使用正则表达式将 URL 提取到新的数据框列中 [英] Extract URLs with regex into a new data frame column
问题描述
我想使用正则表达式将数据框中文本中的所有 URL 提取到新列中.我有一些用于提取关键字的旧代码,因此我希望将代码调整为正则表达式.我想将正则表达式保存为字符串变量并在此处应用:
I want to use a regex to extract all URLs from text in a dataframe, into a new column. I have some older code that I have used to extract keywords, so I'm looking to adapt the code for a regex. I want to save a regex as a string variable and apply here:
data$ContentURL <- apply(sapply(regex, grepl, data$Content, fixed=FALSE), 1, function(x) paste(selection[x], collapse=','))
似乎 fixed=FALSE
应该告诉 grepl
它是一个正则表达式,但 R 不喜欢我试图将正则表达式保存为:>
It seems that fixed=FALSE
should tell grepl
that its a regular expression, but R doesn't like how I am trying to save the regex as:
regex <- "http.*?1-\\d+,\\d+"
我的数据组织成这样的数据框:
My data is organized in a data frame like this:
data <- read.table(text='"Content" "date"
1 "a house a home https://www.foo.com" "12/31/2013"
2 "cabin ideas https://www.example.com in the woods" "5/4/2013"
3 "motel is a hotel" "1/4/2013"', header=TRUE)
希望看起来像:
Content date ContentURL
1 a house a home https://www.foo.com 12/31/2013 https://www.foo.com
2 cabin ideas https://www.example.com in the woods 5/4/2013 https://www.example.com
3 motel is a hotel 1/4/2013
推荐答案
Hadleyverse 解决方案(stringr
包),具有不错的 URL 模式:
Hadleyverse solution (stringr
package) with a decent URL pattern:
library(stringr)
url_pattern <- "http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\\(\\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+"
data$ContentURL <- str_extract(data$Content, url_pattern)
data
## Content date ContentURL
## 1 a house a home https://www.foo.com 12/31/2013 https://www.foo.com
## 2 cabin ideas https://www.example.com in the woods 5/4/2013 https://www.example.com
## 3 motel is a hotel 1/4/2013 <NA>
如果 Content
中有多个,您可以使用 str_extract_all
,但这将涉及到您之后的一些额外处理.
You can use str_extract_all
if there are multiples in Content
, but that will involve some extra processing on your end afterwards.
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