使用大于2GB的数组初始化tensorflow变量

Question

我正在尝试Variable使用预先训练的word2vec嵌入来初始化张量流.

我有以下代码:

import tensorflow as tf
from gensim import models

model = models.Word2Vec.load_word2vec_format('GoogleNews-vectors-negative300.bin', binary=True)
X = model.syn0

embeddings = tf.Variable(tf.random_uniform(X.shape, minval=-0.1, maxval=0.1), trainable=False)

sess.run(tf.initialize_all_variables())

sess.run(embeddings.assign(X))

我收到以下错误:

ValueError: Cannot create an Operation with a NodeDef larger than 2GB.

X我试图分配的数组()是形状(3000000, 300),其大小为3.6GB.

如果我也尝试tf.convert_to_tensor(X),我会得到同样的错误.

我知道它由于阵列大于2GB而失败.但是,我不知道如何将大于2GB的数组分配给张量流Variable

Answer 1

似乎唯一的选择是使用占位符.我能找到的最简洁的方法是直接初始化为占位符:

X_init = tf.placeholder(tf.float32, shape=(3000000, 300))
X = tf.Variable(X_init)
# The rest of the setup...
sess.run(tf.initialize_all_variables(), feed_dict={X_init: model.syn0})

Answer 2

最简单的解决方案是将feed_dict'转换为占位符节点,用于tf.assign到变量.

X = tf.Variable([0.0])
place = tf.placeholder(tf.float32, shape=(3000000, 300))
set_x = X.assign(place)
# set up your session here....
sess.run(set_x, feed_dict={place: model.syn0})

正如Joshua Little在单独的回答中所述,您也可以在初始化程序中使用它:

X = tf.Variable(place)    # place as defined above
...
init = tf.initialize_all_variables()
... create sess ...
sess.run(init, feed_dict={place: model.syn0})