如何将 TensorFlow 张量切成多个

how to slice the tensorflow tensor to multiple

提问人:Rxma 提问时间:2/1/2021 最后编辑:Rxma 更新时间:2/3/2021 访问量:250

问:

张量为: 批处理(3) * 长度 (5) * 暗 (2)

tensor = tf.constant([[[1,1],[2,2],[3,3],[4,4],[5,5]],[[1,1],[2,2],[3,3],[4,4],[5,5]],[[1,1],[2,2],[3,3],[4,4],[5,5]]] )

我想通过 length_index [0,0],[0,1] 获得更多切片......[3,4],[4,4]根据length_axis_index[0,1,2,3,4],操作类似

spans_length=0
with tf.variable_scope("loss_span"):
    output=[]
    for i in range(0,1+n_spans):
        for j in range(1,seq_length):
            if j + i < seq_length:
                res = tf.slice(output_layer_sequence, [0, j, 0], [-1, j+i-j+1, -1])
                res = tf.reduce_sum(res,axis=1)
                output.append(res)
                # output = tf.convert_to_tensor(output)
                spans_length+=1

    output = tf.convert_to_tensor(output)
    vsp = tf.transpose(output, [1,0,2])#batch , spans_length,hidden_size
    vsp = tf.reshape(vsp,[-1,hidden_size])#batch * span_length,hidden_size
    span_logits = tf.matmul(vsp, output_span_weight, transpose_b=True)  # output:[batch * spans_length,class_labels]
    span_logits = tf.nn.bias_add(span_logits, output_span_bias)  # output:[batch * spans_length,class_labels]
    span_matrix = tf.reshape(span_logits,[-1,spans_length,class_labels],name="span_matrix_val")#[batch , spans_length,class_labels]
    
   
    label_span_logists = tf.one_hot(indices=label_span,depth=class_labels, on_value=1, off_value=0, axis=-1, dtype=tf.int32)
    label_span_logists=tf.cast(label_span_logists,tf.int64)
    span_loss = tf.nn.softmax_cross_entropy_with_logits(logits=span_matrix, labels=label_span_logists)
    span_loss = tf.reduce_mean(span_loss, name='loss_span')
               

当我做这样的操作时,训练模型的时间很长;如何加速它.thanks

TensorFlow NLP 切片

评论

0赞 Andrey 2/1/2021
请澄清 - 这是什么意思。是吗?还是 'tensor[:,0:0,:] ?slice(0,0)tensor[:,0,0]
0赞 Rxma 2/1/2021
slice(0,0) 表示 a1=tf.slice(tensor,[0,0,0],[-1,0,-1])) slice(0,1) 表示 a2=tf.slice(tensor,[0,0,0],[-1,1,-1])) slice(0,2) 表示 a3=tf.slice(tensor,[0,0,0],[-1,2,-1])),

答:

2赞 Andrey 2/1/2021 #1

此代码的工作原理:

# tensor = tf.constant([[[1,1],[2,2],[3,3],[4,4],[5,5]],[[1,1],[2,2],[3,3],[4,4],[5,5]],[[1,1],[2,2],[3,3],[4,4],[5,5]]] )
tensor = tf.random.uniform((3, 2000, 2))
length = tf.shape(tensor)[1].numpy()
output = []
for begins in range(length):
  for size in range(length - begins):
    res = tf.slice(tensor, [0, begins, 0], [-1, size + 1, -1])
    res = tf.reduce_sum(res)
    output.append(res)
output = tf.convert_to_tensor(output)

我尝试使用 ,但我没有看到任何好处:tf.scan()

output = tf.constant([], tf.int32)
for begins in range(length):
  t = tensor[:, begins:, :]
  t = tf.transpose(t, (1, 0, 2))
  t = tf.scan(lambda a, x: a + x, t)
  t = tf.transpose(t, (1, 0, 2))
  t = tf.reduce_sum(t, [0, 2])
  output = tf.concat([output, t], 0)

编辑:

尝试在预处理中沿未使用的维度 [0, 2] 应用:reduce_sum()

tensor = tf.reduce_sum(tensor, [0, 2])
output = tf.constant([])
for begins in range(length):
  t = tensor[begins:]
  t = tf.scan(lambda a, x: a + x, t)
  output = tf.concat([output, t], 0)

仍然看不到性能优势。

评论

0赞 Rxma 2/2/2021
当我做这样的操作时,训练时间很长;如何加速它.thanks stackoverflow.com/a/66011043/8375120
1赞 Andrey 2/3/2021
@Colt查看编辑。我认为不可能显着改善 - 瓶颈是,我怀疑是否有可能更快地计算累积总和scan()scan()
0赞 Rxma 2/2/2021 #2
for i in range(0,50):
    for j in range(1,200):
         if j + i < 200:
            res = tf.slice(output_layer_sequence, [0, j, 0], [-1, j+i-j+1, -1])
            res = tf.reduce_sum(res,axis=1)
            output.append(res)
output = tf.convert_to_tensor(output)
                   

当我做这样的操作时,训练时间很长;如何加速它.thanks