方法1: Here'e一種方法化的開展基於linear-indexing
-
def random_block_fill_lidx(a, N, fillval=0):
# a is input array
# N is blocksize
# Store shape info
m,n,r = a.shape
# Get all possible starting linear indices for each 2D slice
possible_start_lidx = (np.arange(n-N+1)[:,None]*r + range(r-N+1)).ravel()
# Get random start indices from all possible ones for all 2D slices
start_lidx = np.random.choice(possible_start_lidx, m)
# Get linear indices for the block of (N,N)
offset_arr = (a.shape[-1]*np.arange(N)[:,None] + range(N)).ravel()
# Add in those random start indices with the offset array
idx = start_lidx[:,None] + offset_arr
# On a 2D view of the input array, use advance-indexing to set fillval.
a.reshape(m,-1)[np.arange(m)[:,None], idx] = fillval
return a
方法2:下面是使用advanced-indexing
另外,可能更有效的一個(對大型2D片) -
def random_block_fill_adv(a, N, fillval=0):
# a is input array
# N is blocksize
# Store shape info
m,n,r = a.shape
# Generate random start indices for second and third axes keeping proper
# distance from the boundaries for the block to be accomodated within.
idx0 = np.random.randint(0,n-N+1,m)
idx1 = np.random.randint(0,r-N+1,m)
# Setup indices for advanced-indexing.
# First axis indices would be simply the range array to select one per elem.
# We need to extend this to 3D so that the latter dim indices could be aligned.
dim0 = np.arange(m)[:,None,None]
# Second axis indices would idx0 with broadcasted additon of blocksized
# range array to cover all block indices along this axis. Repeat for third.
dim1 = idx0[:,None,None] + np.arange(N)[:,None]
dim2 = idx1[:,None,None] + range(N)
a[dim0, dim1, dim2] = fillval
return a
方法3:隨着老信賴環 -
def random_block_fill_loopy(a, N, fillval=0):
# a is input array
# N is blocksize
# Store shape info
m,n,r = a.shape
# Generate random start indices for second and third axes keeping proper
# distance from the boundaries for the block to be accomodated within.
idx0 = np.random.randint(0,n-N+1,m)
idx1 = np.random.randint(0,r-N+1,m)
# Iterate through first and use slicing to assign fillval.
for i in range(m):
a[i, idx0[i]:idx0[i]+N, idx1[i]:idx1[i]+N] = fillval
return a
採樣運行 -
In [357]: a = np.arange(2*4*7).reshape(2,4,7)
In [358]: a
Out[358]:
array([[[ 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, 29, 30, 31, 32, 33, 34],
[35, 36, 37, 38, 39, 40, 41],
[42, 43, 44, 45, 46, 47, 48],
[49, 50, 51, 52, 53, 54, 55]]])
In [359]: random_block_fill_adv(a, N=3, fillval=0)
Out[359]:
array([[[ 0, 0, 0, 0, 4, 5, 6],
[ 7, 0, 0, 0, 11, 12, 13],
[14, 0, 0, 0, 18, 19, 20],
[21, 22, 23, 24, 25, 26, 27]],
[[28, 29, 30, 31, 32, 33, 34],
[35, 36, 37, 38, 0, 0, 0],
[42, 43, 44, 45, 0, 0, 0],
[49, 50, 51, 52, 0, 0, 0]]])
有趣的東西:正在就地灌裝,如果我們繼續運行random_block_fill_adv(a, N=3, fillval=0)
,我們將最終結束了所有的零a
。因此,也驗證了代碼。
運行測試
In [579]: a = np.random.randint(0,9,(10000,4,4))
In [580]: %timeit random_block_fill_lidx(a, N=2, fillval=0)
...: %timeit random_block_fill_adv(a, N=2, fillval=0)
...: %timeit random_block_fill_loopy(a, N=2, fillval=0)
...:
1000 loops, best of 3: 545 µs per loop
1000 loops, best of 3: 891 µs per loop
100 loops, best of 3: 10.6 ms per loop
In [581]: a = np.random.randint(0,9,(1000,40,40))
In [582]: %timeit random_block_fill_lidx(a, N=10, fillval=0)
...: %timeit random_block_fill_adv(a, N=10, fillval=0)
...: %timeit random_block_fill_loopy(a, N=10, fillval=0)
...:
1000 loops, best of 3: 739 µs per loop
1000 loops, best of 3: 671 µs per loop
1000 loops, best of 3: 1.27 ms per loop
因此,要選擇哪一個取決於第一軸長度和塊大小。
@Abdou如前所述 - 「基本上我需要將固定大小的塊清零,但是對於我的數組中的每個平面,在不同的(隨機)位置。所以,不會有一個具體的輸出,但是OP是否可以列出其中一個可能的輸出。 – Divakar
我更新了這個問題,希望更清楚我正在嘗試做什麼。 – MichaelSB
@Divakar,*提取*和*歸零*是兩個不同的東西。這就是爲什麼我要求獲得預期產出。 – Abdou