[R] Converting a list to a data frame or columns at the least

[Glen Sargeant]

Farrel Buchinsky-3 wrote:
> 
> I have a column in which dates and times are specified thus
> m/d/yyyy HH:MM:SS
> Alas, some entries do not include the time and therefore are only
> m/d/yyyy
> so I used read.csv and specified that the relevant column should be read
> as
> is and it remained as a character variable.
> I then split the value on the space
> split.dt.time <-strsplit(teacher$Date.and.Time.of.Lesson," ")
> that gives me a list where each item on the list has two elements if the
> time was specified and only 1 element if the time was not specified.
> How do I take that list and make all the 1st elements go into one column
> and
> all the second elements go into a second column; where there is no time I
> would like the value to be missing (NA)
> I tried playing around with do.call(rbind...
> 
> so I tried the following unsuccessfully
> do.call(rbind,lapply(teacher$Date.and.Time.of.Lesson, function(i)
> strsplit(i," ")) )
> rbind(strsplit (teacher$Date.and.Time.of.Lesson," "))
> do.call(rbind(data.frame(strsplit (teacher$Date.and.Time.of.Lesson," "))))
> 
> 

Farrel,

You cannot create a matrix or data frame
by combining row vectors that are not of
the same length.  Replace the missing 
values before calling rbind() and things
should work fine.

> x.
[[1]]
[1] "(01/02/70" "00:00:00)"

[[2]]
[1] "(01/03/70"

[[3]]
[1] "(01/04/70" "08:00:00)"

> 
> x <- lapply(x.,function(v){
+   if(length(v)<2)v[2] <- NA
+   v})
> 
> do.call(rbind,x)
     [,1]        [,2]       
[1,] "(01/02/70" "00:00:00)"
[2,] "(01/03/70" NA         
[3,] "(01/04/70" "08:00:00)"
> 

You probably will wish to clean up
dangling parentheses and the like,
but I'll leave that to you.

Glen
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