如何从 R 中的 sf 数据框点计算欧氏距离
How to Calculate Euclidean distance from sf data frame points in R
如何从已从几何列转换为数据框的空间点计算以千米为单位的欧氏距离。 (这些点是从空间数据和多边形质心的空间连接派生的点)
我试过了data_sample <- data_sample %>% mutate(distance= distm(cbind(origin_y,origin_x), cbind(dest_y,dest_x),fun = distHaversine)/1000)
我得到的错误是
.pointsToMatrix(x) 错误:经度 > 360
Below is a sample data frame
data_sample <- data.frame(
origin_x = c(
623613.87,
625678.02,
625678.02,
624359.91,
628136.40,
628136.40,
628136.40,
628136.40,
632329.70
),
origin_y = c(
6438093.66,
6455468.02,
6455468.02,
6449819.06,
6462017.42,
6462017.42,
6462017.42,
6462017.42,
6446947.75
),
dest_x = c(
659627.84,
642136.20,
642136.20,
630395.03,
628422.74,
642136.20,
642136.20,
659627.84,
659627.84
),
dest_y = c(
6473200.36,
6456562.78,
6456562.78,
6451979.98,
6459817.02,
6456562.78,
6456562.78,
6473200.36,
6473200.36)
)
我不是 100% 确定,但看起来像是坐标系的问题。我想 distm()
函数采用以度为单位的纬度和经度值。在将几何列转换为数据框之前,更改数据的坐标系。从包 'sf' 使用 st_crs()
找到当前坐标系并使用 st_transform()
.
更改为类似 WGS 84 的坐标系
有什么理由不使用 sf
内置函数 st_distance
??
library(sf)
# Using your data, but separate data.frames
origin_df <- data.frame(
origin_x = c(623613.87,
625678.02,
625678.02,
624359.91,
628136.40,
628136.40,
628136.40,
628136.40,
632329.70
),
origin_y = c(
6438093.66,
6455468.02,
6455468.02,
6449819.06,
6462017.42,
6462017.42,
6462017.42,
6462017.42,
6446947.75)
)
dest_df <- data.frame(
dest_x = c(
659627.84,
642136.20,
642136.20,
630395.03,
628422.74,
642136.20,
642136.20,
659627.84,
659627.84
),
dest_y = c(
6473200.36,
6456562.78,
6456562.78,
6451979.98,
6459817.02,
6456562.78,
6456562.78,
6473200.36,
6473200.36)
)
# Now convert to sf objects
# Just guessing at the CRS. Replace with the correct one
origin_sf = st_as_sf(origin_df, coords=c("origin_x", "origin_y"), crs="EPSG:32632")
dest_sf <- st_as_sf(dest_df, coords=c("dest_x", "dest_y"), crs="EPSG:32632")
# Now get the distance matrix, (will be in meters, since the CRS units are meters
dist_matrix <- st_distance(origin_sf, dest_sf)
# Dispaly in kilometers
(dist_matrix)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 50.29400 26.15693 26.15693 15.453608 22.249261 26.15693 26.15693 50.29400
[2,] 38.30178 16.49455 16.49455 5.866567 5.142693 16.49455 16.49455 38.30178
[3,] 38.30178 16.49455 16.49455 5.866567 5.142693 16.49455 16.49455 38.30178
[4,] 42.31444 19.01248 19.01248 6.410324 10.791932 19.01248 19.01248 42.31444
[5,] 33.41809 15.02490 15.02490 10.288421 2.218953 15.02490 15.02490 33.41809
[6,] 33.41809 15.02490 15.02490 10.288421 2.218953 15.02490 15.02490 33.41809
[7,] 33.41809 15.02490 15.02490 10.288421 2.218953 15.02490 15.02490 33.41809
[8,] 33.41809 15.02490 15.02490 10.288421 2.218953 15.02490 15.02490 33.41809
[9,] 37.87331 13.73376 13.73376 5.391316 13.449255 13.73376 13.73376 37.87331
[,9]
[1,] 50.29400
[2,] 38.30178
[3,] 38.30178
[4,] 42.31444
[5,] 33.41809
[6,] 33.41809
[7,] 33.41809
[8,] 33.41809
[9,] 37.87331
并回应来自 william-g-k 的附加评论:
dist_km = dist_matrix / 1000
dist_km = as.data.frame(matrix(dist_km, nrow=1))
如何从已从几何列转换为数据框的空间点计算以千米为单位的欧氏距离。 (这些点是从空间数据和多边形质心的空间连接派生的点)
我试过了data_sample <- data_sample %>% mutate(distance= distm(cbind(origin_y,origin_x), cbind(dest_y,dest_x),fun = distHaversine)/1000)
我得到的错误是 .pointsToMatrix(x) 错误:经度 > 360
Below is a sample data frame
data_sample <- data.frame(
origin_x = c(
623613.87,
625678.02,
625678.02,
624359.91,
628136.40,
628136.40,
628136.40,
628136.40,
632329.70
),
origin_y = c(
6438093.66,
6455468.02,
6455468.02,
6449819.06,
6462017.42,
6462017.42,
6462017.42,
6462017.42,
6446947.75
),
dest_x = c(
659627.84,
642136.20,
642136.20,
630395.03,
628422.74,
642136.20,
642136.20,
659627.84,
659627.84
),
dest_y = c(
6473200.36,
6456562.78,
6456562.78,
6451979.98,
6459817.02,
6456562.78,
6456562.78,
6473200.36,
6473200.36)
)
我不是 100% 确定,但看起来像是坐标系的问题。我想 distm()
函数采用以度为单位的纬度和经度值。在将几何列转换为数据框之前,更改数据的坐标系。从包 'sf' 使用 st_crs()
找到当前坐标系并使用 st_transform()
.
有什么理由不使用 sf
内置函数 st_distance
??
library(sf)
# Using your data, but separate data.frames
origin_df <- data.frame(
origin_x = c(623613.87,
625678.02,
625678.02,
624359.91,
628136.40,
628136.40,
628136.40,
628136.40,
632329.70
),
origin_y = c(
6438093.66,
6455468.02,
6455468.02,
6449819.06,
6462017.42,
6462017.42,
6462017.42,
6462017.42,
6446947.75)
)
dest_df <- data.frame(
dest_x = c(
659627.84,
642136.20,
642136.20,
630395.03,
628422.74,
642136.20,
642136.20,
659627.84,
659627.84
),
dest_y = c(
6473200.36,
6456562.78,
6456562.78,
6451979.98,
6459817.02,
6456562.78,
6456562.78,
6473200.36,
6473200.36)
)
# Now convert to sf objects
# Just guessing at the CRS. Replace with the correct one
origin_sf = st_as_sf(origin_df, coords=c("origin_x", "origin_y"), crs="EPSG:32632")
dest_sf <- st_as_sf(dest_df, coords=c("dest_x", "dest_y"), crs="EPSG:32632")
# Now get the distance matrix, (will be in meters, since the CRS units are meters
dist_matrix <- st_distance(origin_sf, dest_sf)
# Dispaly in kilometers
(dist_matrix)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 50.29400 26.15693 26.15693 15.453608 22.249261 26.15693 26.15693 50.29400
[2,] 38.30178 16.49455 16.49455 5.866567 5.142693 16.49455 16.49455 38.30178
[3,] 38.30178 16.49455 16.49455 5.866567 5.142693 16.49455 16.49455 38.30178
[4,] 42.31444 19.01248 19.01248 6.410324 10.791932 19.01248 19.01248 42.31444
[5,] 33.41809 15.02490 15.02490 10.288421 2.218953 15.02490 15.02490 33.41809
[6,] 33.41809 15.02490 15.02490 10.288421 2.218953 15.02490 15.02490 33.41809
[7,] 33.41809 15.02490 15.02490 10.288421 2.218953 15.02490 15.02490 33.41809
[8,] 33.41809 15.02490 15.02490 10.288421 2.218953 15.02490 15.02490 33.41809
[9,] 37.87331 13.73376 13.73376 5.391316 13.449255 13.73376 13.73376 37.87331
[,9]
[1,] 50.29400
[2,] 38.30178
[3,] 38.30178
[4,] 42.31444
[5,] 33.41809
[6,] 33.41809
[7,] 33.41809
[8,] 33.41809
[9,] 37.87331
并回应来自 william-g-k 的附加评论:
dist_km = dist_matrix / 1000
dist_km = as.data.frame(matrix(dist_km, nrow=1))