从 R 中的大列表中提取回归系数

Question

我有一个包含大约 100 列的大型数据框，并按年份将其拆分。我想将前一年的 x[i] 作为自变量回归到后一年的 x[i] 作为因变量：xS = a0+ a1xP + e

我的代码如下所示：

     d1 <- structure(list(Date=c("2012-01-01", "2012-06-01",
                            "2013-01-01", "2013-06-01", "2014-01-01", "2014-06-01"),
                     x1=c(NA, NA, 17L, 29L, 27L, 10L), 
                     x2=c(30L, 19L, 22L, 20L, 11L,24L), 
                     x3=c(NA, 23L, 22L, 27L, 21L, 26L),
                     x4=c(30L, 28L, 23L,24L, 10L, 17L), 
                     x5=c(NA, NA, NA, 16L, 30L, 26L)),
                row.names=c(NA, 6L), class="data.frame")
                rownames(d1) <- d1[, "Date"]   
                d1 <- d1[,-1]


df2012 <- d1[1:2,]
df2013 <- d1[3:4,]
df2014 <- d1[4:5,]

condlm <- function(i){    
  if(sum(is.na(df2012[,i]))==dim(df2013)[1]) # ignore the columns     only containing NA's
    return()
  else
    lm.model <- lm(df2013[,i]~df2012[,i])
  summary(lm.model)
}

lms <- lapply(1:dim(df2013)[2], condlm)
lms


zzq <- sapply(lms, coef)
zzq <- do.call(rbind.data.frame, zzq)
zzq <- zzq[grepl("(Intercept)", rownames(zzq)) ,]

编辑 2：

lms 给出以下输出：

[[1]]
NULL

[[2]]

Call:
lm(formula = df2013[, i] ~ df2012[, i])

Residuals:
ALL 2 residuals are 0: no residual degrees of freedom!

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)  16.5455         NA      NA       NA
df2012[, i]   0.1818         NA      NA       NA

Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared:      1, Adjusted R-squared:    NaN 
F-statistic:   NaN on 1 and 0 DF,  p-value: NA


[[3]]

Call:
lm(formula = df2013[, i] ~ df2012[, i])

Residuals:
ALL 1 residuals are 0: no residual degrees of freedom!

Coefficients: (1 not defined because of singularities)
            Estimate Std. Error t value Pr(>|t|)
(Intercept)       27         NA      NA       NA
df2012[, i]       NA         NA      NA       NA

Residual standard error: NaN on 0 degrees of freedom
  (1 observation deleted due to missingness)


[[4]]

Call:
lm(formula = df2013[, i] ~ df2012[, i])

Residuals:
ALL 2 residuals are 0: no residual degrees of freedom!

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)     38.0         NA      NA       NA
df2012[, i]     -0.5         NA      NA       NA

Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared:      1, Adjusted R-squared:    NaN 
F-statistic:   NaN on 1 and 0 DF,  p-value: NA


[[5]]
NULL

[[1]] 和 [[5]] 给我 NULL.

有没有办法修改函数 condlm，它给我一个 NA 而不是 NULL？最后，在使用 zzq <- zzq[grepl("(Intercept)", rownames(zzq)) ,] 提取截距后，我的数据框 zzq 应该如下所示：

             Estimate Std. Error t value Pr(>|t|) 
(Intercept)  NA              NaN     NaN      NaN
(Intercept)2 16.54545        NaN     NaN      NaN
(Intercept)3 27.00000        NaN     NaN      NaN
(Intercept)4 38.00000        NaN     NaN      NaN
(Intercept)5 NA              NaN     NaN      NaN

谢谢

Answer 1

您可以通过以下修改获得标准误差、p 值等：

condlm <- function(i){    
  if(sum(is.na(df2012[,i]))==dim(df2013)[1]) # ignore the columns     only containing NA's
    return()
  else
    lm.model <- lm(df2013[,i]~df2012[,i])
    summary(lm.model)
}


lms <- lapply(1:dim(df2013)[2], condlm)
lms

但是请注意，由于示例中当前数据的结构方式，您没有足够的数据来获取 std 的数值。错误等，因为您的模型拟合不足。

例如，使用您的样本数据，我们将得到以下（部分输出）

> lms
[[1]]
NULL

[[2]]

Call:
lm(formula = df2013[, i] ~ df2012[, i])

Residuals:
ALL 2 residuals are 0: no residual degrees of freedom!

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept)  16.5455         NA      NA       NA
df2012[, i]   0.1818         NA      NA       NA

Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared:      1, Adjusted R-squared:    NaN 
F-statistic:   NaN on 1 and 0 DF,  p-value: NA

从 R 中的大列表中提取回归系数

Extract regression coefficients out of large list in R

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