Error in rrBLUP(., y = pheno_data$y) : could not find function "rrBLUP"

这个错误可能是因为你使用了一个未加载的R软件包。 rrBLUP是一个R软件包，需要首先安装并加载这个软件包，然后才能使用其中的函数。你可以通过运行以下代码来安装并加载rrBLUP软件包：

install.packages("rrBLUP")
library(rrBLUP)

如果你已经安装了rrBLUP软件包，但仍然遇到这个错误，可能是因为你没有正确加载这个软件包。 请确保在使用rrBLUP函数之前首先运行library(rrBLUP)命令以加载软件包。

相关问题

results <- list() for(i in 1:1) { print(paste(i,"fold")) train_index <- unlist(cvlist[-i]) test_index <- cvlist[[i]] m_train = Markers_2[train_index,] m_test = Markers_2[test_index,] Pheno_train = pheno[train_index,] Pheno_test = pheno[test_index,] class(m_test) TS <- as.numeric(Pheno_train) ETA<-list(list(X= m_train ,model='BayesB')) fit_BB=BGLR(y=TS,ETA=ETA,nIter=nIter,burnIn=burnIn,thin=thin,saveAt=saveAt,df0=5,S0=S0,weights=weights,R2=R2) str(fit_BB) pre_eta <- list(list(X= m_test ,model='BayesB')) predictions <- predict( fit_BB, pre_eta ) # predictions[predictions < 0] <- 0.1 obversion <- Pheno_test results[[i]] <- data.frame( test_index,predictions, obversion) }；如该代码所示，predict时候结果是训练集的数量

根据你提供的代码，问题出在predict()函数的调用上。在predict()函数中，你应该使用与训练集相同的ETA对象，而不是使用不同的ETA对象去进行预测。因此，你需要将预测数据的ETA对象设置为与训练数据相同的ETA对象。具体来说，你需要在predict()函数中使用和拟合模型时相同的ETA值。

修改后的代码如下：

results <- list() 
for(i in 1:1) {
  print(paste(i,"fold"))
  train_index <- unlist(cvlist[-i])
  test_index <- cvlist[[i]]
  m_train = Markers_2[train_index,]
  m_test = Markers_2[test_index,]
  Pheno_train = pheno[train_index,]
  Pheno_test = pheno[test_index,]
  TS <- as.numeric(Pheno_train)
  ETA<-list(list(X=  m_train ,model='BayesB'))
  fit_BB=BGLR(y=TS,ETA=ETA,nIter=nIter,burnIn=burnIn,thin=thin,saveAt=saveAt,df0=5,S0=S0,weights=weights,R2=R2)
  str(fit_BB)
  # 使用相同的ETA对象进行预测
  predictions <- predict(fit_BB, list(list(X=m_test, model='BayesB', ETA=ETA)), nIter=nIter, burnIn=burnIn, thin=thin)
  obversion <-  Pheno_test
  results[[i]] <- data.frame(test_index, predictions, obversion)
}

这样修改后，predict()函数将会使用相同的ETA对象去进行预测，而不是使用不同的ETA对象。

用r语言写程序：Calculate polygenic scores with both the marginnal and joint models on testing data. And compare their performance under low and high h^2

以下是一个用R语言写的计算多基因得分（polygenic scores）并分别采用边缘和联合模型在测试数据上评估其性能的示例程序。同时还会比较在低和高h^2情况下的表现，供您参考：

# Load required packages
library(rrBLUP)
library(plyr)

# Load training and testing data
train_data <- read.csv("train_data.csv")
test_data <- read.csv("test_data.csv")

# Define the phenotype and genotypes
pheno <- train_data$pheno
geno <- train_data[,4:ncol(train_data)]

# Define the relationship matrix
A <- A.mat(geno, n.core=4)

# Fit the marginal and joint models
marginal_model <- kin.blup(pheno, A, method="average")
joint_model <- mixed.solve(pheno ~ A, random=1)

# Calculate the polygenic scores
marginal_scores <- as.numeric(marginal_model$pred[test_data[,1]])
joint_scores <- as.numeric(joint_model$u[test_data[,1]])

# Calculate the correlation between true and predicted phenotypes
marginal_corr <- cor(test_data$pheno, marginal_scores)
joint_corr <- cor(test_data$pheno, joint_scores)

# Compare the performance under low and high h^2
h2_range <- seq(0.1, 0.9, by=0.1)
results <- data.frame(h2=NULL, marginal=NULL, joint=NULL)
for (h2 in h2_range) {
  # Simulate a new phenotype with the specified h2
  sim_pheno <- sim.giv(h2, A=A, K=A, n=1)
  
  # Calculate the polygenic scores
  marginal_scores <- as.numeric(marginal_model$pred[sim_pheno$id])
  joint_scores <- as.numeric(joint_model$u[sim_pheno$id])
  
  # Calculate the correlation between true and predicted phenotypes
  marginal_corr <- cor(sim_pheno$pheno, marginal_scores)
  joint_corr <- cor(sim_pheno$pheno, joint_scores)
  
  # Save the results
  results <- rbind(results, data.frame(h2=h2, marginal=marginal_corr, joint=joint_corr))
}

# Plot the results
library(ggplot2)
ggplot(results, aes(x=h2)) +
  geom_line(aes(y=marginal, color="Marginal")) +
  geom_line(aes(y=joint, color="Joint")) +
  scale_color_manual(values=c("blue", "red")) +
  labs(x="h2", y="Correlation") +
  theme_classic()

请注意：这只是一个示例程序，您需要根据您的具体数据和研究问题进行修改和调整。