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<p>我已经编写了一个程序来更新从一个张量到另一个张量的行。下面是我想要达到的一个非常基本的想法</p>
<pre><code>with tf.Session() as sess:
A = tf.Variable(
[[[0.2, 0.8, 0.1], [0.0, 1.0, 3.0], [0.0, 1.0, 3.0]], [[0.0, 1.0, 1.0], [0.0, 1.0, 0.5], [0.0, 1.0, 3.0]],
[[0.0, 1.0, 0.6], [0.0, 1.0, 0.4], [0.0, 1.0, 3.0]]])
B = tf.Variable(
[[[1.2, 1.8, 1.1], [1.1, 1.1, 3.1], [1.1, 1.1, 3.1]], [[1.1, 1.1, 1.1], [1.1 ,1.1, 1.5], [1.1, 1.1, 3.1]],
[[1.0, 1.0, 1.6], [1.1, 1.1, 1.4], [1.1, 1.1, 3.1]]])
sess.run(tf.global_variables_initializer())
parent1 = 0
parent2 = 1
print("\n A")
print(sess.run(A[parent2][0]))
# A=[0. 1. 1.]
print("\n B")
print(sess.run(B[parent1][0]))
# B=[0.2 0.8 0.1]
print("\n Result")
B = B[0,0].assign(A[1,0])
print(sess.run(B[0]))
# Result
# [[0. 1. 1. ]
# [1.1 1.1 3.1]
# [1.1 1.1 3.1]]
</code></pre>
<p>这很管用。添加第二个运算符时出现问题:</p>
<pre><code>print("\n Result")
B = B[0,0].assign(A[1,0])
B = B[1,0].assign(A[0,0])
</code></pre>
<p>然后我被击中(底部的完整堆栈跟踪):</p>
<pre><code>ValueError: None values not supported.
</code></pre>
<p>似乎出于某种原因,它试图做一个张量的内部转换,出于某种原因,一个None类型被传递。我试着在继续之前评估变量,但也不起作用。我的想法是达到一个点,我可以利用这样一个循环:</p>
<pre><code>for i in indexes:
B = B[parent1][i].assign(A[parent2][i])
B = B[parent2][i].assign(A[parent1][i])
</code></pre>
<p>完整堆栈跟踪:</p>
<pre><code>Traceback (most recent call last):
File "tensorflow\python\framework\op_def_library.py", line 527, in _apply_op_helper
preferred_dtype=default_dtype)
File "tensorflow\python\framework\ops.py", line 1224, in internal_convert_to_tensor
ret = conversion_func(value, dtype=dtype, name=name, as_ref=as_ref)
File "tensorflow\python\framework\constant_op.py", line 305, in _constant_tensor_conversion_function
return constant(v, dtype=dtype, name=name)
File "tensorflow\python\framework\constant_op.py", line 246, in constant
allow_broadcast=True)
File "tensorflow\python\framework\constant_op.py", line 284, in _constant_impl
allow_broadcast=allow_broadcast))
File "tensorflow\python\framework\tensor_util.py", line 454, in make_tensor_proto
raise ValueError("None values not supported.")
ValueError: None values not supported.
</code></pre>