如何使用具有 MRF 平滑和邻域结构的 GAM 预测测试数据？

Question

我在对新（测试）数据集上的 mgcv::gam（训练）模型使用 predict() 函数时遇到问题。问题的出现是由于 mrf 平滑我已经整合以解释我的数据的空间性质。

我使用以下调用来创建我的 GAM 模型

## Run GAM with MRF
m <- gam(crime ~ s(district,k=nrow(traindata),
                 bs ='mrf',xt=list(nb=nbtrain)), #define MRF smooth
     data = traindata,
     method = 'REML', 
     family = scat(), #fit scaled t distribution
     gamma = 1.4
)

其中我使用邻域结构预测因变量 crime，在平滑项参数 xt 中解析到模型中。邻域结构作为我使用 poly2nb() 函数创建的 nb 对象出现。

现在，如果我想在新的测试数据集上使用 predict()，我不知道如何将相应的邻域结构传递到调用中。仅提供新数据

pred <- predict.gam(m,newdata=testdata)

抛出以下错误：

Error in predict.gam(m, newdata = testdata) :
7, 16, 20, 28, 35, 36, 37, 43 not in original fit

下面是使用直接从 R 中调用的 Columbus 数据集完整重现的错误：

#ERROR REPRODUCTION

## Load packages
require(mgcv)
require(spdep)
require(dplyr)

## Load Columbus Ohio crime data (see ?columbus for details and credits)
data(columb.polys) #Columbus district shapes list
columb.polys <- lapply(columb.polys,na.omit) #omit NAs (unfortunate problem with the Columbus sample data)
data(columb) #Columbus data frame

df <- data.frame(district=numeric(0),x=numeric(0),y= numeric(0)) #Create empty df to store x, y and IDs for each polygon

## Extract x and y coordinates from each polygon and assign district ID
for (i in 1:length(columb.polys)) {
  district <- i-1
  x <- columb.polys[[i]][,1]
  y <- columb.polys[[i]][,2]
  df <- rbind(df,cbind(district,x,y)) #Save in df data.frame
}

## Convert df into SpatialPolygons
sp <- df %>%
       group_by(district) %>%
       do(poly=select(., x, y) %>%Polygon()) %>%
       rowwise() %>%
       do(polys=Polygons(list(.$poly),.$district)) %>%
       {SpatialPolygons(.$polys)}

## Merge SpatialPolygons with data
spdf <- SpatialPolygonsDataFrame(sp,columb)

## Split into training and test sample (80/20 ratio)
splt <- sample(1:2,size=nrow(spdf),replace=TRUE,prob=c(0.8,0.2))
train <- spdf[splt==1,] 
test <- spdf[splt==2,]

## Prepapre both samples and create NB objects
traindata <- train@data #Extract data from SpatialPolygonsDataFrame
testdata <- test@data
traindata <- droplevels(as(train, 'data.frame')) #Drop levels
testdata <- droplevels(as(test, 'data.frame'))
traindata$district <- as.factor(traindata$district) #Factorize
testdata$district <- as.factor(testdata$district)
nbtrain <- poly2nb(train, row.names=train$Precinct, queen=FALSE) #Create NB objects for training and test sample
nbtest <- poly2nb(test, row.names=test$Precinct, queen=FALSE)
names(nbtrain) <- attr(nbtrain, "region.id") #Set region.id
names(nbtest) <- attr(nbtest, "region.id")

## Run GAM with MRF
m <- gam(crime ~ s(district, k=nrow(traindata), bs = 'mrf',xt = list(nb = nbtrain)), # define MRF smooth
         data = traindata,
         method = 'REML', # fast version of REML smoothness selection; alternatively 'GCV.Cp'
         family = scat(), #fit scaled t distribution
         gamma = 1.4
)

## Run prediction using new testing data
pred <- predict.gam(m,newdata=testdata)

Answer 1

解决方案：

我终于找到时间用解决方案更新此 post。感谢大家帮助我。以下是使用随机训练测试拆分实现 k 折 CV 的代码：

#Apply k-fold cross validation
mses <- data.frame() #Create empty df to store CV squared error values
scores <- data.frame() #Create empty df to store CV R2 values
set.seed(42) #Set seed for reproducibility
k <- 10 #Define number of folds
for (i in 1:k) {
  # Create weighting column
  data$weight <- sample(c(0,1),size=nrow(data),replace=TRUE,prob=c(0.2,0.8)) #0 Indicates testing sample, 1 training sample

  #Run GAM with MRF
  ctrl <- gam.control(nthreads = 6) #Set controls
  m <- gam(crime ~ s(disctrict, k=nrow(data), bs = 'mrf',xt = list(nb = nb)), #define MRF smooth
            data = data,
            weights = data$weight, #Use only weight==1 observations (training)
            method = 'REML', 
            control = ctrl,
            family = scat(), 
            gamma = 1.4
           )
  #Generate test dataset
  testdata <- data[data$weight==0,] #Select test data by weight
  #Predict test data
  pred <- predict(m,newdata=testdata)
  #Extract MSES
  mses[i,1] <- mean((data$R_MeanDiff[data$weight==0] - pred)^2)
  scores[i,1] <- summary(m)$r.sq
}
av.mse.GMRF <- mean(mses$V1)
av.r2.GMRF <- mean(scores$V1)

Answer 2

我对当前的解决方案有一个批评意见，即完整的数据集用于 "train" 模型意味着预测将有偏差，因为测试数据用于训练它。

这只需要一些小的调整就可以解决：

#Apply k-fold cross validation
mses <- data.frame() #Create empty df to store CV squared error values
scores <- data.frame() #Create empty df to store CV R2 values
set.seed(42) #Set seed for reproducibility
k <- 10 #Define number of folds

#For loop for each fold
for (i in 1:k) {

  # Create weighting column
  data$weight <- sample(c(0,1),size=nrow(data),replace=TRUE,prob=c(0.2,0.8)) #0 Indicates testing sample, 1 training sample

  #Generate training dataset
  trainingdata <- data[data$weight == 1, ] #Select test data by weight  

  #Generate test dataset
  testdata <- data[data$weight == 0, ] #Select test data by weight


  #Run GAM with MRF
  ctrl <- gam.control(nthreads = 6) #Set controls
  m <- gam(crime ~ s(disctrict, k=nrow(data), bs = 'mrf',xt = list(nb = nb)), #define MRF smooth
            data    = trainingdata,
            weights = data$weight, #Use only weight==1 observations (training)
            method  = 'REML', 
            control = ctrl,
            family  = scat(), 
            gamma   = 1.4
           )

  #Predict test data
  pred <- predict(m,newdata = testdata)

  #Extract MSES
  mses[i,1] <- mean((data$R_MeanDiff[data$weight==0] - pred)^2)
  scores[i,1] <- summary(m)$r.sq
}

#Get average scores from each k-fold test
av.mse.GMRF <- mean(mses$V1)
av.r2.GMRF <- mean(scores$V1)

如何使用具有 MRF 平滑和邻域结构的 GAM 预测测试数据？

How to predict test data using a GAM with MRF smooth and neighborhood structure?

r

spatial

prediction

smoothing

gam