2016-04-04 16 views
0

我对R和网络抓取相对陌生,因此对任何固有的明显错误表示歉意。R:刮脸网站,在URL中递增循环日期,保存为CSV

我正在寻找从URL 1刮取CSV文件,按日期递增到URL 2,然后保存每个CSV文件。正在输出

startdate <- as.Date("2007-07-01") 
enddate <- as.Date(Sys.Date()) 

for(startdate in enddate){ // Loop through dates on each URL 
    read.csv(url("http://api.foo.com/charts/data?output=csv&data=close&startdate=",startdate,"&enddate=",startdate,"&exchanges=bpi&dev=1")) 
    startdate = startdate + 1 
    startdate <- startdate[-c(1441,1442),] // Irrelevant to question at hand. Removes unwanted information auto-inserted into CSV. 
    write.csv(startdate[-c(1441,1442),], startdate, 'csv', row.names = FALSE) 
} 

以下错误:

read.csv(url("http://api.foo.com/charts/data?output=csv&data=close&startdate=",startdate,"&enddate=",startdate,"&exchanges=bpi&dev=1")) 
// Error in match.arg(method, c("default", "internal", "libcurl", "wininet")) :'arg' should be one of “default”, “internal”, “libcurl”, “wininet” 

和:

write.csv(startdate[c(1441,1442),], startdate, 'csv', row.names = FALSE) 
//Error in charToDate(x) : character string is not in a standard unambiguous format 

就如何解决这些错误有什么建议?

+0

'for(startdate in enddate)'这是试图做什么? – MichaelChirico

+0

'startdate < - startdate [-c(1441,1442),]'这个试图做什么? – MichaelChirico

回答

1

根据您的目标“我正在寻找从URL 1刮取CSV文件,按日期递增到URL 2,然后保存每个CSV文件。”这里是一个示例代码:

startdate <- as.Date("2016-01-01") 
enddate <- as.Date(Sys.Date()) 

geturl <- function(sdt, edt) { 
    paste0("http://api.foo.com/charts/data?output=csv&data=close", 
     "&startdate=",sdt,"&enddate=",edt,"&exchanges=bpi&dev=1") 
} #geturl 

dir.create("data") 
garbage <- lapply(seq.Date(startdate, enddate, by="1 day"), function(dt) { 
    dt <- as.Date(dt) 
    dat <- read.csv(url(geturl(dt, dt))) 
    write.csv(dat, paste0("data/dat-",format(dt, "%Y%m%d"),".csv"), row.names=FALSE) 
}) 

这是你在找什么? 你能提供一个样本链接吗?和一些样本日期?

+0

完美的作品,谢谢! – rsylatian