用于迭代数据帧行以进行谷歌地图距离查询的 R 代码

我正在寻求一些帮助，编写一些 R 代码来迭代数据帧中的行，并将每行中的值传递给函数，并将输出打印到 excel 文件、txt 文件或仅在控制台中。

这样做的目的是使用此网站上的功能自动对谷歌地图进行一系列距离/时间查询（数百个）：http://www.nfactorialanalytics.com/r-vignette-for-the-week-finding-time-distance- Between-two-places/ http://www.nfactorialanalytics.com/r-vignette-for-the-week-finding-time-distance-between-two-places/

该网站的功能如下：

library(XML)
library(RCurl)
distance2Points <- function(origin,destination){
 results <- list();
 xml.url <- paste0('http://maps.googleapis.com/maps/api/distancematrix/xml?origins=',origin,'&destinations=',destination,'&mode=driving&sensor=false')
 xmlfile <- xmlParse(getURL(xml.url))
 dist <- xmlValue(xmlChildren(xpathApply(xmlfile,"//distance")[[1]])$value)
 time <- xmlValue(xmlChildren(xpathApply(xmlfile,"//duration")[[1]])$value)
 distance <- as.numeric(sub(" km","",dist))
 time <- as.numeric(time)/60
 distance <- distance/1000
 results[['time']] <- time
 results[['dist']] <- distance
 return(results)
}

数据框将包含两列：出发地邮政编码和目的地邮政编码（加拿大，是吗？）。我是一名 R 初学者，所以我知道如何使用 read.table 将 txt 文件加载到数据框中。我只是不确定如何迭代数据帧，每次将值传递给 distance2Points 函数并执行。我认为这可以使用 for 循环或 apply 调用之一来完成？

谢谢您的帮助！

edit:

为了简单起见，假设我想将这两个向量转换为数据帧

> a <- c("L5B4P2","L5B4P2")
> b <- c("M5E1E5", "A2N1T3")
> postcodetest <- data.frame(a,b)
> postcodetest
       a      b
1 L5B4P2 M5E1E5
2 L5B4P2 A2N1T3

我应该如何迭代这两行以从 distance2Points 函数返回距离和时间？

这是一种方法，使用lapply生成一个列表，其中包含数据中每一行的结果并使用Reduce(rbind, [yourlist])将该列表连接到一个数据框中，其行与原始数据框中的行相对应。为了完成这项工作，我们还必须调整原始函数中的代码以返回单行数据帧，所以我在这里完成了这一点。

distance2Points <- function(origin,destination){

  require(XML)
  require(RCurl)

  xml.url <- paste0('http://maps.googleapis.com/maps/api/distancematrix/xml?origins=',origin,'&destinations=',destination,'&mode=driving&sensor=false')
  xmlfile <- xmlParse(getURL(xml.url))
  dist <- xmlValue(xmlChildren(xpathApply(xmlfile,"//distance")[[1]])$value)
  time <- xmlValue(xmlChildren(xpathApply(xmlfile,"//duration")[[1]])$value)
  distance <- as.numeric(sub(" km","",dist))
  time <- as.numeric(time)/60
  distance <- distance/1000
  # this gives you a one-row data frame instead of a list, b/c it's easy to rbind
  results <- data.frame(time = time, distance = distance)
  return(results)
}

# now apply that function rowwise to your data, using lapply, and roll the results
# into a single data frame using Reduce(rbind)
results <- Reduce(rbind, lapply(seq(nrow(postcodetest)), function(i)
  distance2Points(postcodetest$a[i], postcodetest$b[i])))

应用于样本数据时的结果：

> results
        time distance
1   27.06667   27.062
2 1797.80000 2369.311

如果您希望在不创建新对象的情况下执行此操作，您还可以编写单独的函数来计算时间和距离 - 或者将这些输出作为选项的单个函数 - 然后使用sapply要不就mutate在原始数据框中创建新列。使用时可能是这样的sapply:

distance2Points <- function(origin, destination, output){

  require(XML)
  require(RCurl)

  xml.url <- paste0('http://maps.googleapis.com/maps/api/distancematrix/xml?origins=',
                    origin, '&destinations=', destination, '&mode=driving&sensor=false')

  xmlfile <- xmlParse(getURL(xml.url))

  if(output == "distance") {

    y <- xmlValue(xmlChildren(xpathApply(xmlfile,"//distance")[[1]])$value)
    y <- as.numeric(sub(" km", "", y))/1000

  } else if(output == "time") {

    y <- xmlValue(xmlChildren(xpathApply(xmlfile,"//duration")[[1]])$value)
    y <- as.numeric(y)/60

  } else {

    y <- NA    

  }

  return(y)

}

postcodetest$distance <- sapply(seq(nrow(postcodetest)), function(i)
  distance2Points(postcodetest$a[i], postcodetest$b[i], "distance"))

postcodetest$time <- sapply(seq(nrow(postcodetest)), function(i)
  distance2Points(postcodetest$a[i], postcodetest$b[i], "time"))

以下是您可以如何做到这一点dplyr管道与mutate:

library(dplyr)

postcodetest <- postcodetest %>%
  mutate(distance = sapply(seq(nrow(postcodetest)), function(i)
           distance2Points(a[i], b[i], "distance")),
         time = sapply(seq(nrow(postcodetest)), function(i)
           distance2Points(a[i], b[i], "time")))