我正在尝试使用Tensorflow 1.9.0重新实现以下函数（以numpy编写）。
def lateral_inhibition2(conv_spikes,SpikesPerNeuronAllowed):
vbn = np.where(SpikesPerNeuronAllowed==0)
conv_spikes[vbn[0],vbn[1],:]=0
return conv_spikes
conv_spikes是等级3的二进制张量，SpikesPerNeuronAllowed是等级2的张量。 conv_spikes是一个变量，用于指示该位置包含1的特定位置的神经元是否已加标，而0则指示该位置的神经元未加标。 SpikesPerNeuronAllowed变量指示是否允许沿X-Y轴的Z位置上的所有神经元都尖峰。 1中的SpikesPerNeuronAllowed表示允许X-Y中相应conv_spikes位置并沿Z轴的神经元尖峰。 0指示不允许在X-Y中的相应conv_spikes位置并沿Z轴的神经元加尖峰。
conv_spikes2 = (np.random.rand(5,5,3)>=0.5).astype(np.int16)
temp2 = np.random.choice([0, 1], size=(25,), p=[3./4, 1./4])
SpikesPerNeuronAllowed2 = temp2.reshape(5,5)
print(conv_spikes2[:,:,0])
print
print(conv_spikes2[:,:,1])
print
print(conv_spikes2[:,:,2])
print
print(SpikesPerNeuronAllowed2)
产生以下输出
##First slice of conv_spikes across Z-axis
[[0 0 1 1 1]
[1 0 0 1 1]
[1 0 1 1 0]
[0 1 0 1 1]
[0 1 0 0 0]]
##Second slice of conv_spikes across Z-axis
[[0 0 1 0 0]
[0 0 1 0 1]
[0 0 1 1 1]
[0 0 0 1 0]
[1 1 1 1 1]]
##Third slice of conv_spikes across Z-axis
[[0 1 1 0 0]
[0 0 1 0 0]
[0 1 1 0 0]
[0 0 0 1 0]
[1 0 1 1 1]]
##SpikesPerNeuronAllowed2
[[0 0 0 0 1]
[0 0 0 0 0]
[0 0 0 0 0]
[1 1 0 0 0]
[0 0 0 1 0]]
现在，当调用函数时
conv_spikes2 = lateral_inhibition2(conv_spikes2,SpikesPerNeuronAllowed2)
print(conv_spikes2[:,:,0])
print
print(conv_spikes2[:,:,1])
print
print(conv_spikes2[:,:,2])
产生以下输出
##First slice of conv_spikes across Z-axis
[[0 0 0 0 1]
[0 0 0 0 0]
[0 0 0 0 0]
[0 1 0 0 0]
[0 0 0 0 0]]
##Second slice of conv_spikes across Z-axis
[[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 1 0]]
##Third slice of conv_spikes across Z-axis
[[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 0 0]
[0 0 0 1 0]]
我试图在Tensorflow中重复如下内容
conv_spikes_tf = tf.Variable((np.random.rand(5,5,3)>=0.5).astype(np.int16))
a_placeholder = tf.placeholder(tf.float32,shape=(5,5))
b_placeholder = tf.placeholder(tf.float32)
inter2 = tf.where(tf.equal(a_placeholder,b_placeholder))
output= sess.run(inter2,feed_dict{a_placeholder:SpikesPerNeuronAllowed2,b_placeholder:0})
print(output)
产生以下输出
[[0 0]
[0 1]
[0 2]
[0 3]
[1 0]
[1 1]
[1 2]
[1 3]
[1 4]
[2 0]
[2 1]
[2 2]
[2 3]
[2 4]
[3 2]
[3 3]
[3 4]
[4 0]
[4 1]
[4 2]
[4 4]]
我尝试用以下代码更新conv_spikes_tf导致错误，我尝试浏览scatter_nd_update的手册，但我认为我不太了解。
update = tf.scatter_nd_update(conv_spikes_tf, output, np.zeros(output.shape[0]))
sess.run(update)
ValueError: The inner 1 dimensions of input.shape=[5,5,3] must match the inner 1 dimensions of updates.shape=[21,2]: Dimension 0 in both shapes must be equal, but are 3 and 2. Shapes are [3] and [2]. for 'ScatterNdUpdate_8' (op: 'ScatterNdUpdate') with input shapes: [5,5,3], [21,2], [21,2].
我不理解错误消息，特别是inner 1 dimensions的含义是什么，如何使用张量流实现上述numpy功能？
答案 0 :(得分：1)
updates中tf.scatter_nd_update的最后一个暗度应为3，等于ref的最后一个暗度。
update = tf.scatter_nd_update(conv_spikes_tf, output, np.zeros(output.shape[0], 3))
如果我的理解正确，您想将SpikesPerNeuronAllowed2（掩码）应用于conv_spikes。一种更简单的方法是将conv_spikes重塑为（3,5,5）并乘以SpikesPerNeuronAllowed2。
我使用一个常量示例来显示结果。您也可以将其更改为tf.Variable。
conv = (np.random.rand(3,5,5)>=0.5).astype(np.int32)
tmp = np.random.choice([0, 1], size=(25,), p=[3./4, 1./4])
mask = tmp.reshape(5,5)
# array([[[1, 1, 0, 0, 0],
# [0, 1, 0, 0, 1],
# [0, 1, 0, 0, 1],
# [1, 0, 0, 0, 1],
# [1, 0, 0, 1, 0]],
# [[1, 0, 0, 0, 1],
# [1, 0, 1, 1, 1],
# [0, 0, 1, 0, 1],
# [0, 0, 0, 1, 1],
# [0, 0, 0, 1, 1]],
# [[0, 0, 0, 1, 0],
# [0, 1, 1, 0, 1],
# [0, 1, 1, 0, 1],
# [1, 1, 1, 1, 0],
# [1, 1, 1, 0, 1]]], dtype=int32)
# array([[0, 0, 0, 1, 1],
# [0, 0, 0, 1, 0],
# [0, 0, 0, 0, 0],
# [0, 1, 0, 1, 0],
# [0, 0, 1, 0, 1]])
tf_conv = tf.constant(conv, dtype=tf.int32)
tf_mask = tf.constant(mask, dtype=tf.int32)
res = tf_conv * tf_mask
sess = tf.InteractiveSession()
sess.run(res)
# array([[[0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0]],
# [[0, 0, 0, 0, 1],
# [0, 0, 0, 1, 0],
# [0, 0, 0, 0, 0],
# [0, 0, 0, 1, 0],
# [0, 0, 0, 0, 1]],
# [[0, 0, 0, 1, 0],
# [0, 0, 0, 0, 0],
# [0, 0, 0, 0, 0],
# [0, 1, 0, 1, 0],
# [0, 0, 1, 0, 1]]], dtype=int32)