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我已经编写了一个函数来从谷歌获取和解析给定股票代码的新闻数据，但我确信有一些方法可以改进它。对于初学者，我的函数返回一个 GMT 时区的对象，而不是用户当前的时区，如果传递的数字大于 299（可能是因为 google 只返回每只股票 300 个故事），它就会失败。这在某种程度上是对我自己关于堆栈溢出的问题的回应，并且很大程度上依赖于这篇博文。

tl;博士：我该如何改进这个功能？

 getNews <- function(symbol, number){

    # Warn about length
    if (number>300) {
        warning("May only get 300 stories from google")
    }

    # load libraries
    require(XML); require(plyr); require(stringr); require(lubridate);
    require(xts); require(RDSTK)

    # construct url to news feed rss and encode it correctly
    url.b1 = 'http://www.google.com/finance/company_news?q='
    url    = paste(url.b1, symbol, '&output=rss', "&start=", 1,
               "&num=", number, sep = '')
    url    = URLencode(url)

    # parse xml tree, get item nodes, extract data and return data frame
    doc   = xmlTreeParse(url, useInternalNodes = TRUE)
    nodes = getNodeSet(doc, "//item")
    mydf  = ldply(nodes, as.data.frame(xmlToList))

    # clean up names of data frame
    names(mydf) = str_replace_all(names(mydf), "value\\.", "")

    # convert pubDate to date-time object and convert time zone
    pubDate = strptime(mydf$pubDate, 
                     format = '%a, %d %b %Y %H:%M:%S', tz = 'GMT')
    pubDate = with_tz(pubDate, tz = 'America/New_york')
    mydf$pubDate = NULL

    #Parse the description field
    mydf$description <- as.character(mydf$description)
    parseDescription <- function(x) {
        out <- html2text(x)$text
        out <- strsplit(out,'\n|--')[[1]]

        #Find Lead
        TextLength <- sapply(out,nchar)
        Lead <- out[TextLength==max(TextLength)]

        #Find Site
        Site <- out[3]

        #Return cleaned fields
        out <- c(Site,Lead)
        names(out) <- c('Site','Lead')
        out
    }
    description <- lapply(mydf$description,parseDescription)
    description <- do.call(rbind,description)
    mydf <- cbind(mydf,description)

    #Format as XTS object
    mydf = xts(mydf,order.by=pubDate)

    # drop Extra attributes that we don't use yet
    mydf$guid.text = mydf$guid..attrs = mydf$description = mydf$link = NULL
    return(mydf) 

}
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1 回答 1

6

这是您的getNews功能的更短（可能更有效）的版本

  getNews2 <- function(symbol, number){

    # load libraries
    require(XML); require(plyr); require(stringr); require(lubridate);  

    # construct url to news feed rss and encode it correctly
    url.b1 = 'http://www.google.com/finance/company_news?q='
    url    = paste(url.b1, symbol, '&output=rss', "&start=", 1,
               "&num=", number, sep = '')
    url    = URLencode(url)

    # parse xml tree, get item nodes, extract data and return data frame
    doc   = xmlTreeParse(url, useInternalNodes = T);
    nodes = getNodeSet(doc, "//item");
    mydf  = ldply(nodes, as.data.frame(xmlToList))

    # clean up names of data frame
    names(mydf) = str_replace_all(names(mydf), "value\\.", "")

    # convert pubDate to date-time object and convert time zone
    mydf$pubDate = strptime(mydf$pubDate, 
                     format = '%a, %d %b %Y %H:%M:%S', tz = 'GMT')
    mydf$pubDate = with_tz(mydf$pubDate, tz = 'America/New_york')

    # drop guid.text and guid..attrs
    mydf$guid.text = mydf$guid..attrs = NULL

    return(mydf)    
}

此外，您的代码中可能存在错误，因为我尝试使用它symbol = 'WMT'并返回错误。我认为getNews2WMT 也适用。检查一下，让我知道它是否适合你。

PS。该description列仍然包含 html 代码。但是从中提取文本应该很容易。我会在有时间时发布更新

于 2011-04-23T03:19:31.843 回答