TensorFlow:有没有一种方法可以测量模型的FLOPS?
发布于 2021-01-29 17:54:54
在此问题中找到了我能找到的最接近的示例:https
:
//github.com/tensorflow/tensorflow/issues/899
使用此最小的可复制代码:
import tensorflow as tf
import tensorflow.python.framework.ops as ops
g = tf.Graph()
with g.as_default():
A = tf.Variable(tf.random_normal( [25,16] ))
B = tf.Variable(tf.random_normal( [16,9] ))
C = tf.matmul(A,B) # shape=[25,9]
for op in g.get_operations():
flops = ops.get_stats_for_node_def(g, op.node_def, 'flops').value
if flops is not None:
print 'Flops should be ~',2*25*16*9
print '25 x 25 x 9 would be',2*25*25*9 # ignores internal dim, repeats first
print 'TF stats gives',flops
但是,返回的FLOPS始终为“无”。有没有一种方法可以具体测量FLOPS,尤其是PB文件?
关注者
0
被浏览
143
1 个回答
-
有点晚了,但也许将来对某些访客有帮助。对于您的示例,我成功测试了以下代码段:
g = tf.Graph() run_meta = tf.RunMetadata() with g.as_default(): A = tf.Variable(tf.random_normal( [25,16] )) B = tf.Variable(tf.random_normal( [16,9] )) C = tf.matmul(A,B) # shape=[25,9] opts = tf.profiler.ProfileOptionBuilder.float_operation() flops = tf.profiler.profile(g, run_meta=run_meta, cmd='op', options=opts) if flops is not None: print('Flops should be ~',2*25*16*9) print('25 x 25 x 9 would be',2*25*25*9) # ignores internal dim, repeats first print('TF stats gives',flops.total_float_ops)
也可以将分析器与
Keras
以下代码段结合使用:import tensorflow as tf import keras.backend as K from keras.applications.mobilenet import MobileNet run_meta = tf.RunMetadata() with tf.Session(graph=tf.Graph()) as sess: K.set_session(sess) net = MobileNet(alpha=.75, input_tensor=tf.placeholder('float32', shape=(1,32,32,3))) opts = tf.profiler.ProfileOptionBuilder.float_operation() flops = tf.profiler.profile(sess.graph, run_meta=run_meta, cmd='op', options=opts) opts = tf.profiler.ProfileOptionBuilder.trainable_variables_parameter() params = tf.profiler.profile(sess.graph, run_meta=run_meta, cmd='op', options=opts) print("{:,} --- {:,}".format(flops.total_float_ops, params.total_parameters))
希望我能帮上忙!