cuffdiff problem--Performed 0 isoform-level transcription difference tests
如何解决cuffdiff运行结果中表达量为0的情况?
cuffdiff -o cdiffout -b ref.fasta -u -p 15 --library-type fr-firststrand \
-L H_3,O_3,F_3 cuffmergeout/merged.gtf \
tophatout/H-1-3/accepted_hits.bam,tophatout/H-2-3/accepted_hits.bam,tophatout/H-3-3/accepted_hits.bam \
tophatout/O-1-3/accepted_hits.bam,tophatout/O-2-3/accepted_h its.bam,tophatout/O-3-3/accepted_hits.bam \
tophatout/F-1-3/accepted_hits.bam,tophatout/F-2-3/accepted_hits.bam,tophatout/F-3-3/accepted
_hits.bam
Processed 33185 loci. [***************] 100%
Performed 0 isoform-level transcription difference tests
Performed 0 tss-level transcription difference tests
Performed 0 gene-level transcription difference tests
Performed 0 CDS-level transcription difference tests
Performed 0 splicing tests
Performed 0 promoter preference tests
Performing 0 relative CDS output tests
Writing isoform-level FPKM tracking
Writing TSS group-level FPKM tracking
Writing gene-level FPKM tracking
Writing CDS-level FPKM tracking
Writing isoform-level count tracking
Writing TSS group-level count tracking
Writing gene-level count tracking
Writing CDS-level count tracking
Writing isoform-level read group tracking
Writing TSS group-level read group tracking
Writing gene-level read group tracking
Writing CDS-level read group tracking
Writing read group info
Writing run info
解决办法:先使用 cuffquant 对每个样本估计 gene 和 transcript 的表达量,然后再用 cuffdiff 进行差异表达的分析。
原理是:distribute your computational load over a cluster;save your time.
cuffdiff problem--Performed 0 isoform-level transcription difference tests的更多相关文章
- [Codechef TASTR] Level of Difference - 后缀数组,容斥原理
[Codechef TASTR] Level of Difference Description 给定两个字符串,求恰好在一个字符串中出现过的本质不同的子串数量. Solution 设 \(U(S)\ ...
- [LeetCode&Python] Problem 107. Binary Tree Level Order Traversal II
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
- [LeetCode&Python] Problem 429. N-ary Tree Level Order Traversal
Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, ...
- UVa 11389 - The Bus Driver Problem 难度:0
题目 https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&a ...
- HDU 2101 A + B Problem Too 分类: ACM 2015-06-16 23:57 18人阅读 评论(0) 收藏
A + B Problem Too Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others ...
- Hibernate Validator 6.0.9.Final - JSR 380 Reference Implementation: Reference Guide
Preface Validating data is a common task that occurs throughout all application layers, from the pre ...
- 【译】Android 6.0 Changes (机翻加轻微人工校对)
Android 6.0 Changes In this document Runtime Permissions Doze and App Standby Apache HTTP Client Rem ...
- HDU4971 A simple brute force problem.(强连通分量缩点 + 最大权闭合子图)
题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=4971 Description There's a company with several ...
- This problem will occur when running in 64 bit mode with the 32 bit Oracle client components installed(在64位模式下运行安装了32位的Oracle客户端组件时,会发生此问题)
部署win服务时出现下面的问题: 在事件查看器中看到如下错误: 日志名称: Application来源: ***调度服务日期: 2014/5/21 12:53:21事件 ID: 0任务类别: 无级别: ...
随机推荐
- php的下载
前言:你的坚持,终将美好! 今天给大家说一下,php的下载,由于php的下载本身就是一个压缩包,解压缩后即可使用.所以,解压缩的过程不再赘述. 第一步:在浏览器的地址栏输入:http://www.ph ...
- 页面加载,使用ajax查询某个类别,并且给它们添加(拼接)连接
直接上代码 $(function(){ var a = ''; $.get("productType1_findAll.action",function(data){ consol ...
- Eclipse: Android Device Chooser - Unknown Target
公司最近所有的项目都使用到了Android开发手机(或PDA)应用.所需要的Android开发技术并不是非常复杂,因为我们的底层方法全部使用WebServcie写好了,做Android开发的人员只需要 ...
- 关于微信小程序下拉出现三个小点
包子这天看美团外卖的小程序,再瞅瞅自己的背景色,发现,美团下拉的时候有三个小点,但是我自己的校车徐下拉的时候没有三个小点,很是郁闷,于是各种的找各种的找,发现,这三个小点是微信小程序自带的,你只需要设 ...
- 【python】-- Ajax
Ajax AJAX,Asynchronous JavaScript and XML (异步的JavaScript和XML),一种创建交互式网页应用的网页开发技术方案. 异步的JavaScript:使用 ...
- 原生JavaScript写AJAX
前端JavaScript: function ajaxGet(url, obj) { var request; if(window.XMLHttpRequest) { request = new XM ...
- unity坑faq
遇到的坑记录下来,大都都是听说,没有实测 1. Graphics.copyTexture,在某些机型上不支持从不同类型拷贝 2. msaa 小米mix2不支持,晓龙845 3. android4.2下 ...
- Django报:builtin_function_or_method' object is not iterable
def detail(request,hero_id): hero=models.HeroInfo.objects.get(id=hero_id) return render_to_response( ...
- python并发编程&IO模型
一 IO模型介绍 为了更好地了解IO模型,可先回顾下:同步.异步.阻塞.非阻塞 同步(synchronous) IO和异步(asynchronous) IO,阻塞(blocking) IO和非阻塞(n ...
- 高斯过程(GP)
随机过程基本概念: 随机过程是一个比随机变量更广泛的概念.在概率论中,通常研究一个或多个这样有限个数的随机变量,即使在大数定律和中心极限定理中考虑了无穷多个随机变量,但也要假设随机变量之间互相独立.而 ...