如何使用tensorflow函数tf.contrib.legacy_seq2seq.sequence_loss_by_example的'weights'参数?
How to use the param of 'weights' of tensorflow function tf.contrib.legacy_seq2seq.sequence_loss_by_example?
代码:
import tensorflow as tf
A = tf.constant([[0.1,0.2,0.3,0.4],[0.2,0.1,0.4,0.3],[0.4,0.3,0.2,0.1],[0.3,0.2,0.1,0.4],[0.1,0.4,0.3,0.2]], dtype=tf.float32)
B = tf.constant([1, 2, 1, 3, 3], dtype=tf.int32)
w_1 = tf.constant(value=[1,1,1,1,1], dtype=tf.float32)
w_2 = tf.constant(value=[1,2,3,4,5], dtype=tf.float32)
D = tf.contrib.legacy_seq2seq.sequence_loss_by_example([A], [B], [w_1])
D_1 = tf.contrib.legacy_seq2seq.sequence_loss_by_example([A], [B], [w_1], average_across_timesteps=False)
D_2 = tf.contrib.legacy_seq2seq.sequence_loss_by_example([A], [B], [w_2])
D_3 = tf.contrib.legacy_seq2seq.sequence_loss_by_example([A], [B], [w_2], average_across_timesteps=False)
with tf.Session() as sess:
print(sess.run(D))
print(sess.run(D_1))
print(sess.run(D_2))
print(sess.run(D_3))
结果是:
[1.4425355 1.2425355 1.3425356 1.2425356 1.4425356]
[1.4425355 1.2425355 1.3425356 1.2425356 1.4425356]
[1.4425355 1.2425355 1.3425356 1.2425356 1.4425356]
[1.4425355 2.485071 4.027607 4.9701424 7.212678 ]
我不明白为什么无论参数average_across_timesteps
设置为'True'还是'False'结果都是一样的。
这是执行平均的源代码:
if average_across_timesteps:
total_size = math_ops.add_n(weights)
total_size += 1e-12 # Just to avoid division by 0 for all-0 weights.
log_perps /= total_size
在你的例子中,weights
是 一个 张量的列表,w_1
或 w_2
,即你有一个时间步长。在这两种情况下,tf.add_n(weights)
都不会改变它,因为它是一个元素的总和(不是 w_1
或 w_2
中元素的总和)。
这解释了结果:D
和 D_1
被评估为相同的数组,因为 D_1 = D * w_1
(逐元素)。 D_2
和 D_3
不同,因为 w_2
不仅包含一个。
代码:
import tensorflow as tf
A = tf.constant([[0.1,0.2,0.3,0.4],[0.2,0.1,0.4,0.3],[0.4,0.3,0.2,0.1],[0.3,0.2,0.1,0.4],[0.1,0.4,0.3,0.2]], dtype=tf.float32)
B = tf.constant([1, 2, 1, 3, 3], dtype=tf.int32)
w_1 = tf.constant(value=[1,1,1,1,1], dtype=tf.float32)
w_2 = tf.constant(value=[1,2,3,4,5], dtype=tf.float32)
D = tf.contrib.legacy_seq2seq.sequence_loss_by_example([A], [B], [w_1])
D_1 = tf.contrib.legacy_seq2seq.sequence_loss_by_example([A], [B], [w_1], average_across_timesteps=False)
D_2 = tf.contrib.legacy_seq2seq.sequence_loss_by_example([A], [B], [w_2])
D_3 = tf.contrib.legacy_seq2seq.sequence_loss_by_example([A], [B], [w_2], average_across_timesteps=False)
with tf.Session() as sess:
print(sess.run(D))
print(sess.run(D_1))
print(sess.run(D_2))
print(sess.run(D_3))
结果是:
[1.4425355 1.2425355 1.3425356 1.2425356 1.4425356]
[1.4425355 1.2425355 1.3425356 1.2425356 1.4425356]
[1.4425355 1.2425355 1.3425356 1.2425356 1.4425356]
[1.4425355 2.485071 4.027607 4.9701424 7.212678 ]
我不明白为什么无论参数average_across_timesteps
设置为'True'还是'False'结果都是一样的。
这是执行平均的源代码:
if average_across_timesteps:
total_size = math_ops.add_n(weights)
total_size += 1e-12 # Just to avoid division by 0 for all-0 weights.
log_perps /= total_size
在你的例子中,weights
是 一个 张量的列表,w_1
或 w_2
,即你有一个时间步长。在这两种情况下,tf.add_n(weights)
都不会改变它,因为它是一个元素的总和(不是 w_1
或 w_2
中元素的总和)。
这解释了结果:D
和 D_1
被评估为相同的数组,因为 D_1 = D * w_1
(逐元素)。 D_2
和 D_3
不同,因为 w_2
不仅包含一个。