如何在传单中的两个标记之间添加路线(折线)

问题描述 投票:1回答:2

我希望在这两点之间添加折线。我怎样才能做到这一点?

m <- leaflet() %>%
     addPolylines() %>%
     addTiles() %>%  # Add default OpenStreetMap map tiles
     addMarkers(lng = -87.6266382, lat = 41.8674336,
                popup = "starting") %>%
     addMarkers(lng = -87.64847, lat = 41.9168862,
                popup = "Destination")

m  # Print the map

数据示例。

| region      | from_lat   | from_long    | to_lat      | to_long    |
|-------------|------------|------------- |-------------|----------- |
| 1           | 41.8674336 | -87.6266382  | 41.887544   | -87.626487 |
| 2           | 41.8674336 | -87.6266382  | 41.9168862  | -87.64847  |
| 3           | 41.8674336 | -87.6266382  | 41.8190937  | -87.6230967|
r google-maps maps leaflet googleway
2个回答
1
投票

这是我的尝试。由于您想在传单地图上绘制一些路线,因此您希望通过googleway包实现此任务。它允许您使用google_directions()decode_pl()提取路径数据。由于您有多个路由,因此您希望使用lapply()并创建数据集。获得路线数据后,您的工作就很简单;你使用addPolylines()中的数据。

library(dplyr)
library(googleway)
library(leaflet)

mydf <- data.frame(region = 1:3,
                   from_lat = 41.8674336,
                   from_long = -87.6266382,
                   to_lat = c(41.887544, 41.9168862, 41.8190937),
                   to_long = c(-87.626487, -87.64847, -87.6230967))

mykey <- "you need to have your API key here"

lapply(1:nrow(mydf), function(x){

    foo <- google_directions(origin = unlist(mydf[x, 2:3]),
                             destination = unlist(mydf[x, 4:5]),
                             key = mykey,
                             mode = "driving",
                             simplify = TRUE)

    pl <- decode_pl(foo$routes$overview_polyline$points)

    return(pl)

        }
    ) %>%
bind_rows(.id = "region") -> temp


m <- leaflet() %>%
     addProviderTiles("OpenStreetMap.Mapnik") %>%
     addPolylines(data = temp, lng = ~lon, lat = ~lat, group = ~region) %>%
     addMarkers(lng = -87.6266382, lat = 41.8674336,
                popup = "starting")%>%
     addMarkers(data = mydf, lng = ~to_long, lat = ~to_lat,
                popup = "Destination")

enter image description here


1
投票

@ jazzurro的答案是现场的,所以应该仍然是公认的答案。

正如他们所提到的,如果你想要并绘制谷歌地图,你可以在googleway包中完成所有这些。使用传单的主要区别在于您不需要解码折线,Google Map可以为您处理编码的字符串。

polylines <- lapply(1:nrow(mydf), function(x){

  foo <- google_directions(origin = unlist(mydf[x, 2:3]),
                           destination = unlist(mydf[x, 4:5]),
                           key = apiKey,
                           mode = "driving",
                           simplify = TRUE)

  ## no need to decode the line, just return the string as-is
  foo$routes$overview_polyline$points
}
)

df <- data.frame(polylines = unlist(polylines), stringsAsFactors = F)

## add some colour values for the markers
mydf$colour_from <- "red"
mydf$colour_to <- "blue"

## plot the polylines and the markers
google_map(key = mapKey) %>%
  add_markers(data = mydf, lat = "from_lat", lon = "from_long", colour = "colour_from") %>%
  add_markers(data = mydf, lat = "to_lat", lon = "to_long", colour = "colour_to") %>%
  add_polylines(data = df, polyline = "polylines")

enter image description here


注意:我是googleway作者。

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