将 googleways 包中的列表合并为 google 方向？

Question

我正在为 googleways api 循环遍历经度和纬度点。我想出了两种方法来访问下面 link:

中显示的点部分

https://cran.r-project.org/web/packages/googleway/vignettes/googleway-vignette.html

不幸的是，因为这使用了唯一的密钥，所以我无法提供可重现的示例，但下面是我的尝试，一个使用 mapply，另一个使用循环。两者都可以产生列表格式的响应，但是我不确定如何像只传递一个点时那样解压它以拉出点路线：

df$routes$overview_polyline$points

有什么建议吗？

library(googleway)

dir_results = mapply(
  myfunction,
  origin = feed$origin,
  destination = feed$destination,
  departure = feed$departure
)

OR

empty_df = NULL
for (i in 1:nrow(feed)) {
  print(i)
  output = google_directions(feed[i,"origin"], 
                             feed[i,"destination"], 
                             mode = c("driving"), 
                    departure_time = feed[i,"departure"], 
                    arrival_time = NULL,
                    waypoints = NULL, alternatives = FALSE, avoid = NULL,
                    units = c("metric"), key = chi_directions, simplify = T)
  empty_df = rbind(empty_df, output)
}

编辑**

预期的输出将是如下所示的数据框：其中 "id" 表示输入的原始行程。

     lat       lon                                                                    id
1  40.71938 -73.99323 40.7193908691406+-73.9932174682617 40.7096214294434+-73.9497909545898
2  40.71992 -73.99292 40.7193908691406+-73.9932174682617 40.7096214294434+-73.9497909545898
3  40.71984 -73.99266 40.7193908691406+-73.9932174682617 40.7096214294434+-73.9497909545898
4  40.71932 -73.99095 40.7193908691406+-73.9932174682617 40.7096214294434+-73.9497909545898
5  40.71896 -73.98981 40.7193908691406+-73.9932174682617 40.7096214294434+-73.9497909545898
6  40.71824 -73.98745 40.7193908691406+-73.9932174682617 40.7096214294434+-73.9497909545898
7  40.71799 -73.98674 40.7193908691406+-73.9932174682617 40.7096214294434+-73.9497909545898
8  40.71763 -73.98582 40.7193908691406+-73.9932174682617 40.7096214294434+-73.9497909545898

编辑**** dput 提供用于回答关于数据框的问题以配对列表：

structure(list(origin = c("40.7193908691406 -73.9932174682617", 
"40.7641792297363 -73.9734268188477", "40.7507591247559 -73.9739990234375"
), destination = c("40.7096214294434-73.9497909545898", "40.7707366943359-73.9031448364258", 
"40.7711143493652-73.9871368408203")), .Names = c("origin", "destination"
), row.names = c(NA, 3L), class = "data.frame")

sql 代码基本是这样的：

feed = sqlQuery(con, paste("select top 10
                             longitude as px,
                             latitude as py,
                             dlongitude as dx ,
                             dlatitude as dy,
                             from mydb"))

然后在提供它之前，我的数据框提要看起来像这样（你可以忽略我在距离 api 中使用的偏离）：

                origin                       destination           departure
1  40.7439613342285 -73.9958724975586  40.716911315918-74.0121383666992 2017-03-03 01:00:32
2  40.7990493774414 -73.9685516357422 40.8066520690918-73.9610137939453 2017-03-03 01:00:33
3  40.7406234741211 -74.0055618286133 40.7496566772461-73.9834671020508 2017-03-03 01:00:33
4  40.7172813415527 -73.9953765869141 40.7503852844238-73.9811019897461 2017-03-03 01:00:33
5  40.7603607177734 -73.9817123413086 40.7416114807129-73.9795761108398 2017-03-03 01:00:34

Answer 1

如您所知，API 查询的结果 return 是一个列表。如果您对 API 进行多次调用，您将 return 多个列表。

因此，要提取感兴趣的数据，您必须对列表执行标准操作。在这个例子中，它可以用几个 *applys

使用 data.frame feed，其中每一行都包含起点 lat/lon（px/py）和终点 lat/lon (dx/dy)

feed <- data.frame(px = c(40.7193, 40.7641),
                   py = c(-73.993, -73.973),
                   dx = c(40.7096, 40.7707),
                   dy = c(-73.949, -73.903))

您可以使用 apply 为 data.frame 的每一行查询 google_directions() API。在同一个 apply 中，您可以随心所欲地执行任何操作，结果 extract/format 随心所欲。

lst <- apply(feed, 1, function(x){
    ## query Google Directions API
    res <- google_directions(key = key, 
                             origin = c(x[['px']], x[['py']]),
                             destination = c(x[['dx']], x[['dy']]))

    ## Decode the polyline
    df_route <- decode_pl(res$routes$overview_polyline$points)

    ## append the original coordinates as an 'id' column
    df_route[, "id"] <- paste0(paste(x[['px']], x[['py']], sep = "+")
                                ," "
                                , paste(x[['dx']], x[['dy']], sep = "+")
                                , collapse = " ")

    ## store the results of the query, the decoded polyline, 
    ## and the original query coordinates in a list
    lst_result <- list(route = df_route, 
                       full_result = res, 
                       origin = c(x[['px']], x[['py']]), 
                       destination = c(x[['dx']],x[['dy']]))

    return(lst_result)

})

所以现在 lst 是一个列表，其中包含每个查询的结果，加上作为 data.frame 的解码折线。要将所有解码的折线作为单个 data.frame，您可以再做一个 lapply，然后 rbind 一起做

## do what we want with the result, for example bind all the route coordinates into one data.frame
df <- do.call(rbind, lapply(lst, function(x) x[['route']]))
head(df)
       lat       lon                              id
1 40.71938 -73.99323 40.7193+-73.993 40.7096+-73.949
2 40.71992 -73.99292 40.7193+-73.993 40.7096+-73.949
3 40.71984 -73.99266 40.7193+-73.993 40.7096+-73.949
4 40.71932 -73.99095 40.7193+-73.993 40.7096+-73.949
5 40.71896 -73.98981 40.7193+-73.993 40.7096+-73.949
6 40.71824 -73.98745 40.7193+-73.993 40.7096+-73.949

将 googleways 包中的列表合并为 google 方向？

uniting lists out of the googleways package for google directions?

json

google-maps

r

mapply

googleway