计算组内所有可能点之间的距离的方法。

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英文:

How to calculate distance between all possible combinations of points within a group

问题

I have a sf object with 68 observation. In each location, I have 4 different points, with different PointIDs(SchlagID) but the same LocationIDs(FieldID). I want to now calculate the distance between each point by location in a way that there are a total of 6 values for each location (including the diagonal distances, assuming the points are in a square).

Till now, I have been able to calculate 4 values but not sure how to calculate the diagonal distances as well.

Here is what the data looks like

Here is the code I am working on:

  1. pdist.df = data.frame(trapID=NA,LS=NA,dist=NA)
  2. list.lsFeat = 1
  3. counter=1
  5. for(i in 1:nrow(raps_filtered_sf)){
  6.   
  7.   this.trap=raps_filtered_sf[i,]
  8.   
  9.   for(l in list.lsFeat){
  10.     
  11.     this.feat = raps_filtered_sf %>%
  12.       filter(FieldID==this.trap$FieldID)
  13.     
  14.     #ind <- st_nearest_feature(this.trap, this.feat)
  15.     dis <- st_distance(this.trap, this.feat)
  16.     print(paste0("trap: ", this.trap$SchlagID, "//LS: ", this.feat$SchlagID, "//Dist: ", dis))
  17.     
  18.     pdist.df[counter,] = c(trapID=this.trap$SchlagID, LS=l, dist=dis)
  19.     counter=counter+1
  20.   }
  21. }
英文:

I have a sf object with 68 observation. In each location I have 4 different points, with different PointIDs(SchlagID) but same LocationIDs(FieldID). I want to now calculate the distance between each point by location in a way that there a total of 6 values for each location (including the diagonal distances, assuming the points are in a square)

Till now I have been able to calculate 4 values but not sure how to calculate the diagonal distances as well.

Here is what the data looks like

Here is the code I am working on

  1. pdist.df = data.frame(trapID=NA,LS=NA,dist=NA)
  2. list.lsFeat = 1
  3. counter=1
  6. for(i in 1:nrow(raps_filtered_sf)){
  7.     
  8.   
  9.   this.trap=raps_filtered_sf[i,]
  10.   
  11.   for(l in list.lsFeat){
  12.     
  13.     this.feat = raps_filtered_sf %&gt;%
  14.       filter(FieldID==this.trap$FieldID)
  15.     
  16.     #ind &lt;- st_nearest_feature(this.trap, this.feat)
  17.     dis &lt;- st_distance(this.trap, this.feat)
  18.     print(paste0(&quot;trap: &quot;, this.trap$SchlagID, &quot;//LS: &quot;, this.feat$SchlagID, &quot;//Dist: &quot;, dis))
  19.     
  20.     pdist.df[counter,] = c(trapID=this.trap$SchlagID, LS=l, dist=dis)
  21.     counter=counter+1
  22.   }
  23. }

答案1

得分: 1

我相信你正在寻找一个 sf::st_distance() 调用。当在单个 sf 数据框上使用它时，它会输出一个距离矩阵。

稍微调整一下名称可以更容易理解，但不是严格要求的。

举个例子，考虑下面这段代码，基于3个半随机的北卡罗来纳州城市（因为我深爱着附带 {sf} 包的 nc.shp 文件）：

  1. library(sf)
  2. library(dplyr)
  4. # 3 semi rancom cities in NC (because I *deeply love* the nc.shp file)
  5. cities <- data.frame(name = c("Raleigh", "Greensboro", "Wilmington"),
  6.                      x = c(-78.633333, -79.819444, -77.912222),
  7.                      y = c(35.766667, 36.08, 34.223333)) %>% 
  8.   st_as_sf(coords = c("x", "y"), crs = 4326)
  10. # prepare a distance matrix
  11. mtx <- st_distance(cities)
  13. # give the rows & cols meaningful names
  14. colnames(mtx) <- cities$name
  15. rownames(mtx) <- cities$name  
  17. # see my work, and find it good...
  18. mtx
  20. # Units: [m]
  21. #             Raleigh Greensboro Wilmington
  22. # Raleigh         0.0   112342.9   183751.2
  23. # Greensboro 112342.9        0.0   269595.9
  24. # Wilmington 183751.2   269595.9        0.0

希望这能帮助你！

英文:

I believe you are looking for a sf::st_distance() call. When used on a single sf data frame it will output a distance matrix.

A little finetuning of names makes it easier to interpret, but is not strictly required.

For an example consider this piece of code, built on top of 3 semi-random North Carolina cities (because I am deeply in love with the nc.shp file that ships with {sf}):

  1. library(sf)
  2. library(dplyr)
  4. # 3 semi rancom cities in NC (because I *deeply love* the nc.shp file)
  5. cities &lt;- data.frame(name = c(&quot;Raleigh&quot;, &quot;Greensboro&quot;, &quot;Wilmington&quot;),
  6.                      x = c(-78.633333, -79.819444, -77.912222),
  7.                      y = c(35.766667, 36.08, 34.223333)) %&gt;% 
  8.   st_as_sf(coords = c(&quot;x&quot;, &quot;y&quot;), crs = 4326)
  10. # prepare a distance matrix
  11. mtx &lt;- st_distance(cities)
  13. # give the rows &amp; cols meaningful names
  14. colnames(mtx) &lt;- cities$name
  15. rownames(mtx) &lt;- cities$name  
  17. # see my work, and find it good...
  18. mtx
  20. # Units: [m]
  21. #             Raleigh Greensboro Wilmington
  22. # Raleigh         0.0   112342.9   183751.2
  23. # Greensboro 112342.9        0.0   269595.9
  24. # Wilmington 183751.2   269595.9        0.0

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  • 本文由 发表于 2023年4月13日 21:48:22
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  • r
  • r-sf
  • r-sp
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