Question

我热衷于将一组jekyll主题的源和演示网址提取到data.frame

library(rvest)

info <- read_html("https://github.com/jekyll/jekyll/wiki/themes")

data <- info %>%
 html_nodes(" #wiki-body li")

data
{xml_nodeset (115)}


[11] <li>Typewriter - (<a href="https://github.com/alixedi/typewriter">source</a>, <a href="http://alixedi.github.io/typewriter">demo</a>)</li>
[12] <li>block-log - (<a href="https://github.com/anandubajith/block-log">source</a>), <a href="https://anandu.net/demo/block-log/">demo</a>)</li>
[13] <li>Otter Pop - (<a href="https://github.com/tybenz/otter-pop">source</a>)</li>

所以我想要一个包含3列的data.frame（df），例如

name        source                                       demo
Typewriter   https://github.com/alixedi/typewriter         http://alixedi.github.io/typewriter

我能够将所有hrefs提取为矢量但是，正如您所看到的，从[13]开始，某些网站没有演示，因此我遇到了困难

有没有一种简单的方法可以从数据创建df？可能使用purrr库

Answer 1

data_out <- c()
for (i in 1:length(data)) {
  row <- data.frame(html_text(data[i]), as.character(html_children(data[[i]]))[1], as.character(html_children(data[[i]]))[2])
  data_out <- rbind(data_out, row)
}
names(data_out) <- c("name", "source", "demo")
data_out$name <- gsub(" - [(]source, demo[)]", "", data_out$name)
data_out$source <- gsub("<a href=\"|\">source</a>", "", data_out$source)
data_out$demo <- gsub("<a href=\"|\">demo</a>", "", data_out$demo)

Answer 2

这是你的purrr - 是的答案：

library(rvest)
library(purrr)
library(dplyr)

info <- read_html("https://github.com/jekyll/jekyll/wiki/themes")

themes <- html_nodes(info, xpath=".//div[@class='markdown-body']/*/li")

zero_to_na <- function(x) { ifelse(length(x)==0, NA, x) }

df <- data_frame(name=gsub(" [- ]*\\(.*$", "", html_text(themes)),
                 source=map_chr(themes, ~html_attr(html_nodes(., xpath=".//a[contains(., 'source')]"), "href")),
                 demo=map_chr(themes, ~zero_to_na(html_attr(html_nodes(., xpath=".//a[contains(., 'demo')]"), "href"))))

glimpse(df)
## Observations: 115
## Variables: 3
## $ name   <chr> "Jalpc", "Pixyll", "Jekyll Metro", "Midnight", "Leap Day", "F...
## $ source <chr> "https://github.com/Jack614/jalpc_jekyll_theme", "https://git...
## $ demo   <chr> "http://www.jack003.com", "http://pixyll.com/", "http://blog-...

交替：

map_df(themes, function(x) {
  data_frame(name=gsub(" [- ]*\\(.*$", "", html_text(x)),
             source=html_attr(html_nodes(x, xpath=".//a[contains(., 'source')]"), "href"),
             demo=zero_to_na(html_attr(html_nodes(x, xpath=".//a[contains(., 'demo')]"), "href")))
})

gsub / sub / etc您不想要的“名称”的任何部分。

Answer 3

您可以使用xpath分别收集带有演示数据的那些和没有演示数据的那些数据来分隔两组：

withDemo <- info %>%
    html_nodes(xpath = "//li[contains(., 'source') and contains(., 'demo')]")

withoutDemo <- info %>%
    html_nodes(xpath = "//li[contains(., 'source') and not(contains(.,'demo'))]")

然后，使用源和演示链接为集合创建数据框：

sourceNdemo <- withDemo %>%
    html_children() %>%              # get all children
    html_attr("href") %>%            # get the href attributes
    matrix(ncol = 2, byrow = TRUE)   # 2 pieces of data for each row

sourceNdemo <- setNames(
    data.frame(html_text(withDemo), sourceNdemo),  # html_text to get "name" column
    c("name", "source", "demo"))

然后，为仅包含源数据的数据框创建数据框

source <- withoutDemo %>% 
    html_children() %>%
    html_attr("href")

# set demo = NA for easy rbind-ing
source <- data.frame(name = html_text(withoutDemo), source = source, demo = NA)

rbind两个数据帧

allInfo <- rbind(sourceNdemo, source)

＆＃34;名称＆＃34;列现在包含＆＃34; Jalpc - （source，demo）＆＃34;等条目。和＆＃34;＆＃34; Bitwiser-Material（来源，演示）＆＃34;。你可以摆脱额外的＆＃34;（来源，演示）＆＃34;使用gsub：

的位

allInfo$name <- sub("\\s(-\\s)?\\(.+$", "", allInfo$name, perl = TRUE)

从R中的列表中提取具有一些缺失值的href

3 个答案: