我正在尝试实现一个自定义Keras层,它将只保留输入的前N个值,并将其余所有值转换为零。我有一个版本,大多数情况下是有效的,但如果有联系,则会留下超过N个值。我想使用一个sort函数总是只留下N个非零值。在
以下是主要工作层,当存在连接时,会留下超过N个值:
def top_n_filter_layer(input_data, n=2, tf_dtype=tf_dtype):
#### Works, but returns more than 2 values if there are ties:
values_to_keep = tf.cast(tf.nn.top_k(input_data, k=n, sorted=True).values, tf_dtype)
min_value_to_keep = tf.cast(tf.math.reduce_min(values_to_keep), tf_dtype)
mask = tf.math.greater_equal(tf.cast(input_data, tf_dtype), min_value_to_keep)
zeros = tf.zeros_like(input_data)
output = tf.where(mask, input_data, zeros)
return output
这是我正在研究的排序方法,但我还是被tf.scatter_更新函数抱怨等级不匹配:
^{pr2}$回溯如下:
---------------------------------------------------------------------------
InvalidArgumentError Traceback (most recent call last)
/opt/conda/lib/python3.6/site-packages/tensorflow/python/framework/ops.py in _create_c_op(graph, node_def, inputs, control_inputs)
1658 try:
-> 1659 c_op = c_api.TF_FinishOperation(op_desc)
1660 except errors.InvalidArgumentError as e:
InvalidArgumentError: Shapes must be equal rank, but are 2 and 3 for 'ScatterUpdate' (op: 'ScatterUpdate') with input shapes: [?,10], [?,2], [?,2].
During handling of the above exception, another exception occurred:
ValueError Traceback (most recent call last)
<ipython-input-10-598e009077f8> in <module>()
27
28 input_layer = Input(shape=(10,))
---> 29 output_data = top_n_filter_layer(input_layer)
30
31 with tf.Session() as sess:
<ipython-input-10-598e009077f8> in top_n_filter_layer(input_data, n, tf_dtype)
18 zeros_variable = tf.assign(zeros_variable, zeros, validate_shape=False)
19
---> 20 output = tf.scatter_update(zeros_variable, indices_to_keep, values_to_keep)
21
22 return output
/opt/conda/lib/python3.6/site-packages/tensorflow/python/ops/state_ops.py in scatter_update(ref, indices, updates, use_locking, name)
297 if ref.dtype._is_ref_dtype:
298 return gen_state_ops.scatter_update(ref, indices, updates,
--> 299 use_locking=use_locking, name=name)
300 return ref._lazy_read(gen_resource_variable_ops.resource_scatter_update( # pylint: disable=protected-access
301 ref.handle, indices, ops.convert_to_tensor(updates, ref.dtype),
/opt/conda/lib/python3.6/site-packages/tensorflow/python/ops/gen_state_ops.py in scatter_update(ref, indices, updates, use_locking, name)
1273 _, _, _op = _op_def_lib._apply_op_helper(
1274 "ScatterUpdate", ref=ref, indices=indices, updates=updates,
-> 1275 use_locking=use_locking, name=name)
1276 _result = _op.outputs[:]
1277 _inputs_flat = _op.inputs
/opt/conda/lib/python3.6/site-packages/tensorflow/python/framework/op_def_library.py in _apply_op_helper(self, op_type_name, name, **keywords)
786 op = g.create_op(op_type_name, inputs, output_types, name=scope,
787 input_types=input_types, attrs=attr_protos,
--> 788 op_def=op_def)
789 return output_structure, op_def.is_stateful, op
790
/opt/conda/lib/python3.6/site-packages/tensorflow/python/util/deprecation.py in new_func(*args, **kwargs)
505 'in a future version' if date is None else ('after %s' % date),
506 instructions)
--> 507 return func(*args, **kwargs)
508
509 doc = _add_deprecated_arg_notice_to_docstring(
/opt/conda/lib/python3.6/site-packages/tensorflow/python/framework/ops.py in create_op(***failed resolving arguments***)
3298 input_types=input_types,
3299 original_op=self._default_original_op,
-> 3300 op_def=op_def)
3301 self._create_op_helper(ret, compute_device=compute_device)
3302 return ret
/opt/conda/lib/python3.6/site-packages/tensorflow/python/framework/ops.py in __init__(self, node_def, g, inputs, output_types, control_inputs, input_types, original_op, op_def)
1821 op_def, inputs, node_def.attr)
1822 self._c_op = _create_c_op(self._graph, node_def, grouped_inputs,
-> 1823 control_input_ops)
1824
1825 # Initialize self._outputs.
/opt/conda/lib/python3.6/site-packages/tensorflow/python/framework/ops.py in _create_c_op(graph, node_def, inputs, control_inputs)
1660 except errors.InvalidArgumentError as e:
1661 # Convert to ValueError for backwards compatibility.
-> 1662 raise ValueError(str(e))
1663
1664 return c_op
ValueError: Shapes must be equal rank, but are 2 and 3 for 'ScatterUpdate' (op: 'ScatterUpdate') with input shapes: [?,10], [?,2], [?,2].
@Vlad下面的答案显示了一种使用一种热编码的工作方法。下面是一个示例,说明它是有效的:
import tensorflow as tf
import numpy as np
tf.reset_default_graph()
model = tf.keras.models.Sequential()
model.add(tf.keras.layers.InputLayer((10,)))
def top_n_filter_layer(input_data, n=2):
topk = tf.nn.top_k(input_data, k=n, sorted=False)
res = tf.reduce_sum(
tf.one_hot(topk.indices,
input_data.get_shape().as_list()[-1]),
axis=1)
res *= input_data
return res
model.add(tf.keras.layers.Lambda(top_n_filter_layer))
x_train = [[1,2,3,4,5,6,7,7,7,7]]
with tf.Session() as sess:
sess.run(tf.global_variables_initializer())
print(model.output.eval({model.inputs[0]:x_train}))
# [[0. 0. 0. 0. 0. 0. 7. 7. 0. 0.]]
让我们一步一步来:
k
处都有一个。然后我们将k
这样的向量求和,得到与k
个正好相同的原始输出形状。在k
位置,我们就用网络的原始softmax
输出进行元素乘法。在top
k=2
值的Tensorflow示例:相关问题 更多 >
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