R 如何将sf空间点列表转换为可路由图
我有一个sf dataframe对象，它有一系列表示公交线路形状的点。我想把这个对象变成一个可路由图，这样我就可以估计从c点到t点的遍历时间 以下是我使用dodgr的尝试，但我不确定我在这里做错了什么：R 如何将sf空间点列表转换为可路由图,r,graph,dodgr,R,Graph,Dodgr,我有一个sf dataframe对象，它有一系列表示公交线路形状的点。我想把这个对象变成一个可路由图，这样我就可以估计从c点到t点的遍历时间 以下是我使用dodgr的尝试，但我不确定我在这里做错了什么： library(dodgr) graph <- weight_streetnet(mydata, wt_profile = "motorcar", type_col="highway" , id_col = "id") 检查osmidx、wt配置文件时出错： 请指定用于加权街道网的类型
library(dodgr)
graph <- weight_streetnet(mydata, wt_profile = "motorcar", type_col="highway" , id_col = "id")
mydata <- structure(list(shape_id = c(52421L, 52421L, 52421L, 52421L, 52421L,
52421L, 52421L, 52421L, 52421L, 52421L, 52421L, 52421L, 52421L,
52421L, 52421L, 52421L, 52421L, 52421L, 52421L, 52421L), length = structure(c(0.191422504106197,
0.191422504106197, 0.191422504106197, 0.191422504106197, 0.191422504106197,
0.191422504106197, 0.191422504106197, 0.191422504106197, 0.191422504106197,
0.191422504106197, 0.191422504106197, 0.191422504106197, 0.191422504106197,
0.191422504106197, 0.191422504106197, 0.191422504106197, 0.191422504106197,
0.191422504106197, 0.191422504106197, 0.191422504106197), units = structure(list(
numerator = "km", denominator = character(0)), class = "symbolic_units"), class = "units"),
geometry = structure(list(structure(c(-46.5623281998182,
-23.5213458001468), class = c("XY", "POINT", "sfg")), structure(c(-46.562221,
-23.52129), class = c("XY", "POINT", "sfg")), structure(c(-46.562121,
-23.521235), class = c("XY", "POINT", "sfg")), structure(c(-46.5620233332577,
-23.5211840000609), class = c("XY", "POINT", "sfg")), structure(c(-46.561925666591,
-23.5211330000609), class = c("XY", "POINT", "sfg")), structure(c(-46.561828,
-23.521082), class = c("XY", "POINT", "sfg")), structure(c(-46.5618098335317,
-23.5212126666783), class = c("XY", "POINT", "sfg")), structure(c(-46.5617916670273,
-23.5213433333544), class = c("XY", "POINT", "sfg")), structure(c(-46.5617735004869,
-23.5214740000284), class = c("XY", "POINT", "sfg")), structure(c(-46.5617553339104,
-23.5216046667004), class = c("XY", "POINT", "sfg")), structure(c(-46.5617371672978,
-23.5217353333702), class = c("XY", "POINT", "sfg")), structure(c(-46.5617190006492,
-23.5218660000379), class = c("XY", "POINT", "sfg")), structure(c(-46.5617008339645,
-23.5219966667036), class = c("XY", "POINT", "sfg")), structure(c(-46.5616826672438,
-23.5221273333671), class = c("XY", "POINT", "sfg")), structure(c(-46.5616645004869,
-23.5222580000284), class = c("XY", "POINT", "sfg")), structure(c(-46.5616463336941,
-23.5223886666877), class = c("XY", "POINT", "sfg")), structure(c(-46.5616281668651,
-23.5225193333449), class = c("XY", "POINT", "sfg")), structure(c(-46.56161,
-23.52265), class = c("XY", "POINT", "sfg")), structure(c(-46.5617355000207,
-23.5226427501509), class = c("XY", "POINT", "sfg")), structure(c(-46.5618610000276,
-23.5226355002012), class = c("XY", "POINT", "sfg"))), class = c("sfc_POINT",
"sfc"), precision = 0, bbox = structure(c(xmin = -46.5623281998182,
ymin = -23.52265, xmax = -46.56161, ymax = -23.521082), class = "bbox"), crs = structure(list(
epsg = 4326L, proj4string = "+proj=longlat +datum=WGS84 +no_defs"), class = "crs"), n_empty = 0L),
id = c("a", "b", "c", "d", "e", "f", "g", "h", "i", "j",
"k", "l", "m", "n", "o", "p", "q", "r", "s", "t"), speed_kmh = c(11,
11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11,
11, 11, 11, 11)), sf_column = "geometry", agr = structure(c(shape_id = NA_integer_,
length = NA_integer_, id = NA_integer_, speed_kmh = NA_integer_
), class = "factor", .Label = c("constant", "aggregate", "identity"
)), row.names = c("1.13", "1.14", "1.15", "1.16", "1.17", "1.18",
"1.19", "1.20", "1.21", "1.22", "1.23", "1.24", "1.25", "1.26",
"1.27", "1.28", "1.29", "1.30", "1.31", "1.32"), class = c("sf",
"data.table", "data.frame"))
我认为您可以通过将数据转换为igraph对象并使用igraph库中的功能来解决此问题。 您需要建立边和顶点以及权重值。 在igraph中，边是表示源和目标两个节点之间连接的链接。在这种情况下，链接是街道，点是节点
library(igraph)
GraphResult <- data.frame(Source = c(NULL),
Target = c(NULL),
weight = c(NULL))
for (i in 1:(dim(mydata)[1] - 1)) {
TempGraphResult <- data.frame(Source = c(0),
Target = c(0),
weight = c(0))
TempGraphResult$Source[1] <- mydata$id[i]
TempGraphResult$Target[1] <- mydata$id[i + 1]
TempGraphResult$weight[1] <- mydata$length[i]
GraphResult <- rbind(GraphResult, TempGraphResult) }
MyIgraph <- graph_from_data_frame(GraphResult)
#In this case works perfectly. But if you have more weight variables and even
#additional variables for the nodes, igraph have functions for constructing the
#igraph object
distances(MyIgraph, "c", "t") #returns 3.254183. Seems correct (0.1914225*17)
SquareMatrix <- distances(MyIgraph)
#*distances() is a function from igraph that performs the routing calculations.
请注意，对于这样的简单问题，igraph通常会更快，因为它使用一组权重计算路线。DoDGR在这个上下文中唯一的真正优势是使用双权重——$DY加权和$D值的能力-这样的路由是根据D$加权计算的，最后的距离根据D.< /P> < P >。如果你想把它包含在一个整洁的工作流程中，你也可以考虑使用SF和TiyGrand之间的混合。后者以tbl_图形类的形式为网络/图形提供了一个整洁的框架，该类是IGRAPHE的子类，因此，您可以在所有IGRAPHE函数中使用tbl_图形对象作为IGRAPHE对象。但是，您可以将节点和边分析为不可分割的，并使用过滤、选择、变异等功能。当然，这些TIBLES还可以包含一个几何体列表列，我们从sf中了解到该列，将地理信息添加到节点和边 这种方法还远远不够完美，改进是非常受欢迎的，但它仍然显示了另一种处理问题的方法 加载库。 图书馆管理员 图书馆 图书馆潮汐记录仪 图书馆记录仪 图书馆单位 与其他答案一样，我们需要在节点之间创建边。现在，我假设这些点只是按字母顺序连接起来的。然而，对于tidygraph方法，我们似乎需要数字ID而不是字符 将数字ID列添加到节点。 节点% 重命名id_chr=id%>% rowid_到_columnid%>% 选择id，id\u chr，所有内容 定义每条边的源节点和每条边的目标节点。 源%slice-n 目标%slice-1 编写一个函数，在源点和目标点的数据帧之间创建线。 pt2l%pullid， 长度=源%>%pulllength， 速度=来源%>%pullspeed\u公里小时， 几何=map2st\U几何资源、st\U几何目标、pt2l %>%st\U as\U sf%>%st\U集合\U crsst\U crsst节点 将时间列添加到边。 边缘% 突变速度=设定单位速度，km/h%>% 突变时间=长度/速度 清理节点数据。 节点% 选择长度、速度和公里小时 使用节点和边创建tbl_图形对象。 不幸的是，将边提供为sf对象对于tidygraph来说是有问题的。 因此，我们必须将其作为一个TIBLE提供。 图表%unlist%>% as_tbl_图 但绘制它也很容易，只需将节点和边导出为sf对象即可保留地理信息
ggplot() +
geom_sf(data = graph %>% activate(edges) %>% as_tibble() %>% st_as_sf(), col = 'darkgrey') +
geom_sf(data = graph %>% activate(nodes) %>% as_tibble() %>% st_as_sf(), col = 'darkgrey', size = 0.5) +
geom_sf(data = path_graph %>% activate(edges) %>% as_tibble() %>% st_as_sf(), lwd = 1, col = 'firebrick') +
geom_sf(data = path_graph %>% activate(nodes) %>% filter(id %in% c(from_node, to_node)) %>% as_tibble() %>% st_as_sf(), size = 2)
一篇r-spatial博客文章可能会出现在这个tidygraph sf方法上 只需一句评论：最少的示例代码使开发人员的工作更轻松。尽管问题很有趣，但期待看到并思考解决方案！“这很有帮助，”奥兰多说。再次感谢！我对这个话题特别感兴趣。更准确地讲，如何使用R计算不同格式的旅行时间和路线。例如，使用与OSM无关的.shp格式的行进行计算会很有用。但这应该是一件容易做的事，而且要有很大的灵活性；许多运输问题和许多运输分析机构都需要这样做。我看到类似ARCGIS的网络分析员在不久的将来为R开发。也许道奇就在那条路上。当你到达一个更高的位置时，请告诉我 不要回答你的问题。
net <- weight_streetnet (x, type_col = "shape_id", id_col = "id", wt_profile = 1)
net$from_id <- mydata$id [as.integer (net$from_id)]
net$to_id <- mydata$id [as.integer (net$to_id)]
net$d_weighted <- as.numeric (mydata$length [1])
dodgr_dists (net, from = "c", to = "t") # 236.0481
net$d <- net$d_weighted
dodgr_dists (net, from = "c", to = "t") # 3.254183
# A tbl_graph: 20 nodes and 19 edges
#
# An undirected simple graph with 1 component
#
# Node Data: 20 x 4 (active)
id id_chr shape_id geometry
<int> <chr> <int> <POINT [°]>
1 1 a 52421 (-46.56233 -23.52135)
2 2 b 52421 (-46.56222 -23.52129)
3 3 c 52421 (-46.56212 -23.52124)
4 4 d 52421 (-46.56202 -23.52118)
5 5 e 52421 (-46.56193 -23.52113)
6 6 f 52421 (-46.56183 -23.52108)
# … with 14 more rows
#
# Edge Data: 19 x 6
from to length speed geometry time
<int> <int> [km] [km/h] <LINESTRING [°]> [h]
1 1 2 0.1914225 11 (-46.56233 -23.52135, -46.56222 -23.5… 0.017402…
2 2 3 0.1914225 11 (-46.56222 -23.52129, -46.56212 -23.5… 0.017402…
3 3 4 0.1914225 11 (-46.56212 -23.52124, -46.56202 -23.5… 0.017402…
# … with 16 more rows
$vpath
$vpath[[1]]
+ 18/20 vertices, from e43a089:
[1] 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
$epath
$epath[[1]]
+ 17/19 edges from e43a089:
[1] 3-- 4 4-- 5 5-- 6 6-- 7 7-- 8 8-- 9 9--10 10--11 11--12 12--13
[11] 13--14 14--15 15--16 16--17 17--18 18--19 19--20
# A tbl_graph: 18 nodes and 17 edges
#
# An undirected simple graph with 1 component
#
# Node Data: 18 x 4 (active)
id id_chr shape_id geometry
<int> <chr> <int> <POINT [°]>
1 3 c 52421 (-46.56212 -23.52124)
2 4 d 52421 (-46.56202 -23.52118)
3 5 e 52421 (-46.56193 -23.52113)
4 6 f 52421 (-46.56183 -23.52108)
5 7 g 52421 (-46.56181 -23.52121)
6 8 h 52421 (-46.56179 -23.52134)
# … with 12 more rows
#
# Edge Data: 17 x 6
from to length speed geometry time
<int> <int> [km] [km/h] <LINESTRING [°]> [h]
1 1 2 0.1914225 11 (-46.56212 -23.52124, -46.56202 -23.5… 0.017402…
2 2 3 0.1914225 11 (-46.56202 -23.52118, -46.56193 -23.5… 0.017402…
3 3 4 0.1914225 11 (-46.56193 -23.52113, -46.56183 -23.5… 0.017402…
# … with 14 more rows
path_graph %>%
activate(edges) %>%
as_tibble() %>%
summarise(total_time = sum(time))
# A tibble: 1 x 1
total_time
[h]
1 0.2958348
ggplot() +
geom_sf(data = graph %>% activate(edges) %>% as_tibble() %>% st_as_sf(), col = 'darkgrey') +
geom_sf(data = graph %>% activate(nodes) %>% as_tibble() %>% st_as_sf(), col = 'darkgrey', size = 0.5) +
geom_sf(data = path_graph %>% activate(edges) %>% as_tibble() %>% st_as_sf(), lwd = 1, col = 'firebrick') +
geom_sf(data = path_graph %>% activate(nodes) %>% filter(id %in% c(from_node, to_node)) %>% as_tibble() %>% st_as_sf(), size = 2)