CODEWITHSUNDEEP
rlistregressioncoefficients
Pogi93
Pogi93

Reputation: 63

Extract regression coefficients out of large list in R

I have a large data frame with about 100 columns and splitted it up by year. I want to regress x[i] from the precedent year as the independent variable on x[i] the subsequent year as the dependent variable: xS = a0+ a1xP + e

My code looks like this:

     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)) ,]

EDIT 2:

lms gives me following Output:

[[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.

Is there a way to modify the function condlm, that gives me a NA instead of NULL? In the End, after extracting the intercepts with zzq <- zzq[grepl("(Intercept)", rownames(zzq)) ,] my Data frame zzq should look like this:

             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

Upvotes: 0

Views: 1157

Answers (1)

NM_
NM_

Reputation: 1999

You can get the std error, p-values, etc. with the following modifications:

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

However please note that due to the way that your data is currently structured in your example, you do not have sufficient data to obtain numeric values for std. error, etc. since you are under-fitting your model.

For example, with your sample data we will get the following (partial output)

> 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

Upvotes: 1

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