张量流索引如何工作 [英] how does tensorflow indexing work
问题描述
我无法理解 tensorflow 的基本概念.张量读/写操作的索引如何工作?为了具体说明，如何将以下 numpy 示例转换为 tensorflow(使用张量来分配数组、索引和值):
I'm having trouble understanding a basic concept with tensorflow. How does indexing work for tensor read/write operations? In order to make this specific, how can the following numpy examples be translated to tensorflow (using tensors for the arrays, indices and values being assigned):
x = np.zeros((3, 4))
row_indices = np.array([1, 1, 2])
col_indices = np.array([0, 2, 3])
x[row_indices, col_indices] = 2
x
带输出:
array([[ 0., 0., 0., 0.],
[ 2., 0., 2., 0.],
[ 0., 0., 0., 2.]])
...和...
x[row_indices, col_indices] = np.array([5, 4, 3])
x
带输出:
array([[ 0., 0., 0., 0.],
[ 5., 0., 4., 0.],
[ 0., 0., 0., 3.]])
...最后...
y = x[row_indices, col_indices]
y
带输出:
array([ 5., 4., 3.])
推荐答案
存在 github 问题 #206 要很好地支持这一点，同时您必须求助于冗长的解决方法
There's github issue #206 to support this nicely, meanwhile you have to resort to verbose work-arounds
第一个例子可以用 tf.select 完成，它通过从一个或另一个中选择每个元素来组合两个形状相同的张量
The first example can be done with tf.select that combines two same-shaped tensors by selecting each element from one or the other
tf.reset_default_graph()
row_indices = tf.constant([1, 1, 2])
col_indices = tf.constant([0, 2, 3])
x = tf.zeros((3, 4))
sess = tf.InteractiveSession()
# get list of ((row1, col1), (row2, col2), ..)
coords = tf.transpose(tf.pack([row_indices, col_indices]))
# get tensor with 1's at positions (row1, col1),...
binary_mask = tf.sparse_to_dense(coords, x.get_shape(), 1)
# convert 1/0 to True/False
binary_mask = tf.cast(binary_mask, tf.bool)
twos = 2*tf.ones(x.get_shape())
# make new x out of old values or 2, depending on mask
x = tf.select(binary_mask, twos, x)
print x.eval()
给予
[[ 0. 0. 0. 0.]
[ 2. 0. 2. 0.]
[ 0. 0. 0. 2.]]
第二个可以用 scatter_update 完成，除了 scatter_update 只支持线性索引和变量.所以你可以创建一个临时变量并像这样使用整形.(为了避免变量，你可以使用 dynamic_stitch，见最后)
The second one could be done with scatter_update, except scatter_update only supports on linear indices and works on variables. So you could create a temporary variable and use reshaping like this. (to avoid variables you could use dynamic_stitch, see the end)
# get linear indices
linear_indices = row_indices*x.get_shape()[1]+col_indices
# turn 'x' into 1d variable since "scatter_update" supports linear indexing only
x_flat = tf.Variable(tf.reshape(x, [-1]))
# no automatic promotion, so make updates float32 to match x
updates = tf.constant([5, 4, 3], dtype=tf.float32)
sess.run(tf.initialize_all_variables())
sess.run(tf.scatter_update(x_flat, linear_indices, updates))
# convert back into original shape
x = tf.reshape(x_flat, x.get_shape())
print x.eval()
给予
[[ 0. 0. 0. 0.]
[ 5. 0. 4. 0.]
[ 0. 0. 0. 3.]]
最后第三个例子已经被gather_nd支持了，你写
Finally the third example is already supported with gather_nd, you write
print tf.gather_nd(x, coords).eval()
得到
[ 5. 4. 3.]
编辑，5 月 6 日
通过使用 select 和 sparse_to_dense，更新 采用稀疏值向量，或依赖 x[cols,rows]=newvals 可以在不使用变量(在会话运行调用之间占用内存)的情况下完成dynamic_stitch
The update x[cols,rows]=newvals can be done without using Variables (which occupy memory between session run calls) by using select with sparse_to_dense that takes vector of sparse values, or relying on dynamic_stitch
sess = tf.InteractiveSession()
x = tf.zeros((3, 4))
row_indices = tf.constant([1, 1, 2])
col_indices = tf.constant([0, 2, 3])
# no automatic promotion, so specify float type
replacement_vals = tf.constant([5, 4, 3], dtype=tf.float32)
# convert to linear indexing in row-major form
linear_indices = row_indices*x.get_shape()[1]+col_indices
x_flat = tf.reshape(x, [-1])
# use dynamic stitch, it merges the array by taking value either
# from array1[index1] or array2[index2], if indices conflict,
# the later one is used
unchanged_indices = tf.range(tf.size(x_flat))
changed_indices = linear_indices
x_flat = tf.dynamic_stitch([unchanged_indices, changed_indices],
[x_flat, replacement_vals])
x = tf.reshape(x_flat, x.get_shape())
print x.eval()
这篇关于张量流索引如何工作的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
