用numpy为RNN准备数据的最快方法是什么？

用numpy为RNN准备数据的最快方法是什么？

这是一种

NumPystrides

用于向量化创建

output_x

-

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

样品运行-

In [83]: input_xOut[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 = 4In [85]: outOut[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]]])

这创建了输入数组的视图，因此在内存方面，我们正在提高效率。在大多数情况下，通过进一步的 *** 作，这也应转化为性能上的好处。让我们验证一下它的视图确实是-

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

验证的另一种可靠方法是在其中设置一些值

output

并检查输入-

In [87]: out[0] = 0In [88]: input_xOut[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]])