如何绘制caffe网络训练曲线

如何绘制caffe网络训练曲线

当我们设计好网络结构后，在神经网络训练的过程中，迭代输出的log信息中，一般包括，迭代次数，训练损失代价，测试损失代价，测试精度等。本文提供一段示例，简单讲述如何绘制训练曲线（training curve）。

首先看一段训练的log输出，网络结构参数的那段忽略，直接跳到训练迭代阶段：
I0627 21:30:06.004370 15558 solver.cpp:242] Iteration 0, loss = 21.6953
I0627 21:30:06.004420 15558 solver.cpp:258] Train net output #0: loss = 21.6953 (* 1 = 21.6953 loss)
I0627 21:30:06.004426 15558 solver.cpp:571] Iteration 0, lr = 0.01
I0627 21:30:28.592690 15558 solver.cpp:242] Iteration 100, loss = 13.6593
I0627 21:30:28.592730 15558 solver.cpp:258] Train net output #0: loss = 13.6593 (* 1 = 13.6593 loss)
I0627 21:30:28.592733 15558 solver.cpp:571] Iteration 100, lr = 0.01

...

I0627 21:37:47.926597 15558 solver.cpp:346] Iteration 2000, Testing net (#0)
I0627 21:37:48.588079 15558 blocking_queue.cpp:50] Data layer prefetch queue empty
I0627 21:40:40.575474 15558 solver.cpp:414] Test net output #0: loss = 13.07728 (* 1 = 13.07728 loss)
I0627 21:40:40.575477 15558 solver.cpp:414] Test net output #1: loss/top-1 = 0.00226
I0627 21:40:40.575487 15558 solver.cpp:414] Test net output #2: loss/top-5 = 0.01204
I0627 21:40:40.708261 15558 solver.cpp:242] Iteration 2000, loss = 13.1739
I0627 21:40:40.708302 15558 solver.cpp:258] Train net output #0: loss = 13.1739 (* 1 = 13.1739 loss)
I0627 21:40:40.708307 15558 solver.cpp:571] Iteration 2000, lr = 0.01

...

I0628 01:28:47.426129 15558 solver.cpp:242] Iteration 49900, loss = 0.960628
I0628 01:28:47.426177 15558 solver.cpp:258] Train net output #0: loss = 0.960628 (* 1 = 0.960628 loss)
I0628 01:28:47.426182 15558 solver.cpp:571] Iteration 49900, lr = 0.01
I0628 01:29:10.084050 15558 solver.cpp:449] Snapshotting to binary proto file train_net/net_iter_50000.caffemodel
I0628 01:29:10.563587 15558 solver.cpp:734] Snapshotting solver state to binary proto filetrain_net/net_iter_50000.solverstate
I0628 01:29:10.692239 15558 solver.cpp:346] Iteration 50000, Testing net (#0)
I0628 01:29:13.192075 15558 blocking_queue.cpp:50] Data layer prefetch queue empty
I0628 01:31:00.595120 15558 solver.cpp:414] Test net output #0: loss = 0.6404232 (* 1 = 0.6404232 loss)
I0628 01:31:00.595124 15558 solver.cpp:414] Test net output #1: loss/top-1 = 0.953861
I0628 01:31:00.595127 15558 solver.cpp:414] Test net output #2: loss/top-5 = 0.998659
I0628 01:31:00.727577 15558 solver.cpp:242] Iteration 50000, loss = 0.680951
I0628 01:31:00.727618 15558 solver.cpp:258] Train net output #0: loss = 0.680951 (* 1 = 0.680951 loss)
I0628 01:31:00.727623 15558 solver.cpp:571] Iteration 50000, lr = 0.0096

这是一个普通的网络训练输出，含有1个loss，可以看出solver.prototxt的部分参数为：
test_interval: 2000
base_lr: 0.01
lr_policy: "step" # or "multistep"
gamma: 0.96
display: 100
stepsize: 50000 # if is "multistep", the first stepvalue is set as 50000
snapshot_prefix: "train_net/net"

当然，上面的分析，即便不理会，对下面的代码也没什么影响，绘制训练曲线本质就是文件操作，从上面的log文件中，我们可以看出：

对于每个出现字段] Iteration和loss =的文本行，含有训练的迭代次数以及损失代价；
对于每个含有字段] Iteration和Testing net (#0)的文本行，含有测试的对应的训练迭代次数；
对于每个含有字段#2:和loss/top-5的文本行，含有测试top-5的精度。

根据这些分析，就可以对文本进行如下处理：
import os
import sys
import numpy as np
import matplotlib.pyplot as plt
import math
import re
import pylab
from pylab import figure, show, legend
from mpl_toolkits.axes_grid1 import host_subplot

# read the log file
fp = open('log.txt', 'r')

train_iterations = []
train_loss = []
test_iterations = []
test_accuracy = []

for ln in fp:
# get train_iterations and train_loss
if '] Iteration ' in ln and 'loss = ' in ln:
arr = re.findall(r'ion \b\d+\b,',ln)
train_iterations.append(int(arr[0].strip(',')[4:]))
train_loss.append(float(ln.strip().split(' = ')[-1]))