.92388329e-02, -1.99117136e+00], 
     [ 1.34895295e-01, -3.64872888e-02, -1.99083793e+00], 
     [ 1.29047096e-01, -3.97952795e-02, -1.99111152e+00]]]) 

この形状：

k_neighboors.shape 
Out[54]: (2999986, 10, 3) 

そして、私は、この関数WICHを有する2次元配列として提供されるいくつかのデータに主成分分析を適用する：

def PCA(data, correlation=False, sort=True): 
    """ Applies Principal Component Analysis to the data 

    Parameters 
    ----------   
    data: array 
     The array containing the data. The array must have NxM dimensions, where each 
     of the N rows represents a different individual record and each of the M columns 
     represents a different variable recorded for that individual record. 
      array([ 
      [V11, ... , V1m], 
      ..., 
      [Vn1, ... , Vnm]]) 

    correlation(Optional) : bool 
      Set the type of matrix to be computed (see Notes): 
       If True compute the correlation matrix. 
       If False(Default) compute the covariance matrix. 

    sort(Optional) : bool 
      Set the order that the eigenvalues/vectors will have 
       If True(Default) they will be sorted (from higher value to less). 
       If False they won't. 
    Returns 
    ------- 
    eigenvalues: (1,M) array 
     The eigenvalues of the corresponding matrix. 

    eigenvector: (M,M) array 
     The eigenvectors of the corresponding matrix. 

    Notes 
    ----- 
    The correlation matrix is a better choice whent there are different magnitudes 
    representing the M variables. Use covariance matrix in any other case. 

    """ 

    #: get the mean of all variables 
    mean = np.mean(data, axis=0, dtype=np.float64) 

    #: adjust the data by substracting the mean to each variable 
    data_adjust = data - mean 

    #: compute the covariance/correlation matrix 
    #: the data is transposed due to np.cov/corrcoef sintaxis 
    if correlation: 
     matrix = np.corrcoef(data_adjust.T) 
    else: 
     matrix = np.cov(data_adjust.T) 

    #: get the eigenvalues and eigenvectors 
    eigenvalues, eigenvectors = np.linalg.eig(matrix) 

    if sort: 
     #: sort eigenvalues and eigenvectors 
     sort = eigenvalues.argsort()[::-1] 
     eigenvalues = eigenvalues[sort] 
     eigenvectors = eigenvectors[:,sort] 

    return eigenvalues, eigenvectors 

そこで質問です：私はこの1つのように永遠に取らない方法で、2999986の10x3アレイの各々の上に上記PCA機能を適用する方法：

data = np.empty((2999986, 3)) 

for i in range(len(k_neighboors)): 
    w, v = PCA(k_neighboors[i]) 
    data[i] = v[:,2] 
    break #: I break the loop in order to don't have to wait for ever. 


data 
Out[64]: 
array([[ 0.10530792, 0.01028906, 0.99438643], 
     [ 0.  , 0.  , 0.  ], 
     [ 0.  , 0.  , 0.  ], 
     ..., 
     [ 0.  , 0.  , 0.  ], 
     [ 0.  , 0.  , 0.  ], 
     [ 0.  , 0.  , 0.  ]]) 

はどうもありがとうございました

Answer 1

@Divakarと@Eelcoのコメントに感謝します。

Divakarがon this answer

def vectorized_app(data):    
     diffs = data - data.mean(1,keepdims=True) 
     return np.einsum('ijk,ijl->ikl',diffs,diffs)/data.shape[1] 

を投稿し、その関数を使用してEelcoは彼のコメントで指摘ものを使用して、私はこれで終わります。

k_neighboors.shape 
Out[48]: (2999986, 10, 3) 

#: THE (ASSUMED)VECTORIZED ANSWER 
data = np.linalg.eig(vectorized_app(k_neighboors))[1][:,:,2] 

data 
Out[50]: 
array([[ 0.10530792, 0.01028906, 0.99438643], 
     [ 0.06462 , 0.00944352, 0.99786526], 
     [ 0.0654035 , 0.00860751, 0.99782177], 
     ..., 
     [-0.0632175 , 0.01613551, 0.99786933], 
     [-0.06449399, 0.00552943, 0.99790278], 
     [-0.06081954, 0.01802078, 0.99798609]]) 

WICHは永遠に取ることなく、forループと同じ結果を与える（althoughtはまだしばらくかかります）：これを行うには良い方法があるかもしれない場合

data2 = np.empty((2999986, 3)) 

for i in range(len(k_neighboors)): 
    if i > 10: 
     break #: I break the loop in order to don't have to wait for ever. 
    w, v = PCA(k_neighboors[i]) 
    data2[i] = v[:,2] 


data2 
Out[52]: 
array([[ 0.10530792, 0.01028906, 0.99438643], 
     [ 0.06462 , 0.00944352, 0.99786526], 
     [ 0.0654035 , 0.00860751, 0.99782177], 
     ..., 
     [ 0.  , 0.  , 0.  ], 
     [ 0.  , 0.  , 0.  ], 
     [ 0.  , 0.  , 0.  ]]) 

は私は知りません私は質問を開いたままにします。