Question

假设我有一个形状为[3, 5, 200, 300, 3]的张量。如何将前两个昏暗合并在一起，以便我具有形状[15, 200, 300, 3]的张量。然后我可以反向合并操作并具有原始形状。

Answer 1

您可以使用tf.reshape

a = tf.random_normal(shape=(3, 5, 200, 300, 3))
b = tf.reshape(a, shape=(15, 200, 300, 3))
...
c = tf.reshape(b, shape=(3, 5, 200, 300, 3))

Answer 2

以下函数合并并拆分张量的前两个维，以推断张量的静态或动态形状。

def infer_shape(x):
    x = tf.convert_to_tensor(x)

    # If unknown rank, return dynamic shape
    if x.shape.dims is None:
        return tf.shape(x)

    static_shape = x.shape.as_list()
    dynamic_shape = tf.shape(x)

    ret = []
    for i in range(len(static_shape)):
        dim = static_shape[i]
        if dim is None:
            dim = dynamic_shape[i]
        ret.append(dim)

    return ret

def merge_first_two_dims(tensor):
    shape = infer_shape(tensor)
    shape[0] *= shape[1]
    shape.pop(1)
    return tf.reshape(tensor, shape)


def split_first_two_dims(tensor, dim_0, dim_1):
    shape = infer_shape(tensor)
    new_shape = [dim_0] + [dim_1] + shape[1:]
    return tf.reshape(tensor, new_shape)

Answer 3

大多数情况下，您可以执行此操作，而这并不依赖于具有静态形状：

a = tf.ones([3, 5, 200, 300, 3])
b = tf.reshape(a, tf.concat([[tf.shape(a)[0] * tf.shape(a)[1]], tf.shape(a)[2:]], axis=0))

Tensorflow合并维度和反向合并

3 个答案: