小编vra*_*js5的帖子

计算矩阵的特征值有多贵？

最佳算法的复杂性是什么？

如果我有一个1000 x 1000的矩阵,在实践中需要多长时间？我认为如果矩阵稀疏会有帮助吗？

是否存在特征值计算不会终止的情况？

在R,我可以计算特征值,如下面的玩具示例所示:

m<-matrix( c(13,2, 5,4), ncol=2, nrow=2 )
eigen(m, only.values=1)
$values
[1] 14  3

有谁知道它使用什么算法？

是否还有其他(开源)软件包可以计算特征值？

在尝试解析xml文件时,我总是遇到此错误.

一个简单的脚本来重现它:

doc <- xmlRoot(xmlTreeParse("http://www.stat.purdue.edu/~mdw/490M/cdcatalog.xml"))

xpathSApply(doc, "//CATALOG/CD/PRICE", xmlValue)

Error in (function (classes, fdef, mtable)  : 
 unable to find an inherited method for function ‘saveXML’ for signature ‘"character"’

加载以下包:

sessionInfo()
 R version 3.1.0 (2014-04-10)
 Platform: x86_64-apple-darwin13.1.0 (64-bit)

 locale:
 [1] de_DE.UTF-8/de_DE.UTF-8/de_DE.UTF-8/C/de_DE.UTF-8/de_DE.UTF-8

 attached base packages:
 [1] stats     graphics  grDevices utils     datasets  methods   base     

 other attached packages:
 [1] XML_3.98-1.1

 loaded via a namespace (and not attached):
 [1] tools_3.1.0

我试图找出,如果在R中有比gsub矢量化函数更快的方法.我有一些"句子"(发送$ words)后面的数据框然后我有从这些句子中删除的单词(存储在wordsForRemoving变量中) ).

sent <- data.frame(words = 
                     c("just right size and i love this notebook", "benefits great laptop",
                       "wouldnt bad notebook", "very good quality", "bad orgtop but great",
                       "great improvement for that bad product but overall is not good", 
                       "notebook is not good but i love batterytop"), 
                   user = c(1,2,3,4,5,6,7),
                   stringsAsFactors=F)

wordsForRemoving <- c("great","improvement","love","great improvement","very good","good",
                      "right", "very","benefits", "extra","benefit","top","extraordinarily",
                      "extraordinary", "super","benefits super","good","benefits great",
                      "wouldnt bad")

然后我将为时间消耗计算创建"大数据"模拟......

df.expanded <- as.data.frame(replicate(1000000,sent$words))
library(zoo)
sent <- coredata(sent)[rep(seq(nrow(sent)),1000000),]
rownames(sent) <- NULL

使用以下gsub …

我正在尝试使用此处的说明从命令行创建OAuth令牌.我可以curl从命令行使用,并获得正确的响应

curl -u 'username:pwd' -d '{"scopes":["user", "gist"]}' \
  https://api.github.com/authorizations

现在,我想在R中使用RCurl或复制相同的内容httr.这是我尝试过的,但两个命令都返回错误.谁能指出我在这里做错了什么？

httr::POST(
  'https://api.github.com/authorizations',
  authenticate('username', 'pwd'),
  body = list(scopes = list("user", "gist"))
)

RCurl::postForm(
  uri = 'https://api.github.com/authorizations',
  .opts = list(
    postFields = '{"scopes": ["user", "gist"]}',
    userpwd = 'username:pwd'
  )
)

我不得不在python中进行一些内省,它并不漂亮:

name = sys._getframe(1).f_code
name = "%s:%d %s()" %(os.path.split(name.co_filename)[1],name.co_firstlineno,name.co_name)

得到类似的东西

foo.py:22 bar() blah blah

在我们的调试输出中.

理想情况下,我希望通过这种信息向stderr添加任何内容 - 是否有可能在python中全局更改print的行为？

我们为我们的应用程序创建了WIX安装程序.我们面临的问题是:我们已经定义了两个不同的自定义操作(比如ActionForInstall和ActionForUninstall),我们希望在以下情况下执行:ActionForInstall:应该在安装,产品升级,维护模式(用于修复和修改)时运行ActionForUninstall:应该运行仅用于卸载.

但我们无法设定适当的条件.你可以参考代码:

<Custom Action=ActionForInstall After='InstallFinalize' > 
    (NOT Installed) OR (Installed AND ((MaintenanceMode = "Modify") OR (MaintenanceMode = "Repair")) AND (NOT (MaintenanceMode = "Remove"))) OR  ((UPGRADINGPRODUCTCODE) AND NOT(REMOVE ~= "ALL"))
</Custom>
<Custom Action=ActionForUninstall Before='InstallFinalize'>
    Installed AND NOT UPGRADINGPRODUCTCODE
</Custom>

请告诉我们我们做错了什么.上面的代码即使是卸载也调用InstallFinalize.

我知道使用以下代码我可以创建闪亮的普通下拉菜单,

selectInput("Input1", "Choose you Input:", choices = c('a1'='1','b2'='2'))

这将创建以下下拉列表

但是我正在使用conditionalPanel,并且我正在填充类似于内联的dropdown菜单

我正在使用以下代码生成这些菜单.

conditionalPanel(condition="input.conditionedPanels==3",
                     div(style="display:inline-block",
                         tags$label('Menu1', `for` = 'Sample'), 
                         tags$select(id = 'Sample', class="input-small")),
                     div(style="display:inline-block",
                         tags$label('Menu2', `for` = 'Sample1'), 
                         tags$select(id = 'Sample1', class="input-small")))

我的问题是我无法在这些下拉菜单中添加项目.我尝试过值或选项,但这并没有改变任何东西.

我希望我提供了足够的信息,如果需要更多信息,请告诉我.

我会从R创建一个tar.gz文件但是tar函数对我不起作用.这是我的示例代码.

writeLines('aaaa', 'tmp.txt')
tar('tmp.tar.gz', 'tmp.txt', compression = 'gzip')

此代码创建tmp.tar.gz,但tmp.txt不包含在gz文件中.

我的代码中有错误吗？

谢谢你的任何建议.

我在windows和linux中运行此代码:

#Windows platform
sessionInfo()
R version 3.1.0 (2014-04-10)
Platform: x86_64-w64-mingw32/x64 (64-bit)

locale:
[1] LC_COLLATE=English_Australia.1252  LC_CTYPE=English_Australia.1252   
[3] LC_MONETARY=English_Australia.1252 LC_NUMERIC=C                      
[5] LC_TIME=English_Australia.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

loaded via a namespace (and not attached):
[1] tools_3.1.0

#Linux platform
sessionInfo()
R version 3.1.0 (2014-04-10)
Platform: x86_64-pc-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C         LC_TIME=C            LC_COLLATE=C        
 [5] LC_MONETARY=C        LC_MESSAGES=C        LC_PAPER=C           LC_NAME=C           
 [9] LC_ADDRESS=C         LC_TELEPHONE=C       LC_MEASUREMENT=C     LC_IDENTIFICATION=C 

attached base packages: …

我正在尝试准备一个神经网络,根据两个参数"no"和"age"预测产品的索赔数量.以下数据集是神经网络的输入.

structure(list(no = c(25305.4104099149, 49282.7650363303, 71596.161588407, 
93100.2399492689, 120575.89348652, 138907.168168798, 152853.150129645, 
164658.048266216, 203323.951054253, 217964.514231364, 232098.010631853, 
245528.300551639, 257729.677107825, 273017.858354583, 289943.942081732, 
307253.529762711, 322779.210756104, 338484.424561413, 354509.62945598, 
376508.167449508, 392559.686167136, 406403.704572922, 418237.95321136, 
428306.956736623, 443032.309329306, 462815.029777392, 483057.035564531, 
501119.337852308, 516468.28989971, 529231.965438745, 546230.529378035), 
age = c(63.5793740593707, 102.316649334314, 139.062062015527, 
159.908188195329, 221.139098010716, 243.371632144127, 255.656705912817, 
321.979062244126, 302.183543005839, 354.719062375634, 369.989444935937, 
415.562730056562, 445.18103403067, 487.443822982359, 522.664771025013, 
531.055799381952, 588.227179384567, 627.155320232965, 631.325866647729, 
656.228738193787, 674.252217317525, 717.171080443709, 741.672049752712, 
788.251261134812, 798.113504685438, 831.731613476353, 834.814816968948, 
868.754851062387, 891.362029551517, 902.022293484355, 940.795814337874), 
claims = c(430.964844652385, 732.996578820216, 1702.3121722574,
2251.25233558302, 2197.47809502525, 2567.04757960458, 3031.86042202782, 
3156.90611199034, 3863.87816105778, 4111.89975688297, …

我试图通过使用内置方法在文本中找到与NLTK的搭配.

现在我有以下示例文本(test和foo相互跟随,但两者之间有一个句子边框):

content_part = """test. foo 0 test. foo 1 test. 
               foo 2 test. foo 3 test. foo 4 test. foo 5"""

标记化的结果collocations()如下:

print nltk.word_tokenize(content_part)
# ['test.', 'foo', 'my', 'test.', 'foo', '1', 'test.',
# 'foo', '2', 'test.', 'foo', '3', 'test.', 'foo', '4', 'test.', 'foo', '5']

print nltk.Text(nltk.word_tokenize(content_part)).collocations()
# test. foo

如何防止NLTK:

1. 在我的标记化中包含点
2. 在句子边框上找不到搭配()？

所以在这个例子中它根本不应该打印任何搭配,但我想你可以设想更复杂的文本,其中句子中也有搭配.

我可以猜测我需要使用Punkt句子分段器,但后来我不知道如何将它们再次组合起来找到与nltk的搭配(collocation()似乎比仅仅计算东西更强大).