Convert a list to a data frame

With rbind

do.call(rbind.data.frame, your_list)

Edit: Previous version return data.frame of list's instead of vectors (as @IanSudbery pointed out in comments).

Update July 2020:

The default for the parameter stringsAsFactors is now default.stringsAsFactors() which in turn yields FALSE as its default.

Assuming your list of lists is called l:

df <- data.frame(matrix(unlist(l), nrow=length(l), byrow=TRUE))

The above will convert all character columns to factors, to avoid this you can add a parameter to the data.frame() call:

df <- data.frame(matrix(unlist(l), nrow=132, byrow=TRUE),stringsAsFactors=FALSE)

You can use the plyr package.For example a nested list of the form

l <- list(a = list(var.1 = 1, var.2 = 2, var.3 = 3)      , b = list(var.1 = 4, var.2 = 5, var.3 = 6)      , c = list(var.1 = 7, var.2 = 8, var.3 = 9)      , d = list(var.1 = 10, var.2 = 11, var.3 = 12)      )

has now a length of 4 and each list in l contains another list of the length 3.Now you can run

  library (plyr)  df <- ldply (l, data.frame)

and should get the same result as in the answer @Marek and @nico.