Question

我有一个1D张量，我希望将其划分为重叠块。我想的是： tensor = tf.constant([1, 2, 3, 4, 5, 6, 7])

overlapping_blocker(tensor,block_size=3,stride=2)
=> [[1 2 3], [3, 4, 5], [5, 6, 7]]

到目前为止，我只找到了将张量分割成非重叠块的方法。谁知道解决这个问题的方法？

这需要适用于任意输入维度（即我的输入类似于tf.placeholder([None])

Answer 1

您可以使用tf.nn.conv2d来提供帮助。基本上，你在输入上使用block_size的滑动过滤器，按步幅步进。要使所有矩阵索引排成一行，您必须进行一些重塑。

通用解决方案

import tensorflow as tf


def overlap(tensor, block_size=3, stride=2):
  reshaped = tf.reshape(tensor, [1,1,-1,1])

  # Construct diagonal identity matrix for conv2d filters.
  ones = tf.ones(block_size, dtype=tf.float32)
  ident = tf.diag(ones)
  filter_dim = [1, block_size, block_size, 1]
  filter_matrix = tf.reshape(ident, filter_dim)

  stride_window = [1, 1, stride, 1]

  # Save the output tensors of the convolutions
  filtered_conv = []
  for f in tf.unstack(filter_matrix, axis=1):
    reshaped_filter = tf.reshape(f, [1, block_size, 1, 1])
    c = tf.nn.conv2d(reshaped, reshaped_filter, stride_window, padding='VALID')
    filtered_conv.append(c)

  # Put the convolutions into a tensor and squeeze to get rid of extra dimensions.
  t = tf.stack(filtered_conv, axis=3)
  return tf.squeeze(t)


# Calculate the overlapping strided slice for the input tensor.
tensor = tf.constant([1, 2, 3, 4, 5, 6, 7], dtype=tf.float32)
overlap_tensor = overlap(tensor, block_size=3, stride=2)

with tf.Session() as sess:
  sess.run(tf.initialize_all_variables())
  in_t, overlap_t = sess.run([tensor, overlap_tensor])
  print 'input tensor:'
  print in_t
  print 'overlapping strided slice:'
  print overlap_t

应该给你输出：

input tensor:
[ 1.  2.  3.  4.  5.  6.  7.]
overlapping strided slice:
[[ 1.  2.  3.]
 [ 3.  4.  5.]
 [ 5.  6.  7.]]

更容易理解解决方案

这是我工作的初始版本，它不允许使用变量block_size，但我认为更容易看到卷积滤波器发生了什么 - 我们采用3个值的向量，每个步幅。

def overlap(tensor, stride=2):
  # Reshape the tensor to allow it to be passed in to conv2d.
  reshaped = tf.reshape(tensor, [1,1,-1,1])

  # Construct the block_size filters.
  filter_dim = [1, -1, 1, 1]
  x_filt = tf.reshape(tf.constant([1., 0., 0.]), filter_dim)
  y_filt = tf.reshape(tf.constant([0., 1., 0.]), filter_dim)
  z_filt = tf.reshape(tf.constant([0., 0., 1.]), filter_dim)

  # Stride along the tensor with the above filters.
  stride_window = [1, 1, stride, 1]
  x = tf.nn.conv2d(reshaped, x_filt, stride_window, padding='VALID')
  y = tf.nn.conv2d(reshaped, y_filt, stride_window, padding='VALID')
  z = tf.nn.conv2d(reshaped, z_filt, stride_window, padding='VALID')

  # Pack the three tensors along 4th dimension.
  result = tf.stack([x, y, z], axis=4)
  # Squeeze to get rid of the extra dimensions.
  result = tf.squeeze(result)
  return result

Answer 2

您可以使用tf.extract_image_patches实现相同的目标。

tensor = tf.placeholder(tf.int32, [None])

def overlapping_blocker(tensor,block_size=3,stride=2):
   return tf.squeeze(tf.extract_image_patches(tensor[None,...,None, None], ksizes=[1, block_size, 1, 1], strides=[1, stride, 1, 1], rates=[1, 1, 1, 1], padding='VALID'))

result = overlapping_blocker(tensor,block_size=3,stride=2)
sess = tf.InteractiveSession()
print(result.eval({tensor:np.array([1, 2, 3, 4, 5, 6, 7], np.int32)}))

#[[1 2 3]
#[3 4 5]
#[5 6 7]]

Answer 3

以下是使用您的示例的相对简单的方法：

MODS: Durgal, Wolfy, Pat

输出：

def overlapping_blocker(tensor,block_size,stride):
    blocks = []
    n = tensor.get_shape().as_list()[0]
    ilo = range(0, n, stride)
    ihi = range(block_size, n+1, stride)
    ilohi = zip(ilo, ihi).
    for ilo, ihi in ilohi:
        blocks.append(tensor[ilo:ihi])
    return(tf.pack(blocks, 0))

with tf.Session() as sess:
    tensor = tf.constant([1., 2., 3., 4., 5., 6., 7.])
    block_tensor = overlapping_blocker(tensor, 3, 2)
    print(sess.run(block_tensor))

Answer 4

如果您正在寻找一种方法将每个滚动窗口作为单个张量（即每次调用tf.FIFOQueue，您的窗口都会移动一个。您可以使用tf.train.range_input_producer以及def window_input_producer(tensor, window_size, capacity=32, num_epochs=None): num_windows = tf.shape(tensor)[0] - window_size range_queue = tf.train.range_input_producer( num_windows, shuffle=False, capacity=capacity, num_epochs=num_epochs ) index = range_queue.dequeue() window = tensor[index:index + window_size] queue = tf.FIFOQueue(capacity=capacity, dtypes=[tensor.dtype.base_dtype], shapes=[window_size]) enq = queue.enqueue(window) tf.train.add_queue_runner( tf.train.QueueRunner(queue, [enq]) ) return queue.dequeue()创建一个执行此操作的队列：

编辑：根据原始答案中的要求更新以使用可变长度张量

        final File test = new File("files/test.txt");
        final File dest = new File("WebContent");
        try {
            FileUtils.copyFileToDirectory(test, dest);
        } catch (final IOException e) {
            e.printStackTrace();
        }

Answer 5

我不确定这个问题是否得到了充分的回答，但您可以使用python生成器函数来创建重叠的窗口：

def gen_batch():

# compute number of batches to emit
num_of_batches = round(((len(sequence) - batch_size) / stride))

# emit batches
for i in range(0, num_of_batches * stride, stride):
    result = np.array(sequence[i:i + batch_size])
    yield result

sequence = np.array[1,2,3,4,5,6,7,8,9]
batch_size = 3
stride = 2
gen = gen_batch()

print(next(gen))
[1,2,3]
print(next(gen))
[3,4,5]
...
print(next(gen))
[7,8,9]

一旦定义了生成器函数，就可以使用TensorFlow的数据集类来调用每个切片：

ds = tf.data.Dataset.from_generator(
gen_batch,
(tf.float64),
(tf.TensorShape([batch_size, dim_width])))

ds_out = ds.make_one_shot_iterator().get_next()

print(sess.run(ds_out)))
[1,2,3]
print(sess.run(ds_out)))
[3,4,5]
...
print(sess.run(ds_out)))
[7,8,9]

Answer 6

我认为该功能

tf.signal.overlap_and_add(
    signal, frame_step, name=None
)

可能是您想要的。它解决了我的问题。已在

中回答

Tensorflow：将Tensor切成重叠块

6 个答案:

通用解决方案

更容易理解解决方案