如何有效地vstack大序列的numpy數組塊？

Question

我生成numpy的陣列序列如下：

def chunker(seq, size): 
    return (seq[pos:pos + size] for pos in range(0, len(seq), size)) 

for i in chunker(X,10000): 
    e = function(i) 
    print('new marix',e) 

new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
... 
new matrix (10000, 3208) 

我想vstack上述n矩陣中的單獨一個。因此，我試過如下：

X = np.vstack(e) 

然而，當我打印X我又收到：

new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
new matrix (10000, 3208) 
... 
new matrix (10000, 3208) 

取而代之的是新的vstacked單個矩陣。任何想法如何vstack這個numpy數組序列？

更新

從傑德沃德的答案我編輯我的代碼如下：

進口numpy的爲NP

def chunker(seq, size): 
    return (seq[pos:pos + size] for pos in range(0, len(seq), size)) 

for (r,i) in enumerate(chunker(X,10000)): 
    e = function(i) 
    print('new matrix',e) 
    X[r,:] = e 

print(X) 

Answer 1

的一種方式，雖然可能不是最有效的，是創建列表中您想要堆疊的列表，然後在循環外堆疊一次。

例如：

import numpy as np 

def chunker(seq, size): 
    return (seq[pos:pos + size] for pos in range(0, len(seq), size)) 

# Some fake function (n.b. this is a silly way to reverse a list) 
def function(arr): 
    arr.reverse() 
    return arr 

# Generate fake X 
X = list(range(100)) 

chunks = [] 
for i in chunker(X,10): 
    e = function(i) 
    print('new matrix',e) 
    chunks.append(e) 

merged = np.vstack(chunks) 
print(merged) 

輸出：

 
new matrix [9, 8, 7, 6, 5, 4, 3, 2, 1, 0] 
new matrix [19, 18, 17, 16, 15, 14, 13, 12, 11, 10] 
new matrix [29, 28, 27, 26, 25, 24, 23, 22, 21, 20] 
new matrix [39, 38, 37, 36, 35, 34, 33, 32, 31, 30] 
new matrix [49, 48, 47, 46, 45, 44, 43, 42, 41, 40] 
new matrix [59, 58, 57, 56, 55, 54, 53, 52, 51, 50] 
new matrix [69, 68, 67, 66, 65, 64, 63, 62, 61, 60] 
new matrix [79, 78, 77, 76, 75, 74, 73, 72, 71, 70] 
new matrix [89, 88, 87, 86, 85, 84, 83, 82, 81, 80] 
new matrix [99, 98, 97, 96, 95, 94, 93, 92, 91, 90] 
[[ 9 8 7 6 5 4 3 2 1 0] 
[19 18 17 16 15 14 13 12 11 10] 
[29 28 27 26 25 24 23 22 21 20] 
[39 38 37 36 35 34 33 32 31 30] 
[49 48 47 46 45 44 43 42 41 40] 
[59 58 57 56 55 54 53 52 51 50] 
[69 68 67 66 65 64 63 62 61 60] 
[79 78 77 76 75 74 73 72 71 70] 
[89 88 87 86 85 84 83 82 81 80] 
[99 98 97 96 95 94 93 92 91 90]] 

或者不創建中間列表：

merged = np.zeros([0,10]) 
for i in chunker(X,10): 
    e = function(i) 
    print('new matrix',e) 
    merged = np.vstack([merged, e]) 

print(merged) 

但最有效的是初始化numpy的陣列在循環之前，然後在內部設置該數組的行循環。你需要首先計算最後的merged數組的尺寸（這裏我只是將它創建爲10x10矩陣，因爲我知道尺寸）。

merged = np.zeros([10,10]) 
for (r,i) in enumerate(chunker(X,10)): 
    e = function(i) 
    print('new matrix',e) 
    merged[r,:] = e 

print(merged)