由于您尚未指定要如何分隔地址,我假设地址块中的每一行都应转到单独的列。
为此,您可以使用足够的 XPath 查询来选择第二个表的 p-tags,例如
//*[@id="DispensingFacilities"]/tbody/tr/td/p/text()
然后遍历结果并从头开始创建表结构。以这种方式生成 CSV 现在有点复杂。有几种方式,如下图一种:
library(xml2)
library(magrittr)
library(rvest)
FLlist <- read_html("http://www.floridahealth.gov/programs-and-services/office-of-medical-marijuana-use/medical-marijuana-treatment-centers/index.html")
FLDispensaries <- as.data.frame(FLlist %>%
html_nodes("table") %>%
.[[2]] %>%
html_table(header = TRUE))
FLTable <- html_nodes(x=FLlist, xpath='//*[@id="DispensingFacilities"]/tbody/tr/td/p/text()')
#helper
trim <- function (x) gsub("^\\s+|\\s+$", "", x)
mat<-matrix(list(), ncol=4)
li <- c()
row <- 0;
col <- 1;
for(i in FLTable){
li <- c(li,trim(html_text(i)));
if(col %% 4 == 0) {
row <- row + 1;
mat[[row]] <- li;
li <- c();
}
col <- col + 1;
}
#to matrix/list to dataframe
library(plyr)
mat.df <- as.data.frame(do.call(rbind, mat))
write.csv(mat.df, "FLTest.csv")
由于您只想获得实际地址,因此我将 XPath 更改为仅从第二个 tr 开始选择第二个 td
FLTable <- html_nodes(x=FLlist, xpath='//*[@id="DispensingFacilities"]/tbody/tr[position()>1]/td[2]/p')
li <- c()
row <- 1;
mat<-matrix(list(), ncol=2)
for (i in seq_along(FLTable)) {
addrlines <- str_split(xml_text(FLTable[[i]]), "\\s\\s")
for (a in seq_along(addrlines[[1]])) {
if(a %% 2 == 0) {
li <- c(li,paste(addrlines[[1]][a], addrlines[[1]][a+1]));
mat[[row]] <- li;
li <- c();
row <- row + 1;
} else if(a %% 3 == 0) {
next; #skip
} else {
li <- c(li,addrlines[[1]][a]);
}
}
}
并创建一个不错的输出
outputFile <- "output.csv"
#to matrix/list to dataframe
mat.df <- as.data.frame(do.call(rbind, mat))
cat(c("No.,", "Name,", "Address"), '\n', file = outputFile)
write.table(mat.df,outputFile,sep=',',append = T, col.names = F)
生成的 CSV 如下所示:
No., Name, Address
"1","AltMed Florida (MuV)","5909 U.S. Hwy 41 N Apollo Beach, FL 33572"
"2","Trulieve","1103 14th Street West Bradenton, FL 34205"
...
"55","Trulieve","1814 Commerce Avenue Vero Beach, FL 32960"