Question

我从SQL数据库导入了一个大表，其结构与此示例表

类似

testData <- data.frame(
  BatchNo = c(1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3),
  Y = c(1,1.247011378,1.340630851,1.319026357,1.41264583,1.093619473,1.38023909,1.473858563,1,1.093619473,1.038888089,1.081833061,1,1.215913383,1.278861891,1.297746443,1.360694952,1.332368123,1.414201183,1,1.081833061,1,1.063661202),
  Categorical1 = c("A9","B5513","B5513","B5514","B5514","A9","B5514","B5514","A9","A9","B1723","A9","A9","B5513","B5514","B5513","B5514","B5514","B5514","A9","A9","A486","B1701"),
  Categorical2 = c("A2793","B5512","B5512","B5512","B5512","B5508","B6623","B6623","B5508","B5508","B5508","A127","A127","B5515","B5515","B5515","B5515","B6623","B6623","A127","A127","A2727","A2727"),
  Categorical3 = c("A5510","B5511","B5511","B5511","B5511","A5510","B5511","B5511","B5511","B5511","B5511","A5518","A5518","B5517","B5517","B5517","B5517","B5517","B5517","B5517","B5517","A2","A2"),
  Categorical4 = c("A5","A5","B649","A5","B649","B649","A5","B649","A5","B649","A5","B649","A5","A5","A5","B649","B649","A5","B649","A5","B649","A649","A649"),
  Binary1 = c(rep(0,times=23)),
  Binary2 = c(rep(0,times=23)),
  Binary3 = c(rep(0,times=23)),
  Binary4 = c(rep(0,times=23))
  )

我想在for循环中做的是：

1.基于BatchNo列（1到2500）创建子集数据帧
2.使用每个子集数据帧的健康线性模型
3.将系数估计列表导出回SQL表

到目前为止，我已经为步骤1＆amp; 2：

n<-max(testData[,1])
for (i in 1:n) {
assign(paste("dat"),droplevels(subset(testData,BatchNo == i, select = 1:10)))
assign(paste("lm.", i, sep =
""),lm(Y~Categorical1+Categorical2+Categorical3+Categorical4+Binary1+Binary2+Binary3+Binary4,data=dat))}

问题在于会创建子集，其中4个分类变量中的至少一个（或者可能所有变量）将具有单个级别（如本示例中的BatchNo = 3），并且R不能在回归中使用这些变量。对于二元预测变量，这不是问题，因为它只会产生N/A系数估计值，并且我会在模型拟合后执行step(backward)删除任何一个。

起初我尝试使用step(forward)在每个循环中仅选择有意义的预测变量，但这并不起作用，因为我必须列出所有潜在的预测变量供选择。

我可以想到两种可能的解决方案：

从每个循环中的＆＃34; dat＆＃34; 中删除单级因子列
或者为每个循环创建一个多级因子名称的向量/列表，并以lm公式

我只是创造了这两个载体：

factors<-dat[,3:6]
f<-names(factors)
levels<-c(length(levels(factors[,1])),length(levels(factors[,2])),length(levels(factors[,3])),length(levels(factors[,4])))

所以现在我只需要从＆＃34; f＆＃34; 中删除第n个元素，其中＆＃34; level＆＃34; 的第n个元素等于1。

Answer 1

最终我已经找到了一种方法来做我想做的事情。可能有一种更简单/更优雅的方式，但我已经使用过：

  l<-nrow(dat)
  a<-length(levels(dat[,3]))
  b<-length(levels(dat[,4]))
  c<-length(levels(dat[,5]))
  d<-length(levels(dat[,6]))
  zeros<-c(rep(0,times=l))
  if (a<2) dat[,2]<-zeros
  if (b<2) dat[,3]<-zeros
  if (c<2) dat[,4]<-zeros
  if (d<2) dat[,5]<-zeros

单级因子被适当长度的每个循环包含零的向量所取代，因此可以运行回归而不会出错。

Answer 2

试试这个：

do.call(rbind,
        lapply(split(testData, testData$BatchNo), function(i){
          #check if factor columns have more than 1 level
          cats <- colnames(i)[c(3:6)[sapply(i[, c(3:6)], function(j){length(unique(j))}) > 1]]
          cats <- paste(cats, collapse = "+")
          fit <- lm(as.formula(paste0("Y~", cats, "+Binary2+Binary3+Binary4")), data = i)
          #return coef as df
          as.data.frame(coef(fit))
          })
        )

输出

#                         coef(fit)
# 1.(Intercept)        1.000000e+00
# 1.Categorical1B1723  3.888809e-02
# 1.Categorical1B5513  3.082241e-01
# 1.Categorical1B5514  3.802391e-01
# 1.Categorical2B5508  5.611389e-16
# 1.Categorical2B5512 -6.121273e-02
# 1.Categorical2B6623            NA
# 1.Categorical3B5511  1.699675e-17
# 1.Categorical4B649   9.361947e-02
# 1.Binary2                      NA
# 1.Binary3                      NA
# 1.Binary4                      NA
# 2.(Intercept)        1.000000e+00
# 2.Categorical1B5513  2.694196e-01
# 2.Categorical1B5514  3.323681e-01
# 2.Categorical2B5515 -5.350623e-02
# 2.Categorical2B6623            NA
# 2.Categorical3B5517  3.289161e-18
# 2.Categorical4B649   8.183306e-02
# 2.Binary2                      NA
# 2.Binary3                      NA
# 2.Binary4                      NA
# 3.(Intercept)        1.000000e+00
# 3.Categorical1B1701  6.366120e-02
# 3.Binary2                      NA
# 3.Binary3                      NA
# 3.Binary4                      NA

在for循环

2 个答案: