shuffle | 易学教程

Randomize Columns

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问题 Is there a way to randomize the values from different columns within a row? Here is an example: Option 1 Option 2 Option 3 Option 4 Gloria Stuart Claire Danes Kim Basinger Kate Winslet Carson Daly Chris Rock Matthew Perry David Arquette Mohawk Bald Mullet Buzz Cut Big Daddy Little Nicky The Waterboy Happy Gilmore Virginia Italy England Germany There are 4 columns. Currently all of the inputs under Option 4 are the correct answer to a question. I want to randomize or shuffle them within their

Better way to shuffle elements of a string in R

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问题 I have to shuffle elements of a string. I wrote a code: sequ <- "GCTTCG" set.seed(2017) i <- sample(1:nchar(sequ)) separate.seq.letters <- unlist(strsplit(sequ, "")) paste(separate.seq.letters[i], collapse = "") [1] "GTCGTC" This code shuffles elements one time. The main question would be is there a better (more effective) way to do that? For very long sequences and huge amount of shuffles strsplit , paste commands takes some extra time. 回答1: Making use of the Rcpp package to handle in C is

Spark 数据倾斜及其解决方案

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【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 本文首发于 vivo互联网技术微信公众号 https://mp.weixin.qq.com/s/lqMu6lfk-Ny1ZHYruEeBdA 作者简介：郑志彬，毕业于华南理工大学计算机科学与技术（双语班）。先后从事过电子商务、开放平台、移动浏览器、推荐广告和大数据、人工智能等相关开发和架构。目前在vivo智能平台中心从事 AI中台建设以及广告推荐业务。擅长各种业务形态的业务架构、平台化以及各种业务解决方案。本文从数据倾斜的危害、现象、原因等方面，由浅入深阐述Spark数据倾斜及其解决方案。一、什么是数据倾斜对 Spark/Hadoop 这样的分布式大数据系统来讲，数据量大并不可怕，可怕的是数据倾斜。对于分布式系统而言，理想情况下，随着系统规模（节点数量）的增加，应用整体耗时线性下降。如果一台机器处理一批大量数据需要120分钟，当机器数量增加到3台时，理想的耗时为120 / 3 = 40分钟。但是，想做到分布式情况下每台机器执行时间是单机时的1 / N，就必须保证每台机器的任务量相等。不幸的是，很多时候，任务的分配是不均匀的，甚至不均匀到大部分任务被分配到个别机器上，其它大部分机器所分配的任务量只占总得的小部分。比如一台机器负责处理 80% 的任务，另外两台机器各处理 10% 的任务。『不患多而患不均

Randomly shuffle a sparse matrix in python

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问题 is there an easy way to shuffle a sparse matrix in python? This is how I shuffle a non-sparse matrix: index = np.arange(np.shape(matrix)[0]) np.random.shuffle(index) return matrix[index] How can I do it with numpy sparse? 回答1: Ok, found it. The sparse format looks a bit confusing in the print-out. index = np.arange(np.shape(matrix)[0]) print index np.random.shuffle(index) return matrix[index, :] 回答2: In case anyone is looking to randomly get a subsample of rows from a sparse matrix, this

Is there a way to rewrite Spark RDD distinct to use mapPartitions instead of distinct?

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问题 I have an RDD that is too large to consistently perform a distinct statement without spurious errors (e.g. SparkException stage failed 4 times, ExecutorLostFailure, HDFS Filesystem closed, Max number of executor failures reached, Stage cancelled because SparkContext was shut down, etc.) I am trying to count distinct IDs in a particular column, for example: print(myRDD.map(a => a._2._1._2).distinct.count()) is there an easy, consistent, less-shuffle-intensive way to do the command above,

numpy random shuffle by row independently

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问题 I have the following array: a= array([[ 1, 2, 3], [ 1, 2, 3], [ 1, 2, 3]) I understand that np.random,shuffle(a.T) will shuffle the array along the row, but what I need is it to shuffe each row idependently. How can this be done in numpy? Speed is critical as there will be several million rows. For this specific problem, each row will contain the same starting population. 回答1: import numpy as np np.random.seed(2018) def scramble(a, axis=-1): """ Return an array with the values of `a`

Controlling distance of shuffling

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问题 I have tried to ask this question before, but have never been able to word it correctly. I hope I have it right this time: I have a list of unique elements. I want to shuffle this list to produce a new list. However, I would like to constrain the shuffle, such that each element's new position is at most d away from its original position in the list. So for example: L = [1,2,3,4] d = 2 answer = magicFunction(L, d) Now, one possible outcome could be: >>> print(answer) [3,1,2,4] Notice that 3

why does this simple shuffle algorithm produce biased results? what is a simple reason?

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问题 it seems that this simple shuffle algorithm will produce biased results: # suppose $arr is filled with 1 to 52 for ($i < 0; $i < 52; $i++) { $j = rand(0, 51); # swap the items $tmp = $arr[j]; $arr[j] = $arr[i]; $arr[i] = $tmp; } you can try it... instead of using 52, use 3 (suppose only 3 cards are used), and run it 10,000 times and tally up the results, you will see that the results are skewed towards certain patterns... the question is... what is a simple explanation that it will happen?

交叉验证

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一、sklearn.cross_validation.cross_val_score sklearn.cross_validation.cross_val_score(estimator, X, y=None, scoring=None,cv=None, n_jobs=1, verbose=0, fit_params=None, pre_dispatch=‘2*n_jobs’) estimator:估计方法对象(分类器) X：数据特征(Features) y：数据标签(Labels) soring：调用方法(包括accuracy和mean_squared_error等等) cv：几折交叉验证 n_jobs：同时工作的cpu个数（-1代表全部）二、sklearn.model_selection.KFold sklearn.model_selection.KFold(n_splits=3, shuffle=False, random_state=None) n_splits：表示划分几等份 shuffle：在每次划分时，是否进行洗牌 random_state：随机种子数属性： ① get_n_splits(X=None, y=None, groups=None)：获取参数n_splits的值 ② split(X, y=None, groups=None)

How do I make shuffle playlist button and repeat button in android studio

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问题 I dont know how to convert this code to android studio i am stuck on it for 2 days and cant figure it out Plz help me btnRepeat.setOnClickListener(new View.OnClickListener() { @Override public void onClick(View arg0) { if(isRepeat){ isRepeat = false; Toast.makeText(getApplicationContext(), "Repeat is OFF", Toast.LENGTH_SHORT).show(); btnRepeat.setImageResource(R.drawable.btn_repeat); }else{ // make repeat to true isRepeat = true; Toast.makeText(getApplicationContext(), "Repeat is ON", Toast