帖子来自[1]

帖主的内容:

With se_resnext50 you can achieve 0.066 on LB. Here is the tips.

  • 5 folds
  • 512x512
  • 3 epochs
  • hflip, crop, brightness, contrast, rotate (here[3] is the details of augmentations)
  • three types of windows, brain, blood and soft tissues.
  • tta (5 times)
  • lb 0.070 to 0.072 for each fold (with tta)
  • cv 0.071 to 0.074 for each fold (without tta)
  • lb 0.066 is achievable by averaging folds

I've also shared the source code on github in case you are interested [2].

The github repo lets you train a basic single model as a baseline which probably scores 0.070 to 0.072 on public LB. This baseline model takes 20 hours to train with a single 1080ti.
If you have any questions please ask. Have fun.

#--------------------------------------------下面是回复---------------------------------------------------------------------

Could you please shed some light on how you preprocess the images? Particularly, when you apply windowing, what do values, that you subtract from image array(0, -20, -150) mean? Are those just image.min() ?
image1 = (image1 - 0) / 80
image2 = (image2 - (-20)) / 200
image3 = (image3 - (-150)) / 380 
Also in your code if policy==1, than you subtract image.min() and divide by (image3.max()-image3.min()) Why if policy==2 you divide by max only? Just trying to grasp the intuition behind this, thank you.

Reply
It does the same thing. After windowing with fixed values you know the theoretical minimum and maximum values and I used these values for min-max normalization.

#-----------------------------------------------------------------------------------------------------------------

hank your sharing. But I have a question on your code, It is about making adjacent labels. Your code is as following.
for j,id in enumerate(group.ID): if j == 0: left = labels[j-1] else: left = '' if j+1 == len(labels): right = '' else: right = labels[j+1] 
The left label won't be " if, only if j == 0. If I modify this code as follows, could it will be better?
for j,id in enumerate(group.ID): if j == 0: left = '' else: left = labels[j-1] if j+1 == len(labels): right = '' else: right = labels[j+1]

Reply
Hi, thank you very much for pointing this out.
I made a simple mistake and your implementation is correct.

LeftLabel and RightLabel are actually not used in the code and does not affect the model. But this feature could be used to improve the traing process by giving some score to its adjacent images and that's why I left them there even though it's not used.

If the target is something like
[1 1 0 1 0 0 1 0 0]

You can spread the score like this
[1 1 0.4 1 0.2 0.2 1 0.2 0]

just an idea.
#----------------------------------------------------------------------------------------------------------------Single FOLD PB 0.074

#----------------------------------------------------------------------------------------------------------------

If you run it without looking at the code, you probably get 0.4, but if you study the code, you may learn a lot.

#----------------------------------------------------------------------------------------------------------------

评论区中提到:

        self.df = apply_dataset_policy(self.df, self.cfg.dataset_policy)
        # self.df = self.df.sample(560)#这个是用来测试整个流程是否能跑通的

#----------------------------------------------------------------------------------------------------------------

代码从复现角度考虑,下面这句话需要设为True,如果不介意复现情况,可以设定为False
torch.backends.cudnn.deterministic = True

Reference:

[1]https://www.kaggle.com/c/rsna-intracranial-hemorrhage-detection/discussion/112819#latest-652212

[2]https://github.com/appian42/kaggle-rsna-intracranial-hemorrhage

[3]https://github.com/appian42/kaggle-rsna-intracranial-hemorrhage/blob/master/conf/model001.py

kaggle颅内出血比赛分析相关推荐

  1. 【干货】Kaggle 数据挖掘比赛经验分享(mark 专业的数据建模过程)

    简介 Kaggle 于 2010 年创立,专注数据科学,机器学习竞赛的举办,是全球最大的数据科学社区和数据竞赛平台.笔者从 2013 年开始,陆续参加了多场 Kaggle上面举办的比赛,相继获得了 C ...

  2. Kaggle 数据挖掘比赛经验分享 (转载)

     [干货]Kaggle 数据挖掘比赛经验分享 (转载) 标签: 数据挖掘数据科学家机器学习kaggle 2017-05-21 19:25 99人阅读 评论(0) 收藏 举报 本文章已收录于: 分类 ...

  3. Kaggle 数据挖掘比赛经验分享(转)

    原作者:陈成龙 简介 Kaggle 于 2010 年创立,专注数据科学,机器学习竞赛的举办,是全球最大的数据科学社区和数据竞赛平台.笔者从 2013 年开始,陆续参加了多场 Kaggle上面举办的比赛 ...

  4. 【干货】Kaggle 数据挖掘比赛经验分享

    [干货]Kaggle 数据挖掘比赛经验分享 Kaggle 于 2010 年创立,专注数据科学,机器学习竞赛的举办,是全球最大的数据科学社区和数据竞赛平台.笔者从 2013 年开始,陆续参加了多场 Ka ...

  5. 自然语言处理NLP星空智能对话机器人系列:NLP on Transformers 101 第16章:Kaggle BERT比赛CommonLit Readability Prize赛题解析

    自然语言处理NLP星空智能对话机器人系列:NLP on Transformers 101 第16章:Kaggle BERT比赛CommonLit Readability Prize赛题解析 第16章: ...

  6. 【干货】Kaggle数据挖掘比赛经验分享,陈成龙博士整理!

    来源:腾讯广告算法大赛(ID:TSA-Contest) 作者简介 陈成龙, 2015 年博士毕业于中山大学,研究图像篡改检测,在图像领域顶级期刊IEEE TIP上发表论文2篇,Kaggle Crowd ...

  7. Kaggle 数据挖掘比赛经验分享

    kaggle历期比赛解决方案汇总 - 简介 Kaggle 于 2010 年创立,专注数据科学,机器学习竞赛的举办,是全球最大的数据科学社区和数据竞赛平台.笔者从 2013 年开始,陆续参加了多场 Ka ...

  8. Kaggle 泰坦尼克号生存分析(数据概览和缺失值处理部分)

    Kaggle 泰坦尼克号生存分析 数据概览 #导入pandas库方便数据读取和预处理,导入os库方便修改工作路径 import os import pandas as pd #读取数据 os.chdi ...

  9. python乒乓球比赛甲乙_用python进行对乒乓球的比赛分析,并且将该程序进行封装...

    2.单打的淘汰赛采用七局四胜制,双打淘汰赛和团体赛采用五局三胜制. 重点: 思维方式:自顶向下即将一个复杂问题分解成几个问题,再细分成一个个具体的小问题,从而来解决复杂问题.自底向上为自顶向下的逆过程 ...

最新文章

  1. 真正的博士是如何参加AAAI, ICML, ICLR等AI顶会的?
  2. 设计模式——装饰者模式
  3. php.ini 配快捷方式,求高手解答!界面保存的网页快捷方式图标默认恢复.
  4. python映射类型-Python中字典映射类型的学习教程
  5. 《七哥说道》第二章:初出茅庐之拜师学艺
  6. 移动app测试的多样性_做移动APP功能测试,这些因素你得都考虑到~
  7. The 2nd AI on Fashion and Textile International Conference 2019
  8. 总结的Server.Mappath的用法
  9. PDMS Pipeline Tool 教程(五):汇料属性设置
  10. 高校后勤管理系统Java项目-基于SSM
  11. Java进阶之路——从初级程序员到架构师,从小工到专家
  12. 卓越产品计划丨神策分析性能优化详解:数据重组织查询优化
  13. 偏心率计算公式matlab,结构计算整体指标(3)——扭转位移比及楼层偏心率
  14. python分解GIF图
  15. Python 金融量化 道路突破策略(唐奇安道路突破策略布林带通道及其市场风险)
  16. easyboot-code-generate 自动生成代码
  17. python循环嵌套打印小星星_python基础:嵌套循环及例子(打印小星星,九九乘法表)...
  18. FFmpeg系列(四)—— mp4音视频流分离
  19. 解决win7(64位)Office(32位)安装64位Access驱动的方法
  20. 2021年电工杯B体详细思路分析

热门文章

  1. Oracle_双机备份_资料
  2. Java 命令行编译项目
  3. 理解向日葵甘特之六——定义数据列
  4. 外部引用CSS中 link与@import的区别
  5. python mock_Python中使用mock.Mock()进行mock测试
  6. artTemplate基本用法
  7. 关于浮点型加减乘除运算不精确的问题
  8. Detectron-MaskRCnn: 用于抠图的FCNN
  9. deeplearning4j – 分布式DL开源项目
  10. Matlab与C++混合编程