Sorting several dates by one observation - r

Question

I am at a loss! I am trying to sort my data by business_id. Each id has several dates associated with it. I am trying to create a new variable that shows the time in days between the first and last date associated with a business_id. Such that
row.names business_id Days
1 x8453 DxUn-ukNL27GOuwjnFGFKA 876
The data currently is structured as:
row.names date business_id
1 X27038 2012-04-21 FV0BkoGOd3Yu_eJnXY15ZA
2 X60951 2012-05-14 Trar_9cFAj6wXiXfKfEqZA
3 X60462 2011-10-05 DxUn-ukNL27GOuwjnFGFKA
4 X2078 2010-12-19 PlcCjELzSI3SqX7mPF5cCw
5 X166883 2011-09-29 pF7uRzygyZsltbmVpjIyvw
6 X177828 2010-09-19 XkNQVTkCEzBrq7OlRHI11Q
7 X128628 2012-05-05 6TWRuHn24DL6vnW8Uyu4Vw
8 X202882 2011-12-10 Xo9Im4LmIhQrzJcO4R3ZbA
9 X64569 2012-02-07 Z67obTep38V9HMtA10yu5A
10 X14667 2009-07-18 xsSnuGCCJD4OgWnOZ0zB4A
11 X17432 2012-08-11 XkNQVTkCEzBrq7OlRHI11Q
Thanks in advance!
Update:
str(data)
'data.frame': 2299 obs. of 2 variables:
$ date :List of 2299
..$ X2736 : chr "2012-05-29"
..$ X160403: chr "2011-08-29"
..$ X19897 : chr "2010-09-27"
..$ X44519 : chr "2012-05-22"
..$ X75910 : chr "2012-10-22"
..$ X13052 : chr "2010-07-14"
$ business_id:List of 2299
..$ X2736 : chr "EFJAVVBQQqftuqY5Wb3WtQ"
..$ X160403: chr "YDlk9buwF8JQE3JgQgraOw"
..$ X19897 : chr "sc1UacpE3cVNJueMdXiCyA"
..$ X44519 : chr "VY_tvNUCCXGXQeSvJl757Q"
..$ X75910 : chr "fowXs9zAM0TQhSfSkPeVuw"
..$ X13052 : chr "xM5F0cLAlKWoB8rOgt5ZOw"
..$ X87807 : chr "nLL0sjLdZ13YdvhXKyss7A"

Answer 1

Edit now that the OP has provided the structure:
Your data is structured quite oddly. A usual structure in R is a data.frame, which is technically a list of vectors where the vectors are all the same length. In your case, you have a list of two (named) lists.
Store the somewhere else for the time being:
old.names <- names(x[[1]])
Then turn the data into an ordinary data.frame, using the handy unlist() function:
x$date <- unlist(x$date)
x$business_id <- unlist(x$business_id)
Use str(x) to see the difference. The names can go back in now, and it's also a good time to turn your "date" column from a character into a proper date, and sort by date order.
x$old.names <- old.names
x$date <- as.POSIXct(x$date)
x <- x[order(x$date), ]
My original answer should now work.
Original answer:
Like agstudy I'd use the plyr package, but if you have the "date" column in a date format and want to keep it that way, you could try:
require(plyr)
ddply(x, "business_id", summarise
, duration = difftime(max(date), min(date), units = "days")
, old.names = old.names[1])
This also gives you flexibility on the units.
With your example data, sorted by date ascending with dat <- dat[order(dat$date), ] means that old.names[1] gives you the name of the earliest row, and old.names[length(old.names)] would give you the name of the most recent row, but I don't know whether that is reliable given the magic inside ddply.
Further edit:
I only showed how to handle the names because they're in your example. They look as though they were originally column headers from imported data, and R has prepended "X" to them because names aren't allowed to begin with numerals.

Answer 2

Using plyr package:
ddply(dat,.(business_id),function(x)
if(length(x$date)>1)
diff(range(as.POSIXct(x$date)))
else 0)
business_id V1
1 6TWRuHn24DL6vnW8Uyu4Vw 0
2 DxUn-ukNL27GOuwjnFGFKA 0
3 FV0BkoGOd3Yu_eJnXY15ZA 0
4 pF7uRzygyZsltbmVpjIyvw 0
5 PlcCjELzSI3SqX7mPF5cCw 0
6 Trar_9cFAj6wXiXfKfEqZA 0
7 XkNQVTkCEzBrq7OlRHI11Q 692
8 Xo9Im4LmIhQrzJcO4R3ZbA 0
9 xsSnuGCCJD4OgWnOZ0zB4A 0
10 Z67obTep38V9HMtA10yu5A 0