Question

我想从http://insideairbnb.com/get-the-data.html下载所有涉及美国城市的名为“ listings.csv.gz”的文件，我可以通过编写每个链接来做到这一点，但是可以循环吗？

最后，我将仅保留每个文件中的几列并将它们合并到一个文件中。

由于使用@CodeNoob解决了问题，所以我想分享所有解决方法：

page <- read_html("http://insideairbnb.com/get-the-data.html")

# Get all hrefs (i.e. all links present on the website)
links <- page %>%
  html_nodes("a") %>%
  html_attr("href")

# Filter for listings.csv.gz, USA cities, data for March 2019
wanted <- grep('listings.csv.gz', links)
USA <- grep('united-states', links)
wanted.USA = wanted[wanted %in% USA]
wanted.links <- links[wanted.USA]
wanted.links = grep('2019-03', wanted.links, value = TRUE)

wanted.cols = c("host_is_superhost", "summary", "host_identity_verified", "street", 
                "city", "property_type", "room_type", "bathrooms", 
                "bedrooms", "beds", "price", "security_deposit", "cleaning_fee", 
                "guests_included", "number_of_reviews", "instant_bookable", 
                "host_response_rate", "host_neighbourhood", 
                "review_scores_rating", "review_scores_accuracy","review_scores_cleanliness",
                "review_scores_checkin" ,"review_scores_communication", 
                "review_scores_location", "review_scores_value", "space", 
                "description", "host_id", "state", "latitude", "longitude")


read.gz.url <- function(link) {
  con <- gzcon(url(link))
  df  <- read.csv(textConnection(readLines(con)))
  close(con)
  df  <- df %>% select(wanted.cols) %>%
    mutate(source.url = link)
  df
}

all.df = list()
for (i in seq_along(wanted.links)) {
  all.df[[i]] = read.gz.url(wanted.links[i])
}

all.df = map(all.df, as_tibble)

Answer 1

您实际上可以提取所有链接，过滤包含listings.csv.gz的链接，然后循环下载：

library(rvest)
library(dplyr)

# Get all download links

page <- read_html("http://insideairbnb.com/get-the-data.html")

# Get all hrefs (i.e. all links present on the website)
links <- page %>%
  html_nodes("a") %>%
  html_attr("href")

# Filter for listings.csv.gz
wanted <- grep('listings.csv.gz', links)
wanted.links <- links[wanted]

for (link in wanted.links) {
  con <- gzcon(url(link))
  txt <- readLines(con)
  df <- read.csv(textConnection(txt))
  # Do what you want
}

示例：下载并合并文件
为了获得所需的结果，我建议编写一个下载函数，该函数对所需的列进行过滤，然后将其合并到单个数据框中，例如：

read.gz.url <- function(url) {
  con <- gzcon(url(link))
  df  <- read.csv(textConnection(readLines(con)))
  close(con)
  df  <- df %>% select(c('calculated_host_listings_count_shared_rooms', 'cancellation_policy' )) %>% # random columns I chose
    mutate(source.url = url) # You may need to remember the origin of each row
  df
}

all.df <- do.call('rbind', lapply(head(wanted.links,2), read.gz.url))

注意，由于它们很大，我仅在前两个文件中进行了测试

如何从HTML页面下载具有相同名称的多个文件？

1 个答案: