Question

我目前有一个(1631160,78) np数组作为神经网络的输入。我想用LSTM尝试一些需要3D结构作为输入数据的东西。我目前正在使用以下代码生成所需的3D结构，但速度非常慢（ETA> 1天）。 numpy有更好的方法吗？

我目前生成数据的代码：

def transform_for_rnn(input_x, input_y, window_size):
    output_x = None
    start_t = time.time()
    for i in range(len(input_x)):
        if i > 100 and i % 100 == 0:
            sys.stdout.write('\rTransform Data: %d/%d\tETA:%s'%(i, len(input_x), str(datetime.timedelta(seconds=(time.time()-start_t)/i * (len(input_x) - i)))))
            sys.stdout.flush()
        if output_x is None:
            output_x = np.array([input_x[i:i+window_size, :]])
        else:
            tmp = np.array([input_x[i:i+window_size, :]])
            output_x = np.concatenate((output_x, tmp))

    print
    output_y = input_y[window_size:]
    assert len(output_x) == len(output_y)
    return output_x, output_y

Answer 1

这是一种使用NumPy strides来矢量化output_x -

的方法的方法

nrows = input_x.shape[0] - window_size + 1
p,q = input_x.shape
m,n = input_x.strides
strided = np.lib.stride_tricks.as_strided
out = strided(input_x,shape=(nrows,window_size,q),strides=(m,m,n))

示例运行 -

In [83]: input_x
Out[83]: 
array([[ 0.73089384,  0.98555845,  0.59818726],
       [ 0.08763718,  0.30853945,  0.77390923],
       [ 0.88835985,  0.90506367,  0.06204614],
       [ 0.21791334,  0.77523643,  0.47313278],
       [ 0.93324799,  0.61507976,  0.40587073],
       [ 0.49462016,  0.00400835,  0.66401908]])

In [84]: window_size = 4

In [85]: out
Out[85]: 
array([[[ 0.73089384,  0.98555845,  0.59818726],
        [ 0.08763718,  0.30853945,  0.77390923],
        [ 0.88835985,  0.90506367,  0.06204614],
        [ 0.21791334,  0.77523643,  0.47313278]],

       [[ 0.08763718,  0.30853945,  0.77390923],
        [ 0.88835985,  0.90506367,  0.06204614],
        [ 0.21791334,  0.77523643,  0.47313278],
        [ 0.93324799,  0.61507976,  0.40587073]],

       [[ 0.88835985,  0.90506367,  0.06204614],
        [ 0.21791334,  0.77523643,  0.47313278],
        [ 0.93324799,  0.61507976,  0.40587073],
        [ 0.49462016,  0.00400835,  0.66401908]]])

这会创建一个输入数组的视图，因此在内存方面我们很有效率。在大多数情况下，这应该转化为对性能的益处以及涉及它的进一步操作。让我们确认它的确是一个观点 -

In [86]: np.may_share_memory(out,input_x)
Out[86]: True   # Doesn't guarantee, but is sufficient in most cases

验证的另一种确定方法是将一些值设置为output并检查输入 -

In [87]: out[0] = 0

In [88]: input_x
Out[88]: 
array([[ 0.        ,  0.        ,  0.        ],
       [ 0.        ,  0.        ,  0.        ],
       [ 0.        ,  0.        ,  0.        ],
       [ 0.        ,  0.        ,  0.        ],
       [ 0.93324799,  0.61507976,  0.40587073],
       [ 0.49462016,  0.00400835,  0.66401908]])

使用numpy为RNN准备数据的最快方法是什么？

1 个答案: