如何在张量流中乘以向量和矩阵而不进行重新整形？

Question

这个：

import numpy as np
a = np.array([1, 2, 1])
w = np.array([[.5, .6], [.7, .8], [.7, .8]])

print(np.dot(a, w))
# [ 2.6  3. ] # plain nice old matrix multiplication n x (n, m) -> m

import tensorflow as tf

a = tf.constant(a, dtype=tf.float64)
w = tf.constant(w)

with tf.Session() as sess:
    print(tf.matmul(a, w).eval())

结果是：

C:\_\Python35\python.exe C:/Users/MrD/.PyCharm2017.1/config/scratches/scratch_31.py
[ 2.6  3. ]
# bunch of errors in windows...
Traceback (most recent call last):
  File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\common_shapes.py", line 671, in _call_cpp_shape_fn_impl
    input_tensors_as_shapes, status)
  File "C:\_\Python35\lib\contextlib.py", line 66, in __exit__
    next(self.gen)
  File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\errors_impl.py", line 466, in raise_exception_on_not_ok_status
    pywrap_tensorflow.TF_GetCode(status))
tensorflow.python.framework.errors_impl.InvalidArgumentError: Shape must be rank 2 but is rank 1 for 'MatMul' (op: 'MatMul') with input shapes: [3], [3,2].

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "C:/Users/MrD/.PyCharm2017.1/config/scratches/scratch_31.py", line 14, in <module>
    print(tf.matmul(a, w).eval())
  File "C:\_\Python35\lib\site-packages\tensorflow\python\ops\math_ops.py", line 1765, in matmul
    a, b, transpose_a=transpose_a, transpose_b=transpose_b, name=name)
  File "C:\_\Python35\lib\site-packages\tensorflow\python\ops\gen_math_ops.py", line 1454, in _mat_mul
    transpose_b=transpose_b, name=name)
  File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\op_def_library.py", line 763, in apply_op
    op_def=op_def)
  File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\ops.py", line 2329, in create_op
    set_shapes_for_outputs(ret)
  File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\ops.py", line 1717, in set_shapes_for_outputs
    shapes = shape_func(op)
  File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\ops.py", line 1667, in call_with_requiring
    return call_cpp_shape_fn(op, require_shape_fn=True)
  File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\common_shapes.py", line 610, in call_cpp_shape_fn
    debug_python_shape_fn, require_shape_fn)
  File "C:\_\Python35\lib\site-packages\tensorflow\python\framework\common_shapes.py", line 676, in _call_cpp_shape_fn_impl
    raise ValueError(err.message)
ValueError: Shape must be rank 2 but is rank 1 for 'MatMul' (op: 'MatMul') with input shapes: [3], [3,2].

Process finished with exit code 1

（不确定为什么在处理过程中会引发同样的异常）

在Tensorflow exception with matmul中建议的解决方案是将矢量重新整形为矩阵，但这导致了不必要的复杂代码 - 是否仍然没有其他方法将矢量与矩阵相乘？

顺便说一句，使用expand_dims（如上面的链接所示）和默认参数会引发一个ValueError - 这在docs中没有提到，并且违背了默认参数的目的。

Answer 1

Matmul编码为二级或更高级别的张量。不知道为什么要诚实，因为numpy有它，它也允许矩阵向量乘法。

import numpy as np
a = np.array([1, 2, 1])
w = np.array([[.5, .6], [.7, .8], [.7, .8]])

print(np.dot(a, w))
# [ 2.6  3. ] # plain nice old matix multiplication n x (n, m) -> m
print(np.sum(np.expand_dims(a, -1) * w , axis=0))
# equivalent result [2.6, 3]

import tensorflow as tf

a = tf.constant(a, dtype=tf.float64)
w = tf.constant(w)

with tf.Session() as sess:
  # they all produce the same result as numpy above
  print(tf.matmul(tf.expand_dims(a,0), w).eval())
  print((tf.reduce_sum(tf.multiply(tf.expand_dims(a,-1), w), axis=0)).eval())
  print((tf.reduce_sum(tf.multiply(a, tf.transpose(w)), axis=1)).eval())

  # Note tf.multiply is equivalent to "*"
  print((tf.reduce_sum(tf.expand_dims(a,-1) * w, axis=0)).eval())
  print((tf.reduce_sum(a * tf.transpose(w), axis=1)).eval())

Answer 2

tf.einsum让您能够以简洁直观的形式完成您所需的工作：

with tf.Session() as sess:
    print(tf.einsum('n,nm->m', a, w).eval())
    # [ 2.6  3. ] 

你甚至可以明确地写下你的评论n x (n, m) -> m。在我看来，它更具可读性和直观性。

我最喜欢的用例是当你想要将一批矩阵与权重向量相乘时：

n_in = 10
n_step = 6
input = tf.placeholder(dtype=tf.float32, shape=(None, n_step, n_in))
weights = tf.Variable(tf.truncated_normal((n_in, 1), stddev=1.0/np.sqrt(n_in)))
Y_predict = tf.einsum('ijk,kl->ijl', input, weights)
print(Y_predict.get_shape())
# (?, 6, 1)

因此，您可以轻松地在所有批次上增加权重，而无需进行转换或重复。这不是通过像其他答案那样扩展尺寸来做到的。因此，您要避免tf.matmul要求具有批次和其他外部尺寸的匹配尺寸：

在任何转置之后，输入必须是秩> = 2的张量，其中内部2维指定有效矩阵乘法参数，并且任何其他外部维度匹配。

Answer 3

你可以使用tf.tensordot并设置axes=1。对于向量乘以矩阵的简单操作，这比tf.einsum更清晰

tf.tensordot(a, w, 1)