Question

我有两个数据帧，两个都包含纬度和经度坐标。第一个数据帧是事件的观察，其中记录了位置和时间。第二个数据框是地理要素，其中记录了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，其中每一行都是df2中最接近的特征。

我处理了这个问题How to assign several names to lat-lon observations，但无法弄清楚如何将其与我的数据相匹配

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

Answer 1

使用st_distance包中的sf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, ])

Answer 2

基于此answer你可以做到

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),])

给出了：

#     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

基于最短地理距离匹配数据帧

2 个答案: