Question

我有两个独立的数据集。一个包含参与者的位置，另一个包含测量站的位置和相应的值，在不同的时间点。下面我生成样本数据集。

# dataset of value
yearmon <- c("Jan 1996","Jan 1996","Jan 1996","Jan 1996","Jan 1996","Jan 1996",
         "Feb 1996","Feb 1996","Feb 1996","Feb 1996","Feb 1996","Feb 1996",
         "Mar 1996","Mar 1996","Mar 1996","Mar 1996","Mar 1996","Mar 1996",
         "Apr 1996","Apr 1996","Apr 1996","Apr 1996","Apr 1996","Apr 1996",
         "May 1996","May 1996","May 1996","May 1996","May 1996","May 1996",
         "Jun 1996","Jun 1996","Jun 1996","Jun 1996","Jun 1996","Jun 1996")

lon <- c(114.1592, 114.1294, 114.1144, 114.0228, 113.9763, 113.9431)

lat <- c(22.35694, 22.31306, 22.33000, 22.37167, 22.37639, 22.45111)

STN <- c("A","B","C","D","E","F")

value <- runif(n=36, min=10, max=20)

df<- data.frame(STN,lon,lat)
df<- rbind(df,df,df,df,df,df)
df <- cbind(df,yearmon,value)
df$value[df$value < 12] <- NA


# dataset of participant location
id <- c(1,2,3,4)
lon.p <- c(114.3608, 114.1850, 114.1581, 114.1683)
lat.p <- c(22.44500, 22.33000, 22.28528, 22.37167)
participant <- data.frame(id,lon.p,lat.p)

＃

样本数据集如下。我想计算每个站点（A-F）和每个参与者（1-4）在每个时间点（yearmon）之间的距离。并将特定时间点的值分配给特定参与者。我无法先将参与者分配到工作站，因为工作站的位置可能会在不同的时间点发生变化（尽管样本数据集中没有变化）

即。如果参与者1在1996年1月离A站最近，那么他/她应该分配值17.03357。

我更喜欢大圆距离，也许可以使用这样的脚本来计算： rdist.earth（location1，location2，miles = FALSE，R = 6371）

head(df,10)
   STN      lon      lat  yearmon    value
1    A 114.1592 22.35694 Jan 1996 17.03357
2    B 114.1294 22.31306 Jan 1996       NA
3    C 114.1144 22.33000 Jan 1996 17.98293
4    D 114.0228 22.37167 Jan 1996 15.98854
5    E 113.9763 22.37639 Jan 1996 16.78647
6    F 113.9431 22.45111 Jan 1996 18.89551
7    A 114.1592 22.35694 Feb 1996       NA
8    B 114.1294 22.31306 Feb 1996 19.90123
9    C 114.1144 22.33000 Feb 1996 17.88482
10   D 114.0228 22.37167 Feb 1996 13.80029

participant
  id    lon.p    lat.p
1  1 114.3608 22.44500
2  2 114.1850 22.33000
3  3 114.1581 22.28528
4  4 114.1683 22.37167

最后，我认为这是我想要回归的。（但填写了值）

   id   lon.p     lat.p Apr 1996 Feb 1996 Jan 1996 Jun 1996 Mar 1996 May 1996
1   1 114.3608 22.44500 
2   2 114.1850 22.33000 
3   3 114.1581 22.28528 
4   4 114.1683 22.37167

谢谢。

Answer 1

这是一种通过几个步骤完成此操作的方法。请注意，我创建了naive_dist函数，就像距离指标的占位符一样。该函数来自here。

naive_dist <- function(long1, lat1, long2, lat2) {
  R <- 6371 # Earth mean radius [km]
  d <- acos(sin(lat1)*sin(lat2) + cos(lat1)*cos(lat2) * cos(long2-long1)) * R
  return(d) # Distance in km
}

dist_by_id <- by(participant, participant$id, FUN = function(x) 
  #you would use your distance metric here
  naive_dist(long1 = x$lon.p, long2 = df$lon, lat1 = x$lat.p, lat2 = df$lat)
  )

#function to find the min for each yearmon, by id
find_min <- function(id, data, by_data){
  data$dist_column = by_data[[id]]
  by(data, data$yearmon, FUN = function(x) x[which.min(x$dist_column),]$value)
}
#initialize
participant[,4:9] = 0
names(participant)[4:9] = as.character(unique(df$yearmon))
#use a for loop to fill in the values
for(i in 1:4){
 participant[i,4:9] = stack(find_min(id = i, data = df, by_data = dist_by_id))[,1] 
}

participant

  id    lon.p    lat.p Jan 1996 Feb 1996 Mar 1996 Apr 1996 May 1996 Jun 1996
1  1 114.3608 22.44500 17.36620 18.88409 19.53951 19.35646 13.00518 18.45556
2  2 114.1850 22.33000 17.36620 18.88409 19.53951 19.35646 13.00518 18.45556
3  3 114.1581 22.28528 18.57447 13.85192 17.52038       NA 16.14562 18.06435
4  4 114.1683 22.37167 17.36620 18.88409 19.53951 19.35646 13.00518 18.45556

显然，一旦你改变距离度量，这些结果可能会改变。

或者，这是一个使用dplyr的选项，我倾向于选择此解决方案，因为它可能更具性能。

library(dplyr)
df2 <- merge(df, participant, all = T) #merge the df's
#calculate distance
df2$distance <- naive_dist(long1 = df2$lon, lat1 = df2$lat,
                           long2 = df2$lon.p, lat2 = df2$lat.p)


df3 <- df2 %>%
  group_by(yearmon, id) %>%
  filter(distance == min(distance)) %>%
  select(id, yearmon, value)

participant2 <- participant
participant2[,4:9] <- 0
names(participant2)[4:9] <- as.character(unique(df$yearmon))

for(i in 1:4){
  participant2[i,4:9] = c(subset(df3, id == i)$value)
}

participant2

  id    lon.p    lat.p Jan 1996 Feb 1996 Mar 1996 Apr 1996 May 1996 Jun 1996
1  1 114.3608 22.44500 19.53951 18.88409 13.00518 17.36620 18.45556 19.35646
2  2 114.1850 22.33000 19.53951 18.88409 13.00518 17.36620 18.45556 19.35646
3  3 114.1581 22.28528 17.52038 13.85192 16.14562 18.57447 18.06435       NA
4  4 114.1683 22.37167 19.53951 18.88409 13.00518 17.36620 18.45556 19.35646

如何计算距离并返回具有最短距离的特定变量的值？

1 个答案: