对列表中的元素进行子集化并将它们放置在数据框中

Question

我有一个看起来像这样的列表（“listanswer”）：

> str(listanswer)
List of 100
 $ : chr [1:3] "" "" "\t\t"
 $ : chr [1:5] "" "Dr. Smith" "123 Fake Street" "New York, ZIPCODE 1" ...
 $ : chr [1:5] "" "Dr. Jones" "124 Fake Street" "New York, ZIPCODE 2" ...


> listanswer
[[1]]
[1] ""   ""   "\t\t"

[[2]]
[1] ""                             "Dr. Smith" "123 Fake Street"         "New York"          
[5] "ZIPCODE 1"    

[[3]]
[1] ""                           "Dr. Jones"   "124 Fake Street,"  "New York"        
[5] "ZIPCODE2"

对于此列表中的每个元素，我注意到子元素中的以下模式：

# first sub-element is always empty
    > listanswer[[2]][[1]]
    [1] ""
# second sub-element is the name
    > listanswer[[2]][[2]]
    [1] "Dr. Smith"
# third sub-element is always the address 
    > listanswer[[2]][[3]]
    [1] "123 Fake Street"
# fourth sub-element is always the city
    > listanswer[[2]][[4]]
    [1] "New York"
# fifth sub-element is always the ZIP
    > listanswer[[2]][[5]]
    [1] "ZIPCODE 1"

我想创建一个数据框，其中包含此列表中的行格式信息。 例如：

  id      name         address     city       ZIP
1  2 Dr. Smith 123 Fake Street New York ZIPCODE 1
2  3 Dr. Jones 124 Fake Street New York ZIPCODE 2

我想到了以下方法来做到这一点：

name = sapply(listanswer,function(x) x[2])
address = sapply(listanswer,function(x) x[3])
city = sapply(listanswer,function(x) x[4])
zip = sapply(listanswer,function(x) x[5])

final_data = data.frame(name, address, city, zip)
id = 1:nrow(final_data)

我的问题：我只是想确认一下 - 这是在列表中引用子元素的正确方法吗？

Answer 1

如果它有效，这是正确的方法，尽管可能有更有效或更易读的方法来做同样的事情。

另一种方法是使用您的列创建一个数据框，并向其中添加行。 IE

#create an empty data frame
df <- data.frame(matrix(ncol = 4, nrow = 0))
colnames(df) <- c("name", "address", "city", "zip")

#add rows
lapply(listanswer, \(x){df[nrow(df) + 1,] <- x[2:5]})

这只是解决同一问题的另一种方法。 可读性是个人喜好，您的解决方案也没有错。

Answer 2

如果这是基于您的大象问题，对于温哥华的企业，那么这主要是可行的。

library(rvest)

url<-"Website/british-columbia/"
page <-read_html(url)

#find the div tab of class=one_third
b = page %>% html_nodes("div.one_third") 

listanswer <- b %>% html_text() %>% strsplit("\\n")
#listanswer2 <- b %>% html_text2() %>% strsplit("\\n")
listanswer[[1]]<-NULL #remove first blank record

rows<-lapply(listanswer, function(element){
   vect<-element[-1] #remove first blank field
   cityindex<-as.integer(grep("Vancouver", vect))  #find city field
   #add some error checking and corrections
   if(length(cityindex)==0) {
      cityindex <- length(vect)-1 }
   else if(length(cityindex)>1) {
      cityindex <- cityindex[2] }

   #get the fields of interest
   address <- vect[cityindex-1]
   city<-vect[cityindex]
   phone <- vect[cityindex+1]
   
  if( cityindex < 3) {
      cityindex <- 3
   }  #error check
   #first groups combine into 1 name
   name <- toString(vect[1:(cityindex-2)])
   data.frame(name, address, city, phone)
})

answer<-bind_rows(rows)
#clean up 
answer$phone <- sub("Website", "", answer$phone)
answer

这仍然需要一些清理来处理不一致，但应该完成 80-90%