R代碼迭代通過數據幀行谷歌地圖距離查詢

Question

我正在尋找一些幫助，編寫一些R代碼遍歷數據框中的行，並將每行中的值傳遞給一個函數，並將輸出打印到一個excel文件，txt文件或只在控制檯中。

這樣做的目的是自動一堆距離/時間查詢（幾百），谷歌使用在這個網站上找到的功能地圖：http://www.nfactorialanalytics.com/r-vignette-for-the-week-finding-time-distance-between-two-places/

該網站上的功能如下：

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) 
} 

數據框將包含兩列：原始郵政編碼和目的地郵政編碼（加拿大，eh？）。我是一名初學R程序員，所以我知道如何使用read.table將txt文件加載到數據框中。我只是不確定如何迭代數據幀，每次將值傳遞給distance2Points函數並執行。我認爲這可以使用for循環或其中一個應用程序調用完成。

感謝您的幫助！

編輯：

爲了簡單起見讓我們假設我想這兩個向量轉換成數據幀

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

我應該如何去遍歷這兩行返回來自距離和時間距離2點函數？

Answer 1

這裏有一種方法：使用lapply產生一個列表，其中包含數據中每行的結果，並使用Reduce(rbind, [yourlist])將該列表連接到一個數據框中，該數據框的行對應於原始列中的行。爲了做到這一點，我們還必須調整原始函數中的代碼以返回一行數據框，所以我在這裏做了。

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]))) 

當適用於您的樣本數據

結果：

> results 
     time distance 
1 27.06667 27.062 
2 1797.80000 2369.311 

如果你寧願做沒有創建新的對象，你也可以寫出計算時間和距離獨立的功能 - 或單一功能與這些輸出作爲選項 - 然後使用sapply或mutate在原始數據框中創建新列。以下是如何可能看起來使用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")) 

而且這裏是你如何能與mutate做一個dplyr管：

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")))