小编pro*_*ian的帖子

我Put (>>)用来存储我在Mathematica中获得的信息.问题是我提出了几个变量.但是,如果我执行以下操作,它会输出到一个名为year而不是我的变量的文件.

例如:

year=64;
sortedTally>>year;

这将导出到名为file的文件year而不是名为的文件64.文档说明expr >> filename等同于expr >> "filename".有没有办法绕过这个并放入一个基于变量更改的文件名？在操作员输入表单(底部)的文档文件中也类似地重申了这一点.

我正在使用各种单词列表进行大量工作.

请考虑我的以下问题:

docText={"settlement", "new", "beginnings", "wildwood", "settlement", "book",
"excerpt", "agnes", "leffler", "perry", "my", "mother", "junetta", 
"hally", "leffler", "brought", "my", "brother", "frank", "and", "me", 
"to", "edmonton", "from", "monmouth", "illinois", "mrs", "matilda", 
"groff", "accompanied", "us", "her", "husband", "joseph", "groff", 
"my", "father", "george", "leffler", "and", "my", "uncle", "andrew", 
"henderson", "were", "already", "in", "edmonton", "they", "came", 
"in", "1910", "we", "arrived", "july", "1", "1911", "the", "sun", 
"was", "shining", "when", "we", "arrived", "however", "it", "had", 
"been", "raining", "for", "days", "and", "it", "was", "very", 
"muddy", "especially", "around", …

我目前正在查看大型文本信息数据库中的单词和短语频率(大约108MB分布在307个文本文件中).我的目标是有一种方法可以快速查看哪些文件最相关并且具有视觉吸引力的格式(尽管这个项目可能也会证明文本表示总是更清晰).

现在我有以下内容:

SetDirectory["/MYMATHEMATICADIRECTORY/"];
filelist = FileNames[];
viewerCount1 = {0};
viewerCount2 = {0};
word1 = "freedom";
word2 = "liberty";
Do[
  searchDB = StringSplit[Import[filename]];
  AppendTo[viewerCount1, Count[searchDB, word1]];
  AppendTo[viewerCount2, Count[searchDB, word2]];
, {filename, filelist}]

list3 = Take[viewerCount1, {2, -1}]
list4 = Take[viewerCount2, {2, -1}]

文件名[]产生一个列表,诸如:{ "001ABbenevolat.txt-cleaned.txt", "002abnature.txt-cleaned.txt", "003aboriginaldocs.txt-cleaned.txt", "004ABpresse.txt-cleaned.txt" ,"005acadian.txt-packaged.txt","006acadiedelile.txt-cleared.txt","007acfa.txt-cleared.txt"} [除了307条目,全部编号].

list3生成一个列表,如:{0,0,10,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 2,0,0,0,2,0,0,0,1,0,0,0,0,0,0,0,1,0,2,0,0,0,10,1,7, 0,0,0,0,23,3,0,0,0,0,0,0,0,0,0,0,0,9,0,1,0,1,0,5, 0,13,0,0,0,0,0,0,0,0,0,1,0,2,0,4,0,0,0,1,11,0,0,0, 0,2,7,1,4,1,0,0,0,0,0,0,0,13,...}等等.

命令:

BarChart3D[{list3, list4}, BarSpacing -> {0.5, 0}, ChartLayout -> "Grid"]

生成接近我想要的东西(将它们想象为文件夹粘贴).但是,我想添加有意义的工具提示.默认情况下,它会出现频率.是否还有一种快速方法可以包括频率附加的文件名以及频率？即一个工具提示,它会调出'007acfa.txt-cleared.txt - 32',其中32个出现在文件7中？

请考虑以下问题.使用此文本文件,我正在尝试设置它,因此它是一个制表符分隔文件.它应该像这样,就像在vim中一样:

请注意,有1,后面是一个选项卡(^ I,上面以浅蓝色显示),然后是网站的文本.

但是当我在TextEdit中打开相同的文件时,我得到了这个:

虚假的断线来自何处？我无法在vim中找到它.

请考虑以下问题.

我正在编写一个快速Manipulate[]程序来显示大量信息,但是我遇到了unicode的问题.这是我目前的输入和输出:

Manipulate[
 request = filenumber <> "*";
 filenames = FileNames[request];
 display = Import[type, "List"];
 Short[display, 25]
 , {filenumber, "001", InputField}, {type, filenames, PopupMenu}]

问题是法语口音出现奇怪.我想到的快速解决方法是将我的代码更改为Import[type,"Plaintext"];有效,但随后以列表形式显示信息,如下所示:

您会建议使用前者的简单列表格式来获得第二个示例的清晰度吗？所以它包装在线上而不是在每次进入后都有换行符.

暂时 - 可能与实际问题本身一样重要 - 任何人都可以解释为什么导入作为"列表"扭曲unicode背后的理由？我在解决这个问题上遇到了很多麻烦,理解潜在的行为可能会帮助我更快地前进.

我有一个大型的文本文件目录,每个目录都很复杂:

说file1.txt:

依靠相信而不是倾听.足够周围去除巴顿同意后悔或它.可以估计的优点是命令提供.今年拍得很好,现在已经过了.挫败站起来嫁给他了.做相关的先生帐户brandon up.从来没有准备好的火腿错了这些诙谐的他.我们的指南针看到年龄不文明的天气禁止她的分钟.准备如何,但真正的儿子新.

他在有利的情况下对比增长,他认为是惊人的.仿佛被拍成一团.通过它足以让山谷渴望做到.主要的伟大的女仆这些火腿匹配她.Abode尝试做女佣.令人怀疑的处置恢复为dashwood而欢欣鼓舞.

和file2.txt:

在做庄园的人中间.是否庆祝它同情考虑.五月欣喜若狂,确实让这个无知的年龄感到惊喜.她最后一次拥有她的小姐.如果他的处理时间过长,那就太多了.怎么,但是儿子们的女士们.她特别令人不愉快的改变继续毫无保留的解决方案.因此希望中国充满希望.它是否认为楼梯分支可以承受三十分钟.

盲目相等,而先生风格.Lain领导,事实没有.一个喜欢解决幸福的运动员继续.高声丰富的真实.哦传达他的直接尖锐.同样欢迎她设置没有任何重力是否派对.肥沃的假设害羞指向坚持尊重.

我需要做的是创建一个新文件,比如allfiles.txt:

Am no an listening depending up believing. Enough around remove to barton agreed regret in or it. Advantage mr estimable be commanded provision. Year well shot deny shew come now had. Shall downs stand marry taken his for out. Do related mr account brandon an up. Wrong for never ready ham these witty him. Our compass see age uncivil matters weather forbade her minutes. Ready how but truth son new under. …