python leave-one-out估計

Question

我想從某個向量x=(x_1,x_2, ..., x_I)獲得矩陣，其中矩陣中的每一行對應於x(i) := (x_1,...,x_{i-1},x_{i+1},...,x_I)。

我知道

from sklearn.cross_validation import LeaveOneOut 
I = 30 
myrowiterator = LeaveOneOut(I) 
for eachrow, _ in myrowiterator: 
    print(eachrow) # prints [1,2,...,29] 
         #  [0,2,...,29] and so on ... 

提供一個例程，以獲得每一行用於上述基質。但我寧願直接在一個步驟中獲得矩陣，直接在這個矩陣上操作，而不是循環遍歷它的行。這會爲我節省一些計算時間。

Answer 1

既然你有numpy的標籤，以下工作：

>>> N = 5 
>>> idx = np.arange(N) 
>>> idx = idx[1:] - (idx[:, None] >= idx[1:]) 
>>> idx 
array([[1, 2, 3, 4], 
     [0, 2, 3, 4], 
     [0, 1, 3, 4], 
     [0, 1, 2, 4], 
     [0, 1, 2, 3]]) 

，現在您可以使用該索引任何其他數組：

>>> a = np.array(['a', 'b', 'c', 'd', 'e']) 
>>> a[idx] 
array([['b', 'c', 'd', 'e'], 
     ['a', 'c', 'd', 'e'], 
     ['a', 'b', 'd', 'e'], 
     ['a', 'b', 'c', 'e'], 
     ['a', 'b', 'c', 'd']], 
     dtype='|S1') 

編輯由於@ user3820991暗示，這可以做成一個

>>> N = 5 
>>> idx = np.arange(1, N) - np.tri(N, N-1, k=-1, dtype=bool) 
>>> idx 
array([[1, 2, 3, 4], 
     [0, 2, 3, 4], 
     [0, 1, 3, 4], 
     [0, 1, 2, 4], 
     [0, 1, 2, 3]]) 

功能np.tri實際上是在此答案的第一版本的神奇比較的高度優化的版本，因爲它使用的最小可能的int類型的數組的大小：通過寫它作爲少隱蔽，因爲numpy的比較使用SIMD進行矢量化，所以類型越小，操作越快。

Answer 2

下面將做到這一點：

In [31]: np.array([row for row, _ in LeaveOneOut(I)]) 
Out[31]: 
array([[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29], 
     [ 0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29], 
     [ 0, 1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29], 
     [ 0, 1, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29], 
     ... 
     [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28]])