使用spatialRF::rf函数时distance.matrix报错，如何正确计算？

问题背景 #

使用spatialRF包执行回归任务时，参考示例预计算distance.matrix并传入spatialRF::rf函数，出现报错：

Error in diag<-(tmp, value = NA): only matrix diagonals can be replaced

自身数据集无x、y空间坐标列，原示例代码及数据集如下：

原示例代码 #

library(spatialRF)

#加载训练数据
data(block.data)

#响应变量和预测变量名称
dependent.variable.name <- "ntl"
predictor.variable.names <- colnames(block.data)[2:4]

#样本坐标
xy <- block.data[, c("x", "y")]

#距离矩阵
distance.matrix <- dist(subset(block.data, select = -c(x, y)))

#用于复现的随机种子
random.seed <- 1

model.non.spatial <- spatialRF::rf(
  data = block.data,
  dependent.variable.name = dependent.variable.name,
  predictor.variable.names = predictor.variable.names,
  distance.matrix = distance.matrix,
  distance.thresholds = 0,
  xy = xy,
  seed = random.seed,
  verbose = FALSE)

自身数据集 #

block.data = structure(list(ntl = c(11.4058170318604, 13.7000455856323, 16.0420398712158, 
17.4475727081299, 26.263370513916, 30.658130645752, 19.8927211761475, 
20.917688369751, 23.7149887084961, 25.2641334533691), pop = c(111.031448364258, 
145.096557617188, 166.351989746094, 193.804962158203, 331.787200927734, 
382.979248046875, 237.971466064453, 276.575958251953, 334.015289306641, 
345.376617431641), tirs = c(35.392936706543, 34.4172630310059, 
33.7765464782715, 35.3224639892578, 40.4262886047363, 39.6619148254395, 
38.6306076049805, 36.752326965332, 37.2010040283203, 36.1100578308105
), agbh = c(1.15364360809326, 0.177780777215958, 0.580717206001282, 
0.647109687328339, 3.84336423873901, 5.6310133934021, 2.10894227027893, 
3.9533429145813, 2.7016019821167, 4.36041164398193), lc = c(40L, 
40L, 40L, 126L, 50L, 50L, 50L, 50L, 40L, 50L)), class = "data.frame", row.names = c(NA, 
-10L))

报错原因 #

dist()函数返回的是dist类的距离对象，而非标准矩阵格式。spatialRF::rf函数内部会尝试修改距离矩阵的对角线（设为NA），但dist对象不支持这种操作，因此触发报错。此外，自身数据集无x、y坐标，不应传入xy参数。

解决方案 #

1. 将dist对象转换为标准矩阵格式，使用as.matrix()函数
2. 移除xy参数（因无空间坐标）

正确计算distance.matrix的方式 #

根据需求选择：

• 特征空间距离：基于数据集的特征变量计算距离，示例：

  # 基于所有非响应变量计算特征距离
  distance.matrix <- as.matrix(dist(subset(block.data, select = -ntl)))
  # 或仅基于预测变量计算
  distance.matrix <- as.matrix(dist(block.data[, predictor.variable.names]))
  

• 空间距离：若后续补充了空间坐标，可基于x、y计算空间距离：

  # 假设已有xy坐标列
  distance.matrix <- as.matrix(dist(block.data[, c("x", "y")]))
  

修正后的完整代码 #

library(spatialRF)

# 加载数据集
block.data = structure(list(ntl = c(11.4058170318604, 13.7000455856323, 16.0420398712158, 
17.4475727081299, 26.263370513916, 30.658130645752, 19.8927211761475, 
20.917688369751, 23.7149887084961, 25.2641334533691), pop = c(111.031448364258, 
145.096557617188, 166.351989746094, 193.804962158203, 331.787200927734, 
382.979248046875, 237.971466064453, 276.575958251953, 334.015289306641, 
345.376617431641), tirs = c(35.392936706543, 34.4172630310059, 
33.7765464782715, 35.3224639892578, 40.4262886047363, 39.6619148254395, 
38.6306076049805, 36.752326965332, 37.2010040283203, 36.1100578308105
), agbh = c(1.15364360809326, 0.177780777215958, 0.580717206001282, 
0.647109687328339, 3.84336423873901, 5.6310133934021, 2.10894227027893, 
3.9533429145813, 2.7016019821167, 4.36041164398193), lc = c(40L, 
40L, 40L, 126L, 50L, 50L, 50L, 50L, 40L, 50L)), class = "data.frame", row.names = c(NA, 
-10L))

# 响应变量和预测变量名称
dependent.variable.name <- "ntl"
predictor.variable.names <- colnames(block.data)[2:4]

# 计算特征空间距离矩阵（转成矩阵格式）
distance.matrix <- as.matrix(dist(block.data[, predictor.variable.names]))

# 随机种子
random.seed <- 1

# 训练模型（移除xy参数）
model.non.spatial <- spatialRF::rf(
  data = block.data,
  dependent.variable.name = dependent.variable.name,
  predictor.variable.names = predictor.variable.names,
  distance.matrix = distance.matrix,
  distance.thresholds = 0,
  seed = random.seed,
  verbose = FALSE
)

# 查看模型结果
summary(model.non.spatial)

内容的提问来源于stack exchange，提问作者Nikos