Question

我正在使用纬度和经度数据在leaflet地图上绘制线（请参见下文）。理想情况下，这些行将存储在数据帧的lat和lng列中。在lat列中，每个起始点lat值后都将跟随一个终点lat值，然后是另一行的起始点lat值（line_id列中可以区分每行）。 lng数据的排列方式类似。理想情况下，数据框应如下所示：

> df.better
  line_id      lat        lng
1     ABC 51.50995 -0.1345093
2     ABC 51.51074 -0.1345093
3     XYZ 51.50991 -0.1345193
4     XYZ 51.51079 -0.1351200

问题在于它以这种格式从数据存储中出来：

> df.wide
  line_id start_lat  end_lat  start_lng  end_lng
1     ABC  51.50995 51.51074 -0.1345093 -0.13519
2     XYZ  51.50991 51.51079  0.1351900  0.13512

这看起来有点像经典的“宽到长”数据争用问题，对此存在很多疑问和答案，但是标准的“长”格式将经纬度和经度数据折叠为一列，我需要两列。我尝试了以下tidyverse解决方案：

df2 <- df.wide %>% pivot_longer(cols = start_lat:end_lng,
                         names_to="variable",
                         values_to="value")

然后我清理variable列：

df2$variable <- gsub(".*_lat","lat",df2$variable)
df2$variable <- gsub(".*_lng","lng",df2$variable)

这是结果，至少看起来数据顺序正确：

> df2
A tibble: 8 x 3
  line_id variable      value
  <fct>   <chr>         <dbl>
1 ABC     lat      51.50995  
2 ABC     lat      51.51074  
3 ABC     lng      -0.1345093
4 ABC     lng      -0.13519  
5 XYZ     lat      51.50991  
6 XYZ     lat      51.51079  
7 XYZ     lng       0.13519  
8 XYZ     lng       0.135120

最后一步似乎涉及再次散布数据，但是使用pivot_wider会导致无法唯一标识值的投诉：

df2 %>% pivot_wider(names_from = variable,values_from = value)
    # A tibble: 2 x 3
      line_id         lat         lng
      <fct>   <list<dbl>> <list<dbl>>
    1 ABC             [2]         [2]
    2 XYZ             [2]         [2]
    Warning message:
Values in `value` are not uniquely identified; output will contain list-cols.

我可以（我认为）理解为什么会发生错误，但是在variable中提供唯一标识符只会使我回到开始的地方。我应该/应该如何处理？

require(magrittr)
require(tidyr)
require(dplyr)

options(pillar.sigfig = 7)

df.better <- data.frame(
  line_id = c("ABC","ABC","XYZ","XYZ"),
  lat = c(51.509950,51.510736,51.509910,51.510786),
  lng = c(-0.1345093,-0.1345093,-0.1345193,-0.135120)
)

df.wide <- data.frame(
  line_id = c("ABC","XYZ"),
  start_lat = c(51.509950,51.509910),
  end_lat = c(51.510736,51.510786),
  start_lng = c(-0.1345093,0.135190),
  end_lng = c(-0.135190,0.135120)
)

df2 <- df.wide %>% pivot_longer(cols = start_lat:end_lng,
                         names_to="variable",
                         values_to="value")

df2$variable <- gsub(".*_lat","lat",df2$variable)
df2$variable <- gsub(".*_lng","lng",df2$variable)

df2 %>% pivot_wider(names_from = variable,values_from = value)

m <- leaflet() %>% setView(lng = -0.1345093, lat = 51.510090, zoom = 18) %>% addTiles()

for (i in unique(df.better$line_id)) { # HT: https://stackoverflow.com/a/44547502/952708
  m <- m %>%
    addPolylines(data = df.better[df.better$line_id == i, ],
                 lng = ~lng, lat = ~lat, color = "Green",
                 opacity = 0.5, weight = 2, dashArray = 5)
}

m

Answer 1

如果我理解正确，那么您正在寻找这样的东西：

df.wide <- data.frame(
  line_id = c("ABC","XYZ"),
  start_lat = c(51.509950,51.509910),
  end_lat = c(51.510736,51.510786),
  start_lng = c(-0.1345093,0.135190),
  end_lng = c(-0.135190,0.135120)
)

df.wide %>% 
  pivot_longer(-line_id,
               names_to = c("set", ".value"),
               names_pattern = "(.+)_(.+)"
  )

#  line_id set        lat        lng
#  <fct>   <chr>    <dbl>      <dbl>
#1 ABC     start 51.50995 -0.1345093
#2 ABC     end   51.51074 -0.13519  
#3 XYZ     start 51.50991  0.13519  
#4 XYZ     end   51.51079  0.135120

Answer 2

可能会成功的方法

library(data.table)
dt <- data.table::fread("line_id start_lat  end_lat  start_lng  end_lng
     ABC  51.50995 51.51074 -0.1345093 -0.13519
     XYZ  51.50991 51.51079  0.1351900  0.13512")

dt.melt <- melt( dt, 
                 id.vars = "line_id", 
                 measure.vars = patterns( lon = "_lng$", 
                                          lat = "_lat$" ), 
                 variable.name = "point_id" )

#    line_id point_id        lon      lat
# 1:     ABC        1 -0.1345093 51.50995
# 2:     XYZ        1  0.1351900 51.50991
# 3:     ABC        2 -0.1351900 51.51074
# 4:     XYZ        2  0.1351200 51.51079

library( sf )
library(dplyr)
library(leaflet)
dt.points <- st_as_sf( dt.melt, coords = c("lon", "lat"), crs = 4326)

dt.lines <- dt.points %>%
  group_by( line_id ) %>%
  summarise( geometry = st_combine( geometry ) ) %>%
  st_cast( "LINESTRING" )

leaflet() %>% addTiles() %>% addPolylines( data = dt.lines, popup = ~line_id )

将四个数据框列折叠为两个交错的列

2 个答案: