Problem: Read in a file of timestamp and other numeric information, i.e. 
              2008-04-20 00:01:27.726517,5,11111                                            
              2008-04-20 00:02:10.170999,4,11011                                            
              2008-04-20 00:03:01.856159,4,11011                                            
              2008-04-20 00:03:33.333176,4,10111                                            
                              .                 
                              .                 
                              .
and create a data frame with columns, year, day, month etc.

Solution: First, using readLines(), we load in the file into an array of character strings, where each line one element in the vector. 

> lines <- readLines("timestampdata.txt")
> lines
[1] "2008-04-20 00:01:27.726517,5,11111" "2008-04-20 00:02:10.170999,4,11011"
[3] "2008-04-20 00:03:01.856159,4,11011" "2008-04-20 00:03:33.333176,4,10111"
[5] "2008-04-20 00:04:06.451844,4,11011" ...
Now we divide and conquer. We create a function, makerow() which converts one line---one character string of the form "2008-04-20 00:01:27.726517,5,11111" into a vector of numbers 2008 4 20 0 1 27.726517 5 11111.

We use strsplit() to break up the string into a vector of substrings on each occurence of a delimiter "-"," ", ":" or ",".
The regular expression "[-: ,]" means any of the set "-"," ", ":" or ",".
The character strings are converted to numeric data type by sapply-ing the function as.numeric() to the character string. Then that result is coerced back into a simple vector. 

> makerow <- function(x) as.vector(sapply(strsplit(x,"[-: ,]"),as.numeric))
Now its just a matter of applying this function to every element in the vector "lines" that we already read in. For example, this produces a matrix close to what we want 
> t(sapply(lines,makerow,USE.NAMES=F))
     [,1] [,2] [,3] [,4] [,5]      [,6] [,7]  [,8]
[1,] 2008    4   20    0    1 27.726517    5 11111
[2,] 2008    4   20    0    2 10.170999    4 11011
[3,] 2008    4   20    0    3  1.856159    4 11011
[4,] 2008    4   20    0    3 33.333176    4 10111
[5,] 2008    4   20    0    4  6.451844    4 11011
...
(sapply has habit of giving us the transpose of what we want, so we just reverse it using transpose t(). Also, USE.NAMES=F prevents sapply from creating unnecessary and ugly names.)

We're almost there - all we need is to make this a data frame name the columns: 

> d <- data.frame( t(sapply(lines,makerow,USE.NAMES=F)) )
> names(d) <- c("year","month","day","hour","min","sec","nfired","which")
> d
  year month day hour min       sec nfired which
1 2008     4  20    0   1 27.726517      5 11111
2 2008     4  20    0   2 10.170999      4 11011
3 2008     4  20    0   3  1.856159      4 11011
4 2008     4  20    0   3 33.333176      4 10111
5 2008     4  20    0   4  6.451844      4 11011
...
and voila, we're ready to start analyzing...