如何在 R 中为一次一个逻辑回归创建循环？

Question

我想对几个（n = 30）SNP（编码为 0,1,2）作为预测变量和一个案例控制变量（0,1）作为结果进行逻辑回归。 由于这些 rs 中很少有相关的，我不能将所有 rs# 放在一个 model 中，但必须一次运行一个回归，即，我不能简单地将它们加到一个 model 中，如 rs1 + rs2 + rs3 等等。 ..我需要像下面一样分别回归；

test1 = glm(casecontrol ~ rs1, data = mydata, family=binomial)

test2 = glm(casecontrol ~ rs2, data = mydata, family=binomial)

test3 = glm(casecontrol ~ rs3, data = mydata, family=binomial)

虽然我可以单独运行上述所有回归，但有没有办法将它们循环在一起，以便我可以在一个 go 中获得所有测试的摘要（）？

我也必须调整年龄和性别，但这将在我运行未调整的循环之后发生。

例如，我的数据头来自dput(head(mydata)) ；

structure(list(ID = 1:6, sex = c(2L, 1L, 2L, 1L, 1L, 1L), age = c(52.4725405022036, 
58.4303618001286, 44.5300065923948, 61.4786037395243, 67.851808819687, 
39.7451378498226), bmi = c(31.4068751083687, 32.0614937413484, 
23.205021363683, 29.1445372393355, 32.6287483051419, 20.5887741968036
), casecontrol = c(0L, 1L, 0L, 1L, 1L, 1L), rs1 = c(1L, 0L, 2L, 
2L, 1L, 2L), rs2 = c(2L, 1L, 2L, 0L, 1L, 1L), rs3 = c(1L, 0L, 
1L, 2L, 2L, 2L), rs4 = c(1L, 1L, 1L, 1L, 0L, 2L), rs5 = c(1L, 
0L, 0L, 0L, 1L, 2L), rs6 = c(1L, 1L, 1L, 1L, 1L, 2L), rs7 = c(0L, 
0L, 0L, 0L, 0L, 0L), rs8 = c(0L, 0L, 1L, 0L, 2L, 1L), rs9 = c(0L, 
0L, 2L, 1L, 1L, 0L), rs10 = c(2L, 0L, 0L, 2L, 2L, 1L), rs11 = c(0L, 
1L, 1L, 0L, 1L, 1L), rs12 = c(1L, 2L, 0L, 1L, 2L, 2L), rs13 = c(0L, 
2L, 0L, 0L, 0L, 0L), rs14 = c(1L, 1L, 1L, 1L, 2L, 2L), rs15 = c(1L, 
2L, 1L, 1L, 0L, 1L), rs16 = c(0L, 2L, 1L, 2L, 2L, 1L), rs17 = c(0L, 
2L, 1L, 1L, 2L, 2L), rs18 = c(1L, 2L, 2L, 1L, 1L, 1L), rs19 = c(1L, 
1L, 0L, 1L, 2L, 2L), rs20 = c(2L, 1L, 0L, 2L, 2L, 1L), rs21 = c(1L, 
2L, 2L, 1L, 1L, 0L), rs22 = c(1L, 1L, 2L, 2L, 0L, 1L), rs23 = c(2L, 
0L, 2L, 1L, 1L, 1L), rs24 = c(0L, 0L, 0L, 2L, 2L, 2L), rs25 = c(2L, 
2L, 1L, 1L, 0L, 1L), rs26 = c(1L, 1L, 0L, 2L, 0L, 1L), rs27 = c(1L, 
1L, 1L, 1L, 0L, 1L), rs28 = c(0L, 1L, 1L, 2L, 0L, 2L), rs29 = c(2L, 
2L, 2L, 2L, 1L, 2L), rs30 = c(0L, 2L, 1L, 2L, 1L, 0L)), row.names = c(NA, 
6L), class = "data.frame")```

Answer 1

可能你想要这样的东西：

lapply(1:30, function(i) glm(as.formula(paste0('casecontrol ~ ', 'rs', i)), data = mydata, family = binomial))

这将使用选定的预测器执行 30 个逻辑回归。

您可以使用sum(grepl('rs', names(mydata))) ，而不是硬编码预测变量的总数，它将返回 30。

您可以使用broom package 中的tidy function 以整洁的格式获取摘要。

purrr::map_dfr(1:30, function(i) data.frame(model = i, tidy(glm(as.formula(paste0('casecontrol ~ ', 'rs', i)), data = mydata, family = binomial))))

或者您可以以更动态的方式执行此操作：

names(mydata)[grepl('rs', names(mydata))] -> pred #get all predictors that contain 'rs'

purrr::map_dfr(1:length(pred), 
               function(i) data.frame(model = i, 
                                      tidy(glm(as.formula(paste0('casecontrol ~ ', pred[i])), data = mydata, family = binomial))))

如果你想包含另一个变量，你只需要调整pred向量。

c(pred, paste0(pred, ' + age')) -> pred #interaction between rs drivers and age

或者

c(pred, paste0(pred, ' + age + sex')) -> pred #interaction between rs drivers age and sex

Answer 2

你可以做这样的事情

outcome<-mydata %>% select("casecontrol")  #select outcome

features <- mydata %>% select("rs1":"rs30") #select features

features_names<-data.frame(colnames(features)) #get feature names

for(i in 1:nrow(features_names))     # loop over length of features_name
{selected=features[,i,drop=FALSE]    #get each feature 
total <- cbind(selected, response)   # combine with outcome
glm.fit <- glm(casecontrol ~ ., data = total, family = binomial("logit")) 
summary(glm.fit)
}