如何從R中的文本數據中刪除「/ url？q =」

我想從R Studio中的文本數據中刪除「/ url？q =」。這是我的谷歌搜索碼：如何從R中的文本數據中刪除「/ url？q =」

## Code for Google Search 
# Enter Search Term Here 
search.term <- "r-project" 
# Creating Function 
getGoogleURL <- function(search.term, domain = '.co.in', quotes=TRUE) 
{ 
    # Getting Search Term 
    search.term <- gsub(' ', '%20', search.term) 
    if(quotes) search.term <- paste('%22', search.term, '%22', sep='') 
    # Putting Search Term in Google Search 
    getGoogleURL <- paste('http://www.google', domain, '/search?q=', search.term, sep='') } 

## Get Links from Google Search 
# Creating Function to Get URLs From Search Results 
getGoogleLinks <- function(google.url) { 
    # Creating a File to Save URLs 
    doc <- getURL(google.url, httpheader = c("User-Agent" = "R(3.4.0)")) 
    # Removing HTML code and Setting Nodes 
    html <- htmlTreeParse(doc, useInternalNodes = TRUE, error=function(...){}) 
    nodes <- getNodeSet(html, "//h3[@class='r']//a") 
    return(sapply(nodes, function(x) x <- xmlAttrs(x)[["href"]])) } 

## Remove quoted text, Create URL List 
    quotes <- "FALSE" 
    search.url <- getGoogleURL(search.term=search.term, quotes=quotes) 
    links <- getGoogleLinks(search.url) 

## Print URL List 
    links

而我的結果是：

[1]？「/ URL Q = https://www.r-project.org/&sa=U&ved=0ahUKEwj78ZWXoabUAhUcTI8KHaTEDTIQFggUMAA&usg=AFQjCNEqtiOAIA7OOTa3meWC8zaTjjTy8A」
[2]「/ URL Q = http://www.cran.r-project.org/&sa=U&ved=0ahUKEwj78ZWXoabUAhUcTI8KHaTEDTIQjBAIGzAB&usg=AFQjCNF8QmYbLzG0c66QZM2wsXF1n1-9tQ 「

如何從上面的鏈接中刪除」/ url？q =「？

來源

2017-06-05 Rugved Modak

您可以使用gsub。

## Code for Google Search 
# Enter Search Term Here 
search.term <- "r-project" 
# Creating Function 
getGoogleURL <- function(search.term, domain = '.co.in', quotes=TRUE) 
{ 
    # Getting Search Term 
    search.term <- gsub(' ', '%20', search.term) 
    if(quotes) search.term <- paste('%22', search.term, '%22', sep='') 
    # Putting Search Term in Google Search 
    getGoogleURL <- paste('http://www.google', domain, '/search?q=', search.term, sep='') } 

## Get Links from Google Search 
# Creating Function to Get URLs From Search Results 
getGoogleLinks <- function(google.url) { 
    # Creating a File to Save URLs 
    doc <- getURL(google.url, httpheader = c("User-Agent" = "R(3.4.0)")) 
    # Removing HTML code and Setting Nodes 
    html <- htmlTreeParse(doc, useInternalNodes = TRUE, error=function(...){}) 
    nodes <- getNodeSet(html, "//h3[@class='r']//a") 
    return(sapply(nodes, function(x) x <- xmlAttrs(x)[["href"]])) } 

## Remove quoted text, Create URL List 
    quotes <- "FALSE" 
    search.url <- getGoogleURL(search.term=search.term, quotes=quotes) 
    links <- getGoogleLinks(search.url) 

## Print URL List 
    gsub("/url?q=", "", links)

來源

2017-06-05 08:49:04 JTeam

謝謝，但我解決了我的問題，而不是'substring'。 –

或者改爲@ JTeam的答案你可以試試這個（中給出的鏈接總是/url?q=開始）：

lapply(links,function(x) paste0(strsplit(x,'=')[[1]][-1],collapse = ''))

這給你乾淨的鏈接一個很好的列表（如果你喜歡一個向量，嘗試sapply）

來源

2017-06-05 08:54:30 Val

謝謝，但我解決了我的問題，而不是'substring'。 –

我這種方式解決它，因爲他們有限的字符數

links <- substring(links,8)

來源

2017-06-05 11:04:51

回答

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