[R] noob requesting help

[capital_P]

Rui Barradas wrote
> 
> Sorry, but the output of dput() starts with 'structure', not like what 
> you've posted.
> And there are much more than 20 dates in the beginning.
> 
> The posting guide is easy to find
> 

apologies for my earlier ignorance.

It seems the output of dput(head(data, 20) is too big to print in full. When
I press stop immediately after pressing enter, I get: 

dput(head(long, 30))
structure(list(tripID = c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L), device_info_serial = c(121L, 121L, 121L, 
121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 
121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 
121L, 121L, 121L, 121L, 121L), mdate = structure(c(33L, 33L, 
33L, 33L, 33L, 33L, 33L, 33L, 33L, 33L, 33L, 33L, 33L, 33L, 33L, 

If I let it run just a little bit longer, it gives the enormous output like
before.

I made a new vector containing only the columns I need for this problem:

> str(dat)
List of 4
 $ device_info_serial: int [1:34773] 121 121 121 121 121 121 121 121 121 121
...
 $ hour              : int [1:34773] 10 11 11 11 11 11 11 11 11 11 ...
 $ time              : POSIXlt[1:34773], format: "2009-05-21 10:59:24" ...
 $ tripID            : int [1:34773] 3 3 3 3 3 3 3 3 3 3 ...

You were right about the factors. I think I solved it using:

dat$time <- strptime(long$date_time.x, format = "%Y-%m-%d %H:%M:%S")

this seems to help, as R doesn't start computing for hours. It does give an
error though:

> departures <- lapply(split(dat, list(dat$device_info_serial, dat$tripID)),
> function(x) x[x$time == min(x$time),])
Error in x[x$time == min(x$time), ] : incorrect number of dimensions
In addition: Warning messages:
1: In split.default(dat, list(dat$device_info_serial, dat$tripID)) :
  data length is not a multiple of split variable
2: In min(x$time) : no non-missing arguments to min; returning Inf

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