n <- 10000000 p <- 10 x <- matrix(rnorm(n*p),ncol = p) # n行p列的随机矩阵x,其中每个元素都服从标准正态分布 beta <- matrix(c(1:p),ncol = 1) # p行1列的系数矩阵beta,其中第i个元素为i z <- x %*% beta # x与beta的矩阵乘积 condprob <- pnorm(z) # 计算二项式分布的概率参数condprob,即每个观测值y为1的概率 y <- matrix(rbinom(n,size = 1,prob = condprob),ncol = 1) # n行1列的随机矩阵y,其中每个元素都服从二项式分布 prob.fit <- glm.fit(x,y,family = binomial(link = "probit"))$coefficients # probit建模 logit.fit <- glm.fit(x,y,family = binomial(link = "logit"))$coefficients # logit建模 linear.fit <- glm.fit(x,y,family = gaussian(link = "identity"))$coefficients # 线性回归建模 coef.mat <- cbind(prob.fit,logit.fit,linear.fit) # 将三个模型的系数矩阵按列合并为一个p行3列的矩阵coef.mat print(coef.mat) prop.mat <- cbind(prob.fit/logit.fit,prob.fit/linear.fit,logit.fit/linear.fit) # 计算三个模型系数之间的比例,并将它们按列合并为一个p行3列的矩阵prop.mat print(prop.mat)
时间: 2024-04-23 13:22:30 浏览: 93
这是一段使用R语言进行二项式回归、logistic回归和线性回归的代码。其中,n为样本量,p为自变量个数,x是从标准正态分布中随机生成的样本数据,beta是一个1到p的向量,z是x和beta的点积,condprob是z的累积分布函数值,y是从二项分布中生成的响应变量。
接下来,使用三种不同的link函数,分别对y和x进行回归拟合,并输出各个系数的值。最后,将三个模型的系数进行比较得到prop.mat。
如果您想使用其他语言或库进行回归分析,可以将代码中的glm函数替换为对应语言或库中的线性回归、逻辑回归或概率模型函数。
相关问题
n <- 10000000 p <- 10 x <- matrix(rnorm(n*p),ncol = p) beta <- matrix(c(1:p),ncol = 1) z <- x %*% beta condprob <- pnorm(z) y <- matrix(rbinom(n,size = 1,prob = condprob),ncol = 1) prob.fit <- glm.fit(x,y,family = binomial(link = "probit"))$coefficients logit.fit <- glm.fit(x,y,family = binomial(link = "logit"))$coefficients linear.fit <- glm.fit(x,y,family = gaussian(link = "identity"))$coefficients coef.mat <- cbind(prob.fit,logit.fit,linear.fit) print(coef.mat) prop.mat <- cbind(prob.fit/logit.fit,prob.fit/linear.fit,logit.fit/linear.fit) print(prop.mat)
这段代码用于生成一个包含n个观测值和p个预测变量的数据集,然后使用不同的广义线性模型(GLM)对y(响应变量)进行建模。具体来说,这里使用logit、probit和线性模型对y进行建模,并输出每个模型的系数矩阵和该矩阵中每个参数的比例矩阵。其中,logit和probit模型是二元响应变量的GLM,而线性模型是用于连续响应变量的GLM。
1 Simulate a dataset containing continuous traits with given ℎ 2 . Split it into training and testing datasets with proportion 4 ∶ 1. set.seed(20132014) n <- 5000 p <- 1000 h <- c(0.2, 0.8)[1] # simulate genotype (not exactly) x_r <- matrix(rnorm(n * p), ncol = p) xmean <- matrix(rep(colMeans(x_r), n), ncol=p, byrow = TRUE) xsd <- matrix(rep(apply(x_r, 2, sd), n), ncol=p, byrow = TRUE) x <- (x_r - xmean)/xsd # ... 2 4 Calculate polygenic scores with both the marginnal and joint models on testing data. And compare their performance under low and high ℎ
Sure, here's the code to calculate polygenic scores with both the marginal and joint models on the testing data:
```R
# perform Lasso on training set
fit_joint <- glmnet(x_train, y_train, alpha = 1)
# calculate polygenic scores on testing set using joint model
ps_joint <- x_test %*% fit_joint$beta[,1]
# perform simple linear regression on training set
fit_marginal <- lm(y_train ~ x_train)
# calculate polygenic scores on testing set using marginal model
ps_marginal <- x_test %*% coef(fit_marginal)[-1]
# compare performance under low and high h^2
h_low <- c(0.2, 0.8)[1]
h_high <- c(0.2, 0.8)[2]
# calculate correlation between true and predicted phenotype for joint model (low h^2)
cor_joint_low <- cor(y_test[h == h_low], ps_joint[h == h_low])
# calculate correlation between true and predicted phenotype for marginal model (low h^2)
cor_marginal_low <- cor(y_test[h == h_low], ps_marginal[h == h_low])
# calculate correlation between true and predicted phenotype for joint model (high h^2)
cor_joint_high <- cor(y_test[h == h_high], ps_joint[h == h_high])
# calculate correlation between true and predicted phenotype for marginal model (high h^2)
cor_marginal_high <- cor(y_test[h == h_high], ps_marginal[h == h_high])
```
To compare the performance of the two models under low and high h^2, we calculated the correlation between the true and predicted phenotype for each model. The correlation for the joint model was calculated using the polygenic scores calculated with the Lasso model, and the correlation for the marginal model was calculated using the polygenic scores calculated with simple linear regression.
You can compare the performance by looking at the values of `cor_joint_low`, `cor_marginal_low`, `cor_joint_high`, and `cor_marginal_high`. The higher the correlation, the better the model's performance at predicting the phenotype.
I hope this helps! Let me know if you have any further questions.
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