【发布时间】:2020-03-08 11:10:02
【问题描述】:
我想从http://insideairbnb.com/get-the-data.html 下载所有名为“listings.csv.gz”的文件,这些文件指的是美国城市,我可以通过编写每个链接来完成,但可以循环执行吗?
最后我将只保留每个文件中的几列并将它们合并到一个文件中。
由于@CodeNoob 解决了问题,我想分享一下它是如何解决的:
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)
【问题讨论】:
-
你自己试过什么代码?您是否尝试过使用stackoverflow.com/questions/50164561/…
标签: r