基于最短地理距离匹配数据帧

Question

我有两个数据帧，它们都包含经度和纬度坐标。第一个数据框是对事件的观察，其中记录了位置和时间。第二个数据框是地理特征，其中记录了关于该特征的位置和信息。

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，我会得到一个新的列添加到df1，其中每行是距离df2最近的功能。

我通过这个问题How to assign several names to lat-lon observations工作，但无法弄清楚如何匹配到我的数据

真正dataframes有行1000，这就是为什么我不能用手工做这个

Answer 1

一该解决方案使用来自sf包的st_distance。 my_df_final是最终输出。在此基础上answer

# 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, ]) 

Answer 2

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),]) 

其中给出：

#  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