如何使用R中的传单将两个坐标与一条线连接

我试图使用R中的传单包绘制一个放大器，并连接下表中给定纬度和经度信息的标记

| Observation | InitialLat | InitialLong | NewLat | NewLong | |-------------|------------|-------------|-----------|-----------| | A | 62.469722 | 6.187194 | 51.4749 | -0.221619 | | B | 48.0975 | 16.3108 | 51.4882 | -0.302621 | | C | 36.84 | -2.435278 | 50.861822 | -0.083278 | | D | 50.834194 | 4.298361 | 54.9756 | -1.62179 | | E | 50.834194 | 4.298361 | 54.9756 | -1.62179 | | F | 50.834194 | 4.298361 | 51.4882 | -0.302621 | | G | 47.460427 | -0.530804 | 51.44 | -2.62021 | | H | 51.5549 | -0.108436 | 53.4281 | -1.36172 | | I | 51.5549 | -0.108436 | 52.9399 | -1.13258 | | J | 51.5549 | -0.108436 | 51.889839 | -0.193608 | | | 51.5549 | -0.108436 | 52.0544 | 1.14554 |

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传单可以使用

addPolylines

功能添加行。问题是它假设每条线都是连接的——你会把它们都连接起来

解决此问题的最佳方法（AFAIK）是使用循环：

library(leaflet)
map3 = leaflet(data) %>% addTiles()
map3 <- map3 %>% addMarkers(~InitialLong,~InitialLat, popup=~Observation)
for(i in 1:nrow(data)){
    map3 <- addPolylines(map3, lat = as.numeric(data[i, c(2, 4)]), 
                               lng = as.numeric(data[i, c(3, 5)]))
}
map3

现在绘制：

map3 = leaflet(data) %>% addTiles()
map3 %>% addMarkers(~InitialLong, ~InitialLat, popup = ~Observation) %>%
         addPolylines(data = y)

Kyle Walker的点到线的来源：

library(sp)
library(maptools)

points_to_line <- function(data, long, lat, id_field = NULL, sort_field = NULL) {

  # Convert to SpatialPointsDataFrame
  coordinates(data) <- c(long, lat)

  # If there is a sort field...
  if (!is.null(sort_field)) {
    if (!is.null(id_field)) {
      data <- data[order(data[[id_field]], data[[sort_field]]), ]
    } else {
      data <- data[order(data[[sort_field]]), ]
    }
  }

  # If there is only one path...
  if (is.null(id_field)) {

    lines <- SpatialLines(list(Lines(list(Line(data)), "id")))

    return(lines)

    # Now, if we have multiple lines...
  } else if (!is.null(id_field)) {  

    # Split into a list by ID field
    paths <- sp::split(data, data[[id_field]])

    sp_lines <- SpatialLines(list(Lines(list(Line(paths[[1]])), "line1")))

    # I like for loops, what can I say...
    for (p in 2:length(paths)) {
      id <- paste0("line", as.character(p))
      l <- SpatialLines(list(Lines(list(Line(paths[[p]])), id)))
      sp_lines <- spRbind(sp_lines, l)
    }

    return(sp_lines)
  }
}

库（sp）
图书馆（地图工具）
点到线这里是使用
传单
包的另一种方法。为了演示，我在你们的数据中取了两个数据点

mydf <- data.frame(Observation = c("A", "B"),
                   InitialLat = c(62.469722,48.0975),
                   InitialLong = c(6.187194, 16.3108),
                   NewLat = c(51.4749, 51.4882),
                   NewLong = c(-0.221619, -0.302621),
                   stringsAsFactors = FALSE)

我修剪了我得到的交互式地图。请看下面的地图。虽然这幅图中有两条线是相连的，但它们是分开的。如果运行代码并放大，您将看到这两行是分开的

我知道这是一年前提出的问题，但我有同样的问题，并在传单中找到了解决方法

首先必须调整数据帧，因为addPolyline只连接序列中的所有坐标。为了本演示的目的，我将制作一个具有4个独立结束位置的数据帧

dest_df <- data.frame (lat = c(41.82, 46.88, 41.48, 39.14),
                   lon = c(-88.32, -124.10, -88.33, -114.90)
                  )


我希望这有助于任何希望在将来做类似事情的人
根据行的用途，另一个很好的选择是gcIntermediate（）。它根据地球的曲率输出弯曲的SpatialLines对象。不过，这对指路不是很好。SpatialLines类对象可以很好地处理传单。请参阅，以获取一个极好的示例。我发布了一个修改过的表单，从Paul Reiners的数据框开始

library(leaflet)
library(geosphere)

mydf <- data.frame(InitialLat = c(62.469722,48.0975), # initial df
               InitialLong = c(6.187194, 16.3108),
               NewLat = c(51.4749, 51.4882),
               NewLong = c(-0.221619, -0.302621))

p1 <- as.matrix(mydf[,c(2,1)]) # it's important to list lng before lat here
p2 <- as.matrix(mydf[,c(4,3)]) # and here

gcIntermediate(p1, p2,  
           n=100, 
           addStartEnd=TRUE,
           sp=TRUE) %>% 
leaflet() %>% 
addTiles() %>% 
addPolylines()

图书馆（传单）
图书馆（地球圈）
mydf认为这是你想要的：

install.packages("leaflet")
library(leaflet)

mydf <- data.frame(Observation = c("A", "B","C","D","E"),
               InitialLat = c(62.469722,48.0975,36.84,50.834194,50.834194),
               InitialLong = c(6.187194, 16.3108,-2.435278,4.298361,4.298361),
               NewLat = c(51.4749, 51.4882,50.861822,54.9756,54.9756),
               NewLong = c(-0.221619, -0.302621,-0.083278,-1.62179,-1.62179),
               stringsAsFactors = FALSE)

mydf
 Observation InitialLat InitialLong   NewLat   NewLong
1           A   62.46972    6.187194 51.47490 -0.221619
2           B   48.09750   16.310800 51.48820 -0.302621
3           C   36.84000   -2.435278 50.86182 -0.083278
4           D   50.83419    4.298361 54.97560 -1.621790
5           E   50.83419    4.298361 54.97560 -1.621790

m<-leaflet(data=mydf)%>%addTiles
for (i in 1:nrow(mydf)) 
m<-m%>%addPolylines(lat=c(mydf[i,]$InitialLat,mydf[i,]$NewLat),lng=c(mydf[i,]$InitialLong,mydf[i,]$NewLong))

install.packages（“传单”）
图书馆（单张）
mydf我想这曾经奏效，但现在真的不行了。即使使用~group，传单也可以连接所有多段线（也可以在“放大此地图”时）。非常感谢David！addPolylines（）中的一个小错误。它应该是lng=~长而不是lon。
mydf2 <- data.frame(group = c("A", "B"),
                    lat = c(mydf$InitialLat, mydf$NewLat),
                    long = c(mydf$InitialLong, mydf$NewLong))

#  group      lat      long
#1     A 62.46972  6.187194
#2     B 48.09750 16.310800
#3     A 51.47490 -0.221619
#4     B 51.48820 -0.302621

library(leaflet)
library(magrittr)

leaflet()%>%
addTiles() %>%
addPolylines(data = mydf2, lng = ~long, lat = ~lat, group = ~group)

dest_df <- data.frame (lat = c(41.82, 46.88, 41.48, 39.14),
                   lon = c(-88.32, -124.10, -88.33, -114.90)
                  )

orig_df <- data.frame (lat = c(rep.int(40.75, nrow(dest_df))),
                   long = c(rep.int(-73.99,nrow(dest_df)))
                  )

orig_df$sequence <- c(sequence = seq(1, length.out = nrow(orig_df), by=2))
dest_df$sequence <- c(sequence = seq(2, length.out = nrow(orig_df), by=2))

library("sqldf")
q <- "
SELECT * FROM orig_df
UNION ALL
SELECT * FROM dest_df
ORDER BY sequence
"
poly_df <- sqldf(q)

library("leaflet")
leaflet() %>%
  addTiles() %>%

  addPolylines(
    data = poly_df,
    lng = ~lon, 
    lat = ~lat,
    weight = 3,
    opacity = 3
  ) 

library(leaflet)
library(geosphere)

mydf <- data.frame(InitialLat = c(62.469722,48.0975), # initial df
               InitialLong = c(6.187194, 16.3108),
               NewLat = c(51.4749, 51.4882),
               NewLong = c(-0.221619, -0.302621))

p1 <- as.matrix(mydf[,c(2,1)]) # it's important to list lng before lat here
p2 <- as.matrix(mydf[,c(4,3)]) # and here

gcIntermediate(p1, p2,  
           n=100, 
           addStartEnd=TRUE,
           sp=TRUE) %>% 
leaflet() %>% 
addTiles() %>% 
addPolylines()

install.packages("leaflet")
library(leaflet)

mydf <- data.frame(Observation = c("A", "B","C","D","E"),
               InitialLat = c(62.469722,48.0975,36.84,50.834194,50.834194),
               InitialLong = c(6.187194, 16.3108,-2.435278,4.298361,4.298361),
               NewLat = c(51.4749, 51.4882,50.861822,54.9756,54.9756),
               NewLong = c(-0.221619, -0.302621,-0.083278,-1.62179,-1.62179),
               stringsAsFactors = FALSE)

mydf
 Observation InitialLat InitialLong   NewLat   NewLong
1           A   62.46972    6.187194 51.47490 -0.221619
2           B   48.09750   16.310800 51.48820 -0.302621
3           C   36.84000   -2.435278 50.86182 -0.083278
4           D   50.83419    4.298361 54.97560 -1.621790
5           E   50.83419    4.298361 54.97560 -1.621790

m<-leaflet(data=mydf)%>%addTiles
for (i in 1:nrow(mydf)) 
m<-m%>%addPolylines(lat=c(mydf[i,]$InitialLat,mydf[i,]$NewLat),lng=c(mydf[i,]$InitialLong,mydf[i,]$NewLong))