Question

我有一个这样的矩阵：

mat1 = tf.Variable([[0.  0.  0.  0. ]
                   [0.7 0.  0.  0. ]
                   [0.  0.  0.  0. ]
                   [0.  0.  0.  0. ]
                   [0.  0.  0.  0. ]
                   [0.  0.  0.  0. ]
                   [0.  0.  0.  0. ]])
mat1 = mat1 + abs(mat1)/2

另外，像这样的索引矩阵：

prob_indice = tf.constant([[0 1]
                           [0 3]
                           [1 1]
                           [1 2]
                           [1 3]
                           [5 0]
                           [5 1]
                           [5 2]
                           [5 3]
                           [6 1]
                           [6 3]])
energy_allocation = 0.05

现在，我想对mat1中energy_allocation中的元素进行归纳，其中索引位于prob_indice中。

所以预期的输出将是：

                  [[0     0.05    0    0.05   ]
                   [0.7   0.05  0.05    0.05 ]
                   [0.    0.      0.    0.   ]
                   [0.    0.      0.    0.   ]
                   [0.    0.      0.    0.   ]
                   [0.05  0.05   0.05   0.05 ]
                   [0.    0.05   0.     0.05 ]]

更新1

mat1是按照mat1 = x + abs(x)/2的方式计算的，这就是为什么如果我使用tf.scatter_nd_add会产生此错误的原因：

返回ref._lazy_read（gen_state_ops.resource_scatter_nd_add（＃ pylint：disable =受保护的访问AttributeError： 'tensorflow.python.framework.ops.EagerTensor'对象没有属性 '_lazy_read'

谢谢！

Answer 1

您需要tf.scatter_nd_add()。

import tensorflow as tf

mat1 = tf.Variable([[0. ,0. ,0. ,0.],
                    [0.7 ,0. , 0.,  0. ],
                    [0., 0., 0., 0.],
                    [0., 0., 0., 0.],
                    [0., 0., 0., 0.],
                    [0., 0., 0., 0.],
                    [0., 0., 0., 0.],])

prob_indice = tf.constant([[0 ,1],
                           [0, 3],
                           [1, 1],
                           [1, 2],
                           [1, 3],
                           [5, 0],
                           [5, 1],
                           [5, 2],
                           [5, 3],
                           [6, 1],
                           [6, 3]])
energy_allocation = 0.05
result = tf.scatter_nd_add(mat1,
                           prob_indice,
                           energy_allocation*tf.ones(shape=(prob_indice.shape[0])))

# if your mat1 is tf.Tensor,you can use tf.scatter_nd to achieve it.
# result = tf.scatter_nd(prob_indice,
#                        energy_allocation * tf.ones(shape=(prob_indice.shape[0])),
#                        mat1.shape) + mat1

with tf.Session() as sess:
    sess.run(tf.global_variables_initializer())
    print(sess.run(result))

# [[0.   0.05 0.   0.05]
#  [0.7  0.05 0.05 0.05]
#  [0.   0.   0.   0.  ]
#  [0.   0.   0.   0.  ]
#  [0.   0.   0.   0.  ]
#  [0.05 0.05 0.05 0.05]
#  [0.   0.05 0.   0.05]]

更新：

您可以在张量流tf.tensor_scatter_nd_add()中使用tf.scatter_nd_add()代替version=2。

如何对张量流中特定索引的值求和

1 个答案: