对于初学者来说，这是我正在使用的表格结构：
df <- structure(list(customer_id = c(353808874L, 69516747L, 357032052L,
307735090L, 307767260L), id = c("8474", "8107",
"1617436", "7698", "1617491"), lon1 = c(-115.032623, -115.155029,
-115.270386, -115.19426, -115.177589), lat1 = c(36.0437202, 36.1366234,
36.1678734, 36.2635803, 36.2218285), lon2 = c(-115.037022076035,
-115.150112230012, -115.27341017806, -115.193072645577, -115.174902476442
), lat2 = c(36.0410001245783, 36.141137860928, 36.1700923382169,
36.2682687632778, 36.2240270452917)), row.names = c(NA, 5L), class = "data.frame")
我尝试使用来自几个不同堆栈溢出问题的信息，但没有一个达到预期的结果。
其中之一是：
earthDist <- function (lon1, lat1, lon2, lat2){
rad <- pi/180
a1 <- lat1 * rad
a2 <- lon1 * rad
b1 <- lat2 * rad
b2 <- lon2 * rad
dlon <- b2 - a2
dlat <- b1 - a1
a <- (sin(dlat/2))^2 + cos(a1) * cos(b1) * (sin(dlon/2))^2
c <- 2 * atan2(sqrt(a), sqrt(1 - a))
R <- 6378.145
d <- R * c
return(d)
}
earthDist(lon[1], lat[1], lon, lat)
但我无法让它产生我正在寻找的输出。我当然不执着于它，所以如果有人有更有效的东西，我全听！
编辑：我预期的结果相当简单。只有三列，distance_between代表lon/lat1和lon/lat2之间的距离：
+-------------+----+------------------+
| customer_id | id | distance_between |
+-------------+----+------------------+
这很容易用 geosphere 包中的distGeo函数(类似于上面的函数(来解决：
library(geosphere)
#calculate distances in meters
df$distance<-distGeo(df[,c("lon1", "lat1")], df[,c("lon2", "lat2")])
#remove columns
df[, -c(3:6)]
customer_id id distance
1 353808874 8474 498.2442
2 69516747 8107 668.4088
3 357032052 1617436 366.9541
4 307735090 7698 531.0785
5 307767260 1617491 343.3051