如何从张量流中的两个张量中创建一个频率张量 [英] how to create a frequency tensor out of two tensor in tensorflow
问题描述
我有一个这样的张量，其中值是频率，行是索引(0 to 6):
I have a tensor like this in which the values are the frequency and the rows are the index(0 to 6):
tf_docs =
[[0, 2],
[1, 2],
[2, 1],
[5, 0],
[0, 1],
[7, 8],
[9, 6]]
我有一个常数张量，其中张量的值是索引:
I have a constant tensor, in which values of the tensor are the index:
tf_topics = tf.constant([[1 2]
[1 3]
[1 0]
[2 3]
[2 0]
[3 0]
[3 4]
[3 2]
[3 1]
[4 2]
[4 1]
[2 1]], shape=(12, 2), dtype=int32)
我需要在 tf_docs 中逐行检查这些索引，结果矩阵将是 tf_docs 中它们不为零的列数(在两个索引).
I need to check these indexes row-wise in tf_docs and the result matrix would be the number of columns in the tf_docs in which they are not zero (in both indexes).
例如，我们在tf_topics中有[1 2].这意味着检查 tf_docs 中行索引 1 和 2 中的值.在 tf_docs 中，第一列和第二列的值都不为零.这就是为什么 [1 2] 的频率是 2.
For example, We have [1 2] in the tf_topics. It means check the values in row index 1 and 2 in tf_docs. In tf_docs the first and second column both values are non-zero. thats why for [1 2] the frequency would be 2.
另一方面，[1,3]得到1作为频率.因为索引 3 的第二列中的值之一为零.
On the other hand, [1,3] get 1 as the frequency. Because one of the value in the second column of the index 3 is zero.
所以结果将是这样的张量(这显然是对称的).对角线将是每个 index 的频率总和:
So the result will be a tensor like this(This is obviously symmetrical). The diagonal will be the sum of frequency of each index:
[[2, 1, 1, 0, null],
[1, 3, 2, 1, 1 ],
[1, 2, 3, 1, 1 ],
[0, 1, 1, 5, 0 ],
[null,1, 1, 0, 1 ]]
到目前为止我所做的:
我决定在两个矩阵上使用 tf.gather 和 tf.count_nonzero.因为我想拆分 topics 中的 index 并查看这些 indexes 是否共同出现在 tf_docs
I decided to use tf.gather and tf.count_nonzero over the two matrices. because I wanted to split the index in the topics and see if these indexes co-occurred in tf_docs
tf.math.count_nonzero(tf.gather(tf_docs, tf_topics, axis=0), axis=1)
虽然，这似乎并没有给我想要的结果.
Though, this seems does not give me the result that I want.
推荐答案
让 nonzero_tf_docs 定义为:
zero_tf_docs = tf.cast(tf.equal(tf_docs, tf.zeros_like(tf_docs)), tf.int32)
nonzero_tf_docs = 1 - tf.reduce_max(zero_tf_docs, axis=-1)
OP 要求计算 tf_topicsi, j 的总和 nonzero_tf_docs[i] + nonzero_tf_docs[j]> 并将结果显示在矩阵中.这可以通过以下方式实现:
The OP is asking to compute the sum nonzero_tf_docs[i] + nonzero_tf_docs[j] for each pair of indices i, j in tf_topics and display the result in a matrix. This can be achieved as follows:
def compute_result(tf_topics_, nonzero_tf_docs, tf_docs):
# Find matrix lower part
values = tf.reduce_sum(tf.gather(nonzero_tf_docs, tf_topics_), axis=-1)
max_index = tf.reduce_max(tf_topics) + 1
out_sparse = tf.sparse.SparseTensor(indices=tf_topics_, values=values, dense_shape=[max_index, max_index])
out_sparse = tf.cast(out_sparse, dtype=tf.int32)
out_sparse = tf.sparse.reorder(out_sparse)
out_dense = tf.sparse.to_dense(out_sparse, default_value=-1)
out_lower = tf.matrix_band_part(out_dense, -1, 0)
# Compute diagonal
diag_values = tf.reduce_sum(tf_docs, axis=-1)
diag = tf.slice(diag_values,
begin=[0],
size=[max_index])
# Construct output matrix
out = out_lower + tf.transpose(out_lower)
mask = tf.eye(max_index, dtype=tf.int32)
out = (1 - mask) * out + mask * diag
return out
# Find docs without zeros
zero_tf_docs = tf.cast(tf.equal(tf_docs, tf.zeros_like(tf_docs)), tf.int32)
nonzero_tf_docs = 1 - tf.reduce_max(zero_tf_docs, axis=-1)
# Transform counts into matrix format
tf_topics = tf.cast(tf_topics, dtype=tf.int64)
tf_topics_reversed = tf.reverse(tf_topics, [-1])
tf_topics_ = tf_topics_reversed
out_1 = compute_result(tf_topics_, nonzero_tf_docs, tf_docs)
out_2 = compute_result(tf_topics, nonzero_tf_docs, tf_docs)
out = tf.maximum(out_1, out_2)
with tf.Session() as sess:
r = sess.run(out)
print(r) # prints [[ 2 1 1 0 -1]
# [ 1 3 2 1 1]
# [ 1 2 3 1 1]
# [ 0 1 1 5 0]
# [-1 1 1 0 1]]
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