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我有 2 个数据框（df1 和 df2），由三列组成；x 坐标，y 坐标，类别（有 5 个级别 AE）。所以我基本上有两组点数据，每个点都分配给一个类别

例如

X    Y    Cat
1    1.5  A
2    1.5  B
3.3  1.9  C

等等......（虽然我的两个数据框都有100个点）

我想从第二个数据帧（df2）中为我的第一个数据帧（df1）中的每个点找到相同类别的最近邻居。

我在包 spatstat 中使用了 nncross 来为 df1 中的每个点与 df2 找到最近的邻居，然后列出这些距离中的每一个，如下所示；

# Convert the dataframes to ppp objects

df1.ppp <- ppp(df1$X,df1$Y,c(0,10),c(0,10),marks=df1$Cat)
df2.ppp <- ppp(df2$X,df2$Y,c(0,10),c(0,10),marks=df2$Cat)

# Produce anfrom output that lists the distance from each point in df1 to its nearest neighbour in df2

out<-nncross(X=df1.ppp,Y=df2.ppp,what=c("dist","which"))

但是我正在努力弄清楚如何使用存储在 ppp 对象中的类别标签（由标记定义）来从同一类别中找到最近的邻居。我相信它应该是相当直截了当的，但如果有人有任何建议或任何替代方法来达到相同的结果，我将不胜感激。

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2 回答 2

0

首先要使用一些人工数据：

library(spatstat)

# Artificial data similar to the question
set.seed(42)
X1 <- rmpoint(100, win = square(10), types = factor(LETTERS[1:5]))
X2 <- rmpoint(100, win = square(10), types = factor(LETTERS[1:5]))

然后是一个简单的解决方案（但它会丢失 id 信息）：

# Separate patterns for each type:
X1list <- split(X1)
X2list <- split(X2)

# For each point in X1 find nearest neighbour of same type in X2:
out <- list()
for(i in 1:5){
  out[[i]] <- nncross(X1list[[i]], X2list[[i]], what=c("dist","which"))
}

最后，一个丑陋的解决方案可以恢复邻居的 id：

# Make separate marks for pattern 1 and 2 and collect into one pattern
marks(X1) <- factor(paste0(marks(X1), "1"))
marks(X2) <- factor(paste0(marks(X2), "2"))
X <- superimpose(X1, X2)

# For each point get the nearest neighbour of each type from both X1 and X2
# (both dist and index)
nnd <- nndist(X, by = marks(X))
nnw <- nnwhich(X, by = marks(X))

# Type to look for. I.e. the mark with 1 and 2 swapped
# (with 0 as intermediate step)
type <- marks(X)
type <- gsub("1", "0", type)
type <- gsub("2", "1", type)
type <- gsub("0", "2", type)

# Result
rslt <- cbind(as.data.frame(X), dist = 0, which = 0)
for(i in 1:nrow(rslt)){
  rslt$dist[i] <- nnd[i, type[i]]
  rslt$which[i] <- nnw[i, type[i]]
}

# Separate results
rslt1 <- rslt[1:npoints(X1),]
rslt2 <- rslt[npoints(X1) + 1:npoints(X2),]
rslt1$which <- rslt1$which - npoints(X1)
于 2016-02-04T22:13:39.130 回答
0

我还尝试了另一种方法来解决这个问题，但是通过使用 geosphere 包从我的原始数据帧创建一个距离矩阵，并找到了一种非常简单的方法来解决这个问题。

# load geosphere library 
library("geosphere")

#create a distance matrix between all points in the 2 dataframes
dist<-distm(df1[,c('X','Y')],df2[,c('X','Y')])

# find the nearest neighbour to each point
df1$nearestneighbor <- apply(dist,1,min)

# create a distance matrix where only the distances between points of the same category are recorded
sameCat <- outer(df1$Cat, df2$Cat, "!=")
dist2 <- dist + ifelse(sameCat, Inf, 0)

# find the nearest neighbour of the same category
df1$closestmatch <- apply(dist2,1,min)
于 2016-02-05T10:07:29.057 回答