我在r中编写一个脚本来确定距离物体最近的气象站。除此之外，当运行我的代码时，它会自动返回气象站3（由Table Station中的Cam指示）最接近所有不是这种情况的对象。任何想法为什么会这样？如果可能的话，我也想获得电台名称，而不仅仅是索引。
代码：
earth.dist <- function (long1, lat1, long2, lat2)
{
rad <- pi/180
a1 <- lat1 * rad
a2 <- long1 * rad
b1 <- lat2 * rad
b2 <- long2 * 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)
}
for (i in 1:length(Object$Lat))
{
for (j in 1:length(Station$Lat))
{
a[j] <- earth.dist(Station$Lon[j], Station$Lat[j], Object$Lon[i], Object$Lat[i])
}
index <- which(min(a) %in% a)
Object$Station[i] = Station$Station[index]
}
站台：
Station Lat Lon
SF 37.7749 -122.4194
CH 41.8781 -87.6298
Cam 52.2053 -0.1218
对象表：
Object Lat Lon
1 38.983 -123.092
2 36.941 -121.767
3 36.121 -121.084
4 38.415 -121.787
5 36.854 -121.362
6 38.651 -121.218
7 37.314 -120.386
8 36.158 -119.8514
9 38.599 -121.54
10 35.335 -120.734
11 34.841 -120.212
12 38.004 -122.02
13 37.599 -122.052
14 38.0903 -122.5267
15 37.664 -121.885
16 51.50853 -0.076132
答案 0 :(得分：1)
我们不需要重新发明带有r / t地球距离计算的车轮。
数据：
read.table(text="Station Lat Lon
SF 37.7749 -122.4194
CH 41.8781 -87.6298
Cam 52.2053 -0.1218", stringsAsFactors=FALSE, header=TRUE) -> stations
read.table(text="Object Lat Lon
1 38.983 -123.092
2 36.941 -121.767
3 36.121 -121.084
4 38.415 -121.787
5 36.854 -121.362
6 38.651 -121.218
7 37.314 -120.386
8 36.158 -119.8514
9 38.599 -121.54
10 35.335 -120.734
11 34.841 -120.212
12 38.004 -122.02
13 37.599 -122.052
14 38.0903 -122.5267
15 37.664 -121.885
16 51.50853 -0.076132", stringsAsFactors=FALSE, header=TRUE) -> objs
代码：
library(geosphere)
library(tidyverse)
find_closest_station <- function(lon, lat) {
mutate(stations, dist=map2_dbl(Lon, Lat, ~distHaversine(c(lon, lat), c(.x, .y)))) %>%
top_n(-1, wt=dist) %>%
.$Station
}
mutate(objs, wx_st=map2_chr(Lon, Lat, find_closest_station))
## Object Lat Lon wx_st
## 1 1 38.98300 -123.092000 SF
## 2 2 36.94100 -121.767000 SF
## 3 3 36.12100 -121.084000 SF
## 4 4 38.41500 -121.787000 SF
## 5 5 36.85400 -121.362000 SF
## 6 6 38.65100 -121.218000 SF
## 7 7 37.31400 -120.386000 SF
## 8 8 36.15800 -119.851400 SF
## 9 9 38.59900 -121.540000 SF
## 10 10 35.33500 -120.734000 SF
## 11 11 34.84100 -120.212000 SF
## 12 12 38.00400 -122.020000 SF
## 13 13 37.59900 -122.052000 SF
## 14 14 38.09030 -122.526700 SF
## 15 15 37.66400 -121.885000 SF
## 16 16 51.50853 -0.076132 Cam
答案 1 :(得分：0)
基本上，你的which()逻辑有点偏差。为了说明，下面始终返回TRUE，因此which()返回1.
which("R" %in% LETTERS)
# [1] 1
但是，在第18个字母下面只有TRUE：
which(LETTERS == "R")
# [1] 18
所以简单地替换
index <- which(min(a) %in% a)
使用：
index <- which(a == min(a))
此外，考虑应用解决方案，以防您需要保留在基础R中，因为您可以将嵌套for替换为sapply（传入两个向量），并使用apply生成索引：
dist.matrix <- sapply(seq(nrow(Station)), function(x, y)
earth.dist(Station$Lon[x], Station$Lat[x],
Object$Lon[y], Object$Lat[y]),
seq(nrow(Object)))
Object$Station <- apply(dist.matrix, 1, function(i) Station$Station[which(i == min(i))])
Object
# Object Lat Lon Station
# 1 1 38.98300 -123.092000 SF
# 2 2 36.94100 -121.767000 SF
# 3 3 36.12100 -121.084000 SF
# 4 4 38.41500 -121.787000 SF
# 5 5 36.85400 -121.362000 SF
# 6 6 38.65100 -121.218000 SF
# 7 7 37.31400 -120.386000 SF
# 8 8 36.15800 -119.851400 SF
# 9 9 38.59900 -121.540000 SF
# 10 10 35.33500 -120.734000 SF
# 11 11 34.84100 -120.212000 SF
# 12 12 38.00400 -122.020000 SF
# 13 13 37.59900 -122.052000 SF
# 14 14 38.09030 -122.526700 SF
# 15 15 37.66400 -121.885000 SF
# 16 16 51.50853 -0.076132 Cam
甚至基本R的outer也可以使用转置的 dist.matrix 与上面的sapply输出相比较，因此后者apply按列边距运行：
dist.mat <- outer(seq(nrow(Station)), seq(nrow(Object)),
function(x, y) earth.dist(Station$Lon[x], Station$Lat[x],
Object$Lon[y], Object$Lat[y]))
Object$Station <- apply(dist.mat, 2, function(i) Station$Station[which(i == min(i))])