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如何修改函数以提取R中的某些回归系数

[英]How to modify a function to extract certain regression coefficients in R

I have some trouble with extracting some coefficients out of multiple linear regressions. 从多个线性回归中提取一些系数时遇到一些麻烦。 This is a simple replicable version of my code: 这是我的代码的简单可复制版本:

     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)) ,] 

lms gives me following Output: lms给我以下输出:

[[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]] and [[5]] gives me NULL . [[1]][[5]]给我NULL

Is there a way to modify the function condlm, that gives me a NA instead of NULL ? 有没有办法修改函数condlm,这给了我NA而不是NULL吗? In the End, after extracting the intercepts with zzq <- zzq[grepl("(Intercept)", rownames(zzq)) ,] my Data frame zzq should look like this: 最后,用zzq <- zzq[grepl("(Intercept)", rownames(zzq)) ,]提取截距后zzq <- zzq[grepl("(Intercept)", rownames(zzq)) ,]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

Thanks 谢谢

purrr:map_dfr with broom::tidy seems to be what you want purrr:map_dfrbroom::tidy似乎是您想要的

purrr::map_dfr(lms, ~ broom::tidy(.x)[1,])

# # A tibble: 5 x 5
#   term        estimate std.error statistic p.value
#   <chr>          <dbl>     <dbl>     <dbl>   <dbl>
# 1 NA              NA          NA        NA      NA
# 2 (Intercept)     16.5       NaN       NaN     NaN
# 3 (Intercept)     27         NaN       NaN     NaN
# 4 (Intercept)     38.        NaN       NaN     NaN
# 5 NA              NA          NA        NA      NA

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