On Jun 17, 2012, at 7:19 AM, capital_P wrote:
Rui Barradas wrote
It's stringsAsFactors = FALSE, just one '='.
sapply(dat, length)
It should return 4 times the value 34773.
use dput()
it worked!
dat2 <- data.frame(dat, stringsAsFactors = FALSE)
sapply(dat2, length)
device_info_serial hour time
34773 34773 34773
tripID
34773
dput(head(dat2, 20))
structure(list(device_info_serial = c(121L, 121L, 121L, 121L,
121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L, 121L,
121L, 121L, 121L, 121L, 121L), hour = c(10L, 11L, 11L, 11L, 11L,
11L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 12L, 12L, 12L, 12L, 12L,
12L, 13L), time = structure(c(1242896364, 1242896671, 1242897111,
1242897399, 1242897686, 1242897978, 1242898281, 1242898589,
1242898916,
1242899253, 1242899542, 1242899835, 1242901122, 1242901902,
1242902199,
1242902489, 1242902807, 1242903096, 1242903402, 1242903708), class =
c("POSIXct",
"POSIXt"), tzone = ""), tripID = c(3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L)), .Names =
c("device_info_serial",
"hour", "time", "tripID"), row.names = c(NA, 20L), class =
"data.frame")
departures <- lapply(split(dat2, list(dat2$device_info_serial,
dat2$tripID)), function(x) x[x$time == min(x$time),])
That's going to remove a lot of rows.
Now I have a new problem:
I need to make a histogram of the hours (in this case, their time of
departure), for the whole population and for each bird seperately.
Since you are not telling us what columns correspond to those
concepts, I'm going to use the terms you are. Something along the
lines of:
require(lattice)
histogram( ~ hours | bird, data=dat2, nint =24 , xlim=c(0,24))
histogram( ~ hours , data=dat2, nint =24 , xlim=c(0,24))
Before, I used this:
hist(active$hour, breaks = 24, xlim=c(0,24))
and:
b121 <- active[which(active$device_info_serial == 121),]
b130 <- active[which(active$device_info_serial == 130),]
b132 <- active[which(active$device_info_serial == 132),]
b133 <- active[which(active$device_info_serial == 133),]
etc.
But now there are no departures$device_info_serial or departures$hour,
because departures is a list, not a data frame.
The result on that example was just a list with one element, a
dataframe.
> str(departures)
List of 1
$ 121.3:'data.frame': 1 obs. of 4 variables:
..$ device_info_serial: int 121
..$ hour : int 10
..$ time : POSIXct[1:1], format: "2009-05-21 04:59:24"
..$ tripID : int 3
So to get what you might have expected
> dep <- departures[[1]]
> dep
device_info_serial hour time tripID
1 121 10 2009-05-21 04:59:24 3
But since you didn't actually describe what you did want (or if you
did you left upthread), its only a guess.
I've tried:
dep <- data.frame(departures, stringsAsFactors = FALSE)
Error in data.frame(`121.1` = list(device_info_serial = integer(0),
hour =
integer(0), :
arguments imply differing number of rows: 0, 1, 2
Learn to use str().
dep <- data.frame(departures)
Error in data.frame(`121.1` = list(device_info_serial = integer(0),
hour =
integer(0), :
arguments imply differing number of rows: 0, 1, 2
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