r语言 - 基于最短地理距离匹配数据框



我有两个数据帧,都包含纬度和经度坐标。第一个数据帧是对事件的观测,其中记录了位置和时间。第二个数据帧是地理要素,用于记录有关要素的位置和信息。

my_df_1 <- structure(list(START_LAT = c(-33.15, -35.6, -34.08333, -34.13333, 
-34.31667, -47.38333, -47.53333, -34.08333, -47.38333, -47.15
), START_LONG = c(163, 165.18333, 162.88333, 162.58333, 162.76667, 
148.98333, 148.66667, 162.9, 148.98333, 148.71667)), row.names = c(1175L, 
528L, 1328L, 870L, 672L, 707L, 506L, 981L, 756L, 210L), class = "data.frame", .Names = c("START_LAT", 
"START_LONG"))
my_df_2 <- structure(list(latitude = c(-42.7984, -34.195, -49.81, -35.417, 
-28.1487, -44.657, -42.7898, -36.245, -39.1335, -31.8482), longitude = c(179.9874, 
179.526, -176.68, 178.765, -168.0314, 174.695, -179.9873, 177.7873, 
-170.0583, 173.2424), depth_top = c(935L, 2204L, 869L, 1973L, 
4750L, 555L, 894L, 1500L, 4299L, 1303L)), row.names = c(580L, 
1306L, 926L, 1102L, 60L, 1481L, 574L, 454L, 1168L, 144L), class = "data.frame", .Names = c("latitude", 
"longitude", "depth_top"))

我需要做的是,对于 df1 中的每个观察,我需要找出 df2 中的哪个特征在地理上最接近。理想情况下,我会在 df1 上附加一个新列,每一行都是 df2 中最接近的特征。

我解决了这个问题如何为纬度观测值分配多个名称,但无法弄清楚如何将其与我的数据匹配

真正的数据帧有 1000 行,这就是为什么我不能手动执行此操作的原因

使用sf包中的st_distance的解决方案。my_df_final是最终输出。

# Load packages
library(tidyverse)
library(sp)
library(sf)
# Create ID for my_df_1 and my_df_2 based on row id
# This step is not required, just help me to better distinguish each point
my_df_1 <- my_df_1 %>% mutate(ID1 = row.names(.))
my_df_2 <- my_df_2 %>% mutate(ID2 = row.names(.))
# Create spatial point data frame
my_df_1_sp <- my_df_1
coordinates(my_df_1_sp) <- ~START_LONG + START_LAT
my_df_2_sp <- my_df_2
coordinates(my_df_2_sp) <- ~longitude + latitude
# Convert to simple feature
my_df_1_sf <- st_as_sf(my_df_1_sp)
my_df_2_sf <- st_as_sf(my_df_2_sp)
# Set projection based on the epsg code
st_crs(my_df_1_sf) <- 4326
st_crs(my_df_2_sf) <- 4326
# Calculate the distance
m_dist <- st_distance(my_df_1_sf, my_df_2_sf)
# Filter for the nearest
near_index <- apply(m_dist, 1, order)[1, ]
# Based on the index in near_index to select the rows in my_df_2
# Combine with my_df_1
my_df_final <- cbind(my_df_1, my_df_2[near_index, ])

基于这个答案,你可以做

library(geosphere)
mat <- distm(my_df_1[2:1], my_df_2[2:1], fun = distVincentyEllipsoid)
cbind(my_df_1, my_df_2[max.col(-mat),])

这给了:

#     START_LAT START_LONG  ID1 latitude longitude depth_top  ID2
#10   -33.15000   163.0000 1175 -31.8482  173.2424      1303  144
#10.1 -35.60000   165.1833  528 -31.8482  173.2424      1303  144
#10.2 -34.08333   162.8833 1328 -31.8482  173.2424      1303  144
#10.3 -34.13333   162.5833  870 -31.8482  173.2424      1303  144
#10.4 -34.31667   162.7667  672 -31.8482  173.2424      1303  144
#6    -47.38333   148.9833  707 -44.6570  174.6950       555 1481
#6.1  -47.53333   148.6667  506 -44.6570  174.6950       555 1481
#10.5 -34.08333   162.9000  981 -31.8482  173.2424      1303  144
#6.2  -47.38333   148.9833  756 -44.6570  174.6950       555 1481
#6.3  -47.15000   148.7167  210 -44.6570  174.6950       555 1481

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