问题描述
已编辑：
我有一个时间序列，例如ts = [[0 0][1 1][2 2][3 3][4 4][5 5][6 6][7 7][8 8]]，我想按以下两个顺序进行划分：
X = [[[[0][1]][[1][2]][[2][3]]] [[[1][2]][[2][3]][[3][4]]] [[[2][3]][[3][4]][[4][5]]] [[[3][4]][[4][5]][[5][6]]] [[[4][5]][[5][6]][[6][7]]] [[[5][6]][[6][7]][[7][8]]]]
y = [[3][4][5][6][7][8]]
X是三个两步滑动窗口的大块序列，而y是它的特征。 我的策略是首先采用以下方法：
def split_sequences(sequences,n_steps):
X,y = list(),list()
for i in range(len(sequences)):
# find the end of this pattern
end_ix = i + n_steps
prev_end_ix = end_ix - 1
# check if we are beyond the dataset
if end_ix > len(sequences):
break
# gather input and output parts of the pattern
seq_x,seq_y = sequences[i:end_ix,:-1],sequences[prev_end_ix:end_ix,-1]
X.append(seq_x)
y.append(seq_y)
return np.array(X),np.array(y)
哪些内容会被撤销
：X =[[[0][1]] [[1][2]] [[2][3]] [[3][4]] [[4][5]] [[5][6]] [[6][7]] [[7][8]]]
y = [[1][2][3][4][5][6][7][8]]
def separar_uni_X(sequencia,n_passos):
X = list()
for i in range(len(sequencia)):
# find the end of this pattern
end_ix = i + n_passos
# check if we are beyond the sequence
if end_ix > len(sequencia):
break
# gather input and output parts of the pattern
seq_x = sequencia[i:end_ix,:]
X.append(seq_x)
return np.array(X)
def separar_uni_y(sequencia,n_passos):
y = list()
for i in range(len(sequencia)):
# find the end of this pattern
end_ix = i + n_passos
# check if we are beyond the sequence
if end_ix > len(sequencia):
break
# gather input and output parts of the pattern
seq_y = sequencia[i:end_ix,:]
y.append(seq_y[-1])
return np.array(y)
问题：问题在于，为了获得所需的输出，它必须存储从第一种方法到第二种方法的数据，并且当序列太长时，它将超过存储容量。为了解决此缺点，我在子流程中分解了流程：
def split_sequence_3D(sequences,n_steps,batch_size):
X,list()
for i in range(len(sequences)):
# find the end of this pattern
end_ix = i + n_steps
prev_end_ix = end_ix - 1
# check if we are beyond the dataset
if end_ix > len(sequences):
break
# gather input and output parts of the pattern
seq_x,-1]
sub_X,sub_y = [],[]
for j in range(batch_size):
sub_X.append(seq_x)
sub_y.append(seq_y)
X.append(sub_X)
y.append(sub_y[-1])
return np.array(X),np.array(y)
X = [[[[0][1]][[0][1]][[0][1]]] [[[1][2]][[1][2]][[1][2]]] [[[2][3]][[2][3]][[2] [3]]] [[[3][4]][[3][4]][[3][4]]] [[[4][5]][[4][5]][[4][5]]] [[[5][6]][[5][6]][[5 [6]]] [[[6][7]][[6][7]][[6][7]]] [[[7][8]][[7][8]][[7][8]]]]
y = [[1][2][3][4][5][6][7][8]]
我已经在广泛寻找替代品，但没有找到。
解决方法
好吧，我真的很想解决您的问题，这也是我的。但是最终，解决方案变得很简单。我的解决方案是使滑动窗口迭代器也能滑动。
def input_3D(sequencia,lote,janela):
if lote > len(sequencia):
raise ValueError('Tamanho do lote maior que o conjunto dos dados')
if janela > len(sequencia):
raise ValueError('Tamanho da janela maior que o conjunto dos dados')
X_,y_ = [],[]
for j in range (len(sequencia)):
if j+lote+janela > len(sequencia):
break
X,y = [],[]
for i in range (j,j+lote,1):
end_ix = i+janela
prev_end_ix = end_ix - 1
seq_x,seq_y = sequencia[i:end_ix,:-1],sequencia[prev_end_ix:end_ix,-1]
X.append(np.array(seq_x))
y.append(np.array(seq_y[-1]))
X_.append(np.array(X))
y_.append(np.array(y[-1]))
return np.array(X_),np.array(y_)
假设您输入的是：
arr_x = list(range(0,100))
arr_y = list(range(0,100))
arr = np.stack([arr_x,arr_y])
arr = arr.T
那么您的输出将是：
[[[[ 0]
[ 1]
[ 2]
...
[ 7]
[ 8]
[ 9]]
[[ 1]
[ 2]
[ 3]
...
[ 8]
[ 9]
[10]]
[[ 2]
[ 3]
[ 4]
...
[ 9]
[10]
[11]]
[[ 3]
[ 4]
[ 5]
...
[10]
[11]
[12]]]
...
[[[86]
[87]
[88]
...
[93]
[94]
[95]]
[[87]
[88]
[89]
...
[94]
[95]
[96]]
[[88]
[89]
[90]
...
[95]
[96]
[97]]
[[89]
[90]
[91]
...
[96]
[97]
[98]]]] [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 56 57 58 59
60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83
84 85 86 87 88 89 90 91 92 93 94 95 96 97 98]