cat

I am trying to use cat() as functions inside apply(). I can almost make R do what I want, but I'm getting some very confusing (to me) NULLS at the end of the return. Here is a silly example, to highlight what I'm getting. val1 <- 1:10 val2 <- 25:34 values <- data.frame(val1, val2) apply(values, 1, function(x) cat(x[1], x[2], fill=TRUE)) This "works" in that R accepts it and it runs, but I don't understand the results. > apply(values, 1, function(x) cat(x[1], x[2], fill=TRUE)) 1 25 2 26 3 27 4 28 5 29 6 30 7 31 8 32 9 33 10 34 NULL But, I want to get: > apply(values, 1, function(x) cat(x[1], x

I have several logs containing lines all starting with a timestamp, so that the following works as expected to merge them: cat myLog1.txt myLog2.txt | sort -n > combined.txt Problem is, that myLog2.txt can also contain lines without a timestamp (e.g. java stack traces). Is there an easy way without any custom scripts to still merge them and preserve the multiline content? Example myLog1.txt 11:48:18.825 [main] INFO org.hibernate.cfg.Environment - HHH000206: hibernate.properties not found 11:48:55.784 [main] INFO o.h.tool.hbm2ddl.SchemaUpdate - HHH000396: Updating schema Example myLog2.txt 11

Linux 命令笔记 一、目录/文件 1.1 目录文件日常操作 . -> 当前目录 .. -> 上一级目录 .file/.dir -> 隐藏文件/文件夹 【ls】 查看指定目录文件 ls 命令是最常用的linux命令，要配合着选项使用。 常用命令 ls -lh # 方便查看文件大小 ls -AsSh 【pwd】:查看当前目录 pwd等价于/bin/pwd -L，显示当前绝对路径，如果是链接，则显示链接路径 pwd -P等价于/bin/pwd -P，显示实际路径，而非链接路径 如果文件夹被临时删除，pwd还是显示未删除文件夹的那个路径，这时候可以使用/bin/pwd，就会提示当前路径不存在 【cd】：切换目录 cd - # 上一目录 cd ~ # 家目录 cd # 家目录 cd .. # 上一目录 cd !$ # 将上命令的参数做为cd 参数 【touch】：新建文件 1. 创建新文件同时可以指定一些时间参数 $ touch newfile 2. 一次性创建多个文件 $ touch {1..10}.txt 可以对已有文件修改时间戳（ll显示的时间） 【1】 touch -d [[CC]YY]MMDD text touch -t [CC[YY]MMDDhhmm[.SS] text CC为年数中的前两位，即”世纪数”；YY为年数的后两位，即某世纪中的年数．如果不给出CC的值 touch

I have around 30 K files. I want to merge them into one. I used CAT but I am getting this error. cat *.n3 > merged.n3 -bash: /usr/bin/xargs: Argument list too long How to increase the limit of using the "cat" command? Please help me if there is any iterative method to merge a large number of files. Here's a safe way to do it, without the need for find : printf '%s\0' *.n3 | xargs -0 cat > merged.txt (I've also chosen merged.txt as the output file, as @MichaelDautermann soundly advises; rename to merged.n3 afterward). Note: The reason this works is: printf is a bash shell builtin , whose

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I want to connect some object's signals derived from an interface class. The connection is done in QWidget::listenToAnimal(AnimalInterface*) . This does not work because qt_metacall is not a member of 'AnimalInterface' and static assertion failed: No Q_OBJECT in the class with the signal . Of course AnimalInterface does not have the Q_OBJECT macro and does not inherit QObject because it is an interface... I want to connect through the interface class because I do not want to manually retype the same code for Cat and for Dog . Is it possible

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: How can I suppress this compiler warning: Meta method 'prefix' in category from '...soap+prefix.o' conflicts with same method from another category ? here is the category soap+Prefix.h: @interface Soap (Prefix) +(NSString*)prefix; @end and soap+prefix.m: #import "Soap.h" #import "Soap+Prefix.h" @implementation Soap (Prefix) +(NSString*)prefix { return @"EInspector"; } @end and these two files by the way are automatically generated with SudZc wrapper for web services. p.s. this warning is issued ONLY in XCode 4.4 thank you so much in advance.

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Possible Duplicate: How to use std::sort with a vector of structures and compare function? I have a cat object (what?) and a catSort object which obviously sorts the cat objects. Below is the classes class cat { public: int age; }; class catSorter { public: vector< cat > cats; vector< cat > SortCatsByAge(); void AddCat( cat new_cat ); }; void catSorter::AddCat(cat new_cat){ this->cats.push_back(new_cat) } vector< cat > catSorter::SortCatsByAge(){ // Sort cats here by age! } cat tim; tim.age = 10; cat mark; mark.age = 20 cat phil; phil.age =

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have a pyspark.sql.dataframe where each row is a news article. I then have a RDD that represents the words contained in each article. I want to add the RDD of words as a column named 'words' to my dataframe of new articles. I tried df.withColumn('words', words_rdd ) but I get the error AssertionError: col should be Column The DataFrame looks something like this Articles the cat and dog ran we went to the park today it will rain but I have 3k news articles. I applied a function to clean the text such as remove stop words and I have a RDD

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have a question in Jekyll Liquid. I have layout, where I want to show pages from category. To show category I use page.categories variable. When I show in bracket {{page.categories}} is correct. but I don't know, how to pass to loop? {% for post in site.categories[page.categories] %} <li><a href="{{ post.url }}">{{ post.title }}</a></li> {% endfor %} {% for post in site.categories[{{page.categories}}] %} <li><a href="{{ post.url }}">{{ post.title }}</a></li> {% endfor %} Don't work. If I passed explicite: {% for post in site.categories[