Re: [R] Need a faster function to replace missing data
Many thanks to Jim, Bill, and Carl. Using indexes instead of the for loop gave me my answer in minutes instead of hours! Thanks for all of your great suggestions! Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 5/22/09, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] Need a faster function to replace missing data To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, May 22, 2009, 4:59 PM Here is a modification that should now find the closest: myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00), + format=%H:%M:%S)) # convert to numeric names(myvscan)-c(Latitude,DateTime) myvscan$tn - as.numeric(myvscan$DateTime) # numeric for findInterval mygarmin-data.frame(c(20,30,40),as.POSIXct(c(12:00:00,12:10:00,12:15:00), + format=%H:%M:%S)) names(mygarmin)-c(Latitude,DateTime) mygarmin$tn - as.numeric(mygarmin$DateTime) # use 'findInterval' na.indx - which(is.na(myvscan$Latitude)) # find NAs # create matrix of values to test the range indices - findInterval(myvscan$tn[na.indx],mygarmin$tn) x - cbind(indices, + abs(myvscan$tn[na.indx] - mygarmin$tn[indices]), # lower + abs(myvscan$tn[na.indx] - mygarmin$tn[indices + 1])) #higher # now determine which index is closer closest - x[,1] + (x[,2] x[,3]) # determine the proper index # replace with garmin latitude myvscan$Latitude[na.indx] - mygarmin$Latitude[closest] myvscan Latitude DateTime tn 1 1.0 2009-05-23 12:00:00 124308 2 40.0 2009-05-23 12:14:00 1243080840 3 1.5 2009-05-23 12:20:00 1243081200 On Fri, May 22, 2009 at 7:39 PM, Tim Clark mudiver1...@yahoo.com wrote: Jim, Thanks! I like the way you use indexing instead of the loops. However, the find.Interval function does not give the right result. I have been playing with it and it seems to give the closest number that is less than the one of interest. In this case, the correct replacement should have been 40, not 30, since 12:15 from mygarmin is closer to 12:14 in myvscan than 12:10. Is there a way to get the function to find the closest in value instead of the next smaller value? I was trying to use which.min to get the closet date but can't seem to get it to work right either. Aloha, Tim Tim Clark Department of Zoology University of Hawaii --- On Fri, 5/22/09, jim holtman jholt...@gmail.com wrote: From: jim holtman jholt...@gmail.com Subject: Re: [R] Need a faster function to replace missing data To: Tim Clark mudiver1...@yahoo.com Cc: r-help@r-project.org Date: Friday, May 22, 2009, 7:24 AM I think this does what you want. It uses 'findInterval' to determine where a possible match is: myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00), format=%H:%M:%S)) # convert to numeric names(myvscan)-c(Latitude,DateTime) myvscan$tn - as.numeric(myvscan$DateTime) # numeric for findInterval mygarmin-data.frame(c(20,30,40),as.POSIXct(c(12:00:00,12:10:00,12:15:00), format=%H:%M:%S)) names(mygarmin)-c(Latitude,DateTime) mygarmin$tn - as.numeric(mygarmin$DateTime) # use 'findInterval' na.indx - which(is.na(myvscan$Latitude)) # find NAs # replace with garmin latitude myvscan$Latitude[na.indx] - mygarmin$Latitude[findInterval(myvscan$tn[na.indx], mygarmin$tn)] myvscan Latitude DateTime tn 1 1.0 2009-05-22 12:00:00 1243008000 2 30.0 2009-05-22 12:14:00 1243008840 3 1.5 2009-05-22 12:20:00 1243009200 On Fri, May 22, 2009 at 12:45 AM, Tim Clark mudiver1...@yahoo.com wrote: Dear List, I need some help in coming up with a function that will take two data sets, determine if a value is missing in one, find a value in the second that was taken at about the same time, and substitute the second value in for where the first should have been. My problem is from a fish tracking study. We put acoustic tags in fish and track them for several days. Location data is supposed to be automatically recorded every time we detect a ping from the fish. Unfortunately the GPS had some problems and sometimes the fishes depth was recorded but not its location. I fortunately had a back-up GPS that was taking location data every five minutes. I would like to merge the two files, replacing the missing value in the vscan (automatic) file with the location from the garmin file. Since we were getting vscan records every 1-2 seconds and garmin records every 5 minutes, I need to find the right place in the vscan file to place the garmin record - i.e. the closest in time, but not greater than 5 minutes. I have written a function
Re: [R] Need a faster function to replace missing data
Tim Clark mudiver1200 at yahoo.com writes: I need some help in coming up with a function that will take two data sets, determine if a value is missing in one, find a value in the second that was taken at about the same time, and substitute the second value in for where the first should have been. This is the type of job I would do with a database, not R (alone). The main advantage is that you have to do the cleanup job only once and can retrieve the data in a rather well-documented way later (it's possible with R, I know). Put the 5 minutes data into one table. I would two new columns giving the delta to the next value for easier linear interpolation, but that's secondary. Make sure to index the table. Put the 1 seconds data into another table, adding values rounded to 5 seconds, and giving these an index. From R/ODBC or with RSQLite, make a Join of all values in Table 1 that do have NULL values in the coordinates. If you do not want to do a linear interpolation, you could even do this within the database and SQL alone. Compute the linear interpolation, and write the data back into the database. If you want to be careful, you might mark the interpolated values in a separate field as computed When at a later time new data come in, you can run the procedure again without penalty. Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need a faster function to replace missing data
Here are 2 functions, which.just.above and which.just.below,
which may help you. They will tell which element in a reference
dataset is the first just above (or just below) each element
in the main dataset (x). They return NA if there is no reference
element above (or below) an element of x. The strict argument
lets you say if the inequalities are strict or if equality is
acceptable.
They are vectorized so are pretty quick.
E.g.,
which.just.below(c(14,14.5), 11:15, strict=TRUE)
[1] 3 4
which.just.above(c(14,14.5), 11:15, strict=FALSE)
[1] 4 5
They should work with any class of data that order() and sort()
work on. In particular, POSIXct times work. The attached file
has a demonstration function called 'test' with some examples.
In your case the 'reference' data would be the times at which your
backup measurements were taken and the 'x' data would be the
times of the pings. You can look at the elements of 'reference' just
before and just after each ping (or just the pings that are missing
locations) and decide how to combine the data from the bracketing
reference elements to inpute a location for the ping.
Here are the functions, in case the attachment doesn't make it
through. I'm sure some mailer will throw in some newlines so
it will be corrupted.
which.just.above -
function(x, reference, strict = T)
{
# output[k] will be index of smallest value in reference vector
# larger than x[k]. If strict=F, replace 'larger than' by
# 'larger than or equal to'.
# We should allow NA's in x (but we don't). NA's in reference
# should not be allowed.
if(any(is.na(x)) || any(is.na(reference))) stop(NA's in input)
if(strict)
i - c(rep(T, length(reference)), rep(F,
length(x)))[order(
c(reference, x))]
else i - c(rep(F, length(x)), rep(T,
length(reference)))[order(c(
x, reference))]
i - cumsum(i)[!i] + 1.
i[i length(reference)] - NA
# i is length of x and has values in range 1:length(reference)
or NA
# following needed if reference is not sorted
i - order(reference)[i]
# following needed if x is not sorted
i[order(order(x))]
}
which.just.below -
function(x, reference, strict = T)
{
# output[k] will be index of largest value in reference vector
# less than x[k]. If strict=F, replace 'less than' by
# 'less than or equal to'. Neither x nor reference need be
# sorted, although they should not have NA's (in theory, NA's
# in x are ok, but not in reference).
if(any(is.na(x)) || any(is.na(reference))) stop(NA's in input)
if(!strict)
i - c(rep(T, length(reference)), rep(F,
length(x)))[order(
c(reference, x))]
else i - c(rep(F, length(x)), rep(T,
length(reference)))[order(c(
x, reference))]
i - cumsum(i)[!i]
i[i = 0] - NA
# i is length of x and has values in range 1:length(reference)
or NA
# following needed if reference is not sorted
i - order(reference)[i]
# following needed if x is not sorted
i[order(order(x))]
}
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Tim Clark
Sent: Thursday, May 21, 2009 9:45 PM
To: r-help@r-project.org
Subject: [R] Need a faster function to replace missing data
Dear List,
I need some help in coming up with a function that will take
two data sets, determine if a value is missing in one, find a
value in the second that was taken at about the same time,
and substitute the second value in for where the first should
have been. My problem is from a fish tracking study. We put
acoustic tags in fish and track them for several days.
Location data is supposed to be automatically recorded every
time we detect a ping from the fish. Unfortunately the GPS
had some problems and sometimes the fishes depth was recorded
but not its location. I fortunately had a back-up GPS that
was taking location data every five minutes. I would like to
merge the two files, replacing the missing value in the vscan
(automatic) file with the location from the garmin file.
Since we were getting vscan records every 1-2 seconds and
garmin records every 5 minutes, I need to find the right
place in the vscan file to place the garmin record - i.e. the
closest in time, but not greater than 5 minutes. I have
written a function that does this. However, it works with my
test data but locks up my computer with my real data. I have
several million vscan records and several thousand garmin
records. Is there a better way to do this?
My function and test data:
myvscan-data.frame(c(1,NA,1.5),times(c(12:00:00,12:14:00,
12:20:00)))
names(myvscan)-c
Re: [R] Need a faster function to replace missing data
I think this does what you want. It uses 'findInterval' to determine where
a possible match is:
myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00),
format=%H:%M:%S))
# convert to numeric
names(myvscan)-c(Latitude,DateTime)
myvscan$tn - as.numeric(myvscan$DateTime) # numeric for findInterval
mygarmin-data.frame(c(20,30,40),as.POSIXct(c(12:00:00,12:10:00,12:15:00),
format=%H:%M:%S))
names(mygarmin)-c(Latitude,DateTime)
mygarmin$tn - as.numeric(mygarmin$DateTime)
# use 'findInterval'
na.indx - which(is.na(myvscan$Latitude)) # find NAs
# replace with garmin latitude
myvscan$Latitude[na.indx] -
mygarmin$Latitude[findInterval(myvscan$tn[na.indx], mygarmin$tn)]
myvscan
LatitudeDateTime tn
1 1.0 2009-05-22 12:00:00 1243008000
2 30.0 2009-05-22 12:14:00 1243008840
3 1.5 2009-05-22 12:20:00 1243009200
On Fri, May 22, 2009 at 12:45 AM, Tim Clark mudiver1...@yahoo.com wrote:
Dear List,
I need some help in coming up with a function that will take two data sets,
determine if a value is missing in one, find a value in the second that was
taken at about the same time, and substitute the second value in for where
the first should have been. My problem is from a fish tracking study. We
put acoustic tags in fish and track them for several days. Location data is
supposed to be automatically recorded every time we detect a ping from the
fish. Unfortunately the GPS had some problems and sometimes the fishes
depth was recorded but not its location. I fortunately had a back-up GPS
that was taking location data every five minutes. I would like to merge the
two files, replacing the missing value in the vscan (automatic) file with
the location from the garmin file. Since we were getting vscan records
every 1-2 seconds and garmin records every 5 minutes, I need to find the
right place in the vscan file to place the garmin record - i.e. the
closest in time, but not greater than 5 minutes. I have written a
function that does this. However, it works with my test data but locks up my
computer with my real data. I have several million vscan records and
several thousand garmin records. Is there a better way to do this?
My function and test data:
myvscan-data.frame(c(1,NA,1.5),times(c(12:00:00,12:14:00,12:20:00)))
names(myvscan)-c(Latitude,DateTime)
mygarmin-data.frame(c(20,30,40),times((12:00:00,12:10:00,12:15:00)))
names(mygarmin)-c(Latitude,DateTime)
minute.diff-1/24/12 #Time diff is in days, so this is 5 minutes
for (k in 1:nrow(myvscan))
{
if (is.na(myvscan$Latitude[k]))
{
if ((min(abs(mygarmin$DateTime-myvscan$DateTime[k]))) minute.diff )
{
index.min.date-which.min(abs(mygarmin$DateTime-myvscan$DateTime[k]))
myvscan$Latitude[k]-mygarmin$Latitude[index.min.date]
}}}
I appreciate your help and advice.
Aloha,
Tim
Tim Clark
Department of Zoology
University of Hawaii
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need a faster function to replace missing data
Jim,
Thanks! I like the way you use indexing instead of the loops. However, the
find.Interval function does not give the right result. I have been playing
with it and it seems to give the closest number that is less than the one of
interest. In this case, the correct replacement should have been 40, not 30,
since 12:15 from mygarmin is closer to 12:14 in myvscan than 12:10. Is there a
way to get the function to find the closest in value instead of the next
smaller value? I was trying to use which.min to get the closet date but can't
seem to get it to work right either.
Aloha,
Tim
Tim Clark
Department of Zoology
University of Hawaii
--- On Fri, 5/22/09, jim holtman jholt...@gmail.com wrote:
From: jim holtman jholt...@gmail.com
Subject: Re: [R] Need a faster function to replace missing data
To: Tim Clark mudiver1...@yahoo.com
Cc: r-help@r-project.org
Date: Friday, May 22, 2009, 7:24 AM
I think this does what you
want. It uses 'findInterval' to determine where a
possible match is:
myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00),
format=%H:%M:%S))
# convert to numeric
names(myvscan)-c(Latitude,DateTime)
myvscan$tn - as.numeric(myvscan$DateTime) #
numeric for findInterval
mygarmin-data.frame(c(20,30,40),as.POSIXct(c(12:00:00,12:10:00,12:15:00),
format=%H:%M:%S))
names(mygarmin)-c(Latitude,DateTime)
mygarmin$tn - as.numeric(mygarmin$DateTime)
# use 'findInterval'
na.indx - which(is.na(myvscan$Latitude)) # find
NAs
# replace with garmin latitude
myvscan$Latitude[na.indx] -
mygarmin$Latitude[findInterval(myvscan$tn[na.indx],
mygarmin$tn)]
myvscan
Latitude DateTime
tn
1 1.0 2009-05-22 12:00:00 1243008000
2 30.0 2009-05-22 12:14:00 1243008840
3 1.5 2009-05-22 12:20:00 1243009200
On Fri, May 22, 2009 at 12:45 AM,
Tim Clark mudiver1...@yahoo.com
wrote:
Dear List,
I need some help in coming up with a function that will
take two data sets, determine if a value is missing in one,
find a value in the second that was taken at about the same
time, and substitute the second value in for where the first
should have been. My problem is from a fish tracking
study. We put acoustic tags in fish and track them for
several days. Location data is supposed to be
automatically recorded every time we detect a
ping from the fish. Unfortunately the GPS had
some problems and sometimes the fishes depth was recorded
but not its location. I fortunately had a back-up GPS that
was taking location data every five minutes. I would like
to merge the two files, replacing the missing value in the
vscan (automatic) file with the location from the garmin
file. Since we were getting vscan records every 1-2
seconds and garmin records every 5 minutes, I need to find
the right place in the vscan file to place the garmin record
- i.e. the
closest in time, but not greater than 5 minutes. I have
written a function that does this. However, it works with my
test data but locks up my computer with my real data. I
have several million vscan records and several thousand
garmin records. Is there a better way to do this?
My function and test data:
myvscan-data.frame(c(1,NA,1.5),times(c(12:00:00,12:14:00,12:20:00)))
names(myvscan)-c(Latitude,DateTime)
mygarmin-data.frame(c(20,30,40),times((12:00:00,12:10:00,12:15:00)))
names(mygarmin)-c(Latitude,DateTime)
minute.diff-1/24/12 #Time diff is in days, so this
is 5 minutes
for (k in 1:nrow(myvscan))
{
if (is.na(myvscan$Latitude[k]))
{
if ((min(abs(mygarmin$DateTime-myvscan$DateTime[k])))
minute.diff )
{
index.min.date-which.min(abs(mygarmin$DateTime-myvscan$DateTime[k]))
myvscan$Latitude[k]-mygarmin$Latitude[index.min.date]
}}}
I appreciate your help and advice.
Aloha,
Tim
Tim Clark
Department of Zoology
University of Hawaii
__
R-help@r-project.org
mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need a faster function to replace missing data
Here is a modification that should now find the closest:
myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00),
+ format=%H:%M:%S))
# convert to numeric
names(myvscan)-c(Latitude,DateTime)
myvscan$tn - as.numeric(myvscan$DateTime) # numeric for findInterval
mygarmin-data.frame(c(20,30,40),as.POSIXct(c(12:00:00,12:10:00,12:15:00),
+ format=%H:%M:%S))
names(mygarmin)-c(Latitude,DateTime)
mygarmin$tn - as.numeric(mygarmin$DateTime)
# use 'findInterval'
na.indx - which(is.na(myvscan$Latitude)) # find NAs
# create matrix of values to test the range
indices - findInterval(myvscan$tn[na.indx],mygarmin$tn)
x - cbind(indices,
+abs(myvscan$tn[na.indx] - mygarmin$tn[indices]), # lower
+abs(myvscan$tn[na.indx] - mygarmin$tn[indices + 1])) #higher
# now determine which index is closer
closest - x[,1] + (x[,2] x[,3]) # determine the proper index
# replace with garmin latitude
myvscan$Latitude[na.indx] - mygarmin$Latitude[closest]
myvscan
LatitudeDateTime tn
1 1.0 2009-05-23 12:00:00 124308
2 40.0 2009-05-23 12:14:00 1243080840
3 1.5 2009-05-23 12:20:00 1243081200
On Fri, May 22, 2009 at 7:39 PM, Tim Clark mudiver1...@yahoo.com wrote:
Jim,
Thanks! I like the way you use indexing instead of the loops. However,
the find.Interval function does not give the right result. I have been
playing with it and it seems to give the closest number that is less than
the one of interest. In this case, the correct replacement should have been
40, not 30, since 12:15 from mygarmin is closer to 12:14 in myvscan than
12:10. Is there a way to get the function to find the closest in value
instead of the next smaller value? I was trying to use which.min to get the
closet date but can't seem to get it to work right either.
Aloha,
Tim
Tim Clark
Department of Zoology
University of Hawaii
--- On Fri, 5/22/09, jim holtman jholt...@gmail.com wrote:
From: jim holtman jholt...@gmail.com
Subject: Re: [R] Need a faster function to replace missing data
To: Tim Clark mudiver1...@yahoo.com
Cc: r-help@r-project.org
Date: Friday, May 22, 2009, 7:24 AM
I think this does what you
want. It uses 'findInterval' to determine where a
possible match is:
myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00),
format=%H:%M:%S))
# convert to numeric
names(myvscan)-c(Latitude,DateTime)
myvscan$tn - as.numeric(myvscan$DateTime) #
numeric for findInterval
mygarmin-data.frame(c(20,30,40),as.POSIXct(c(12:00:00,12:10:00,12:15:00),
format=%H:%M:%S))
names(mygarmin)-c(Latitude,DateTime)
mygarmin$tn - as.numeric(mygarmin$DateTime)
# use 'findInterval'
na.indx - which(is.na(myvscan$Latitude)) # find
NAs
# replace with garmin latitude
myvscan$Latitude[na.indx] -
mygarmin$Latitude[findInterval(myvscan$tn[na.indx],
mygarmin$tn)]
myvscan
LatitudeDateTime
tn
1 1.0 2009-05-22 12:00:00 1243008000
2 30.0 2009-05-22 12:14:00 1243008840
3 1.5 2009-05-22 12:20:00 1243009200
On Fri, May 22, 2009 at 12:45 AM,
Tim Clark mudiver1...@yahoo.com
wrote:
Dear List,
I need some help in coming up with a function that will
take two data sets, determine if a value is missing in one,
find a value in the second that was taken at about the same
time, and substitute the second value in for where the first
should have been. My problem is from a fish tracking
study. We put acoustic tags in fish and track them for
several days. Location data is supposed to be
automatically recorded every time we detect a
ping from the fish. Unfortunately the GPS had
some problems and sometimes the fishes depth was recorded
but not its location. I fortunately had a back-up GPS that
was taking location data every five minutes. I would like
to merge the two files, replacing the missing value in the
vscan (automatic) file with the location from the garmin
file. Since we were getting vscan records every 1-2
seconds and garmin records every 5 minutes, I need to find
the right place in the vscan file to place the garmin record
- i.e. the
closest in time, but not greater than 5 minutes. I have
written a function that does this. However, it works with my
test data but locks up my computer with my real data. I
have several million vscan records and several thousand
garmin records. Is there a better way to do this?
My function and test data:
myvscan-data.frame(c(1,NA,1.5),times(c(12:00:00,12:14:00,12:20:00)))
names(myvscan)-c(Latitude,DateTime)
mygarmin-data.frame(c(20,30,40),times((12:00:00,12:10:00,12:15:00)))
names(mygarmin)-c(Latitude,DateTime)
minute.diff-1/24/12 #Time diff is in days, so this
is 5 minutes
for (k in 1:nrow(myvscan))
{
if (is.na(myvscan$Latitude[k]))
{
if ((min(abs(mygarmin$DateTime-myvscan
[R] Need a faster function to replace missing data
Dear List,
I need some help in coming up with a function that will take two data sets,
determine if a value is missing in one, find a value in the second that was
taken at about the same time, and substitute the second value in for where the
first should have been. My problem is from a fish tracking study. We put
acoustic tags in fish and track them for several days. Location data is
supposed to be automatically recorded every time we detect a ping from the
fish. Unfortunately the GPS had some problems and sometimes the fishes depth
was recorded but not its location. I fortunately had a back-up GPS that was
taking location data every five minutes. I would like to merge the two files,
replacing the missing value in the vscan (automatic) file with the location
from the garmin file. Since we were getting vscan records every 1-2 seconds
and garmin records every 5 minutes, I need to find the right place in the vscan
file to place the garmin record - i.e. the
closest in time, but not greater than 5 minutes. I have written a function
that does this. However, it works with my test data but locks up my computer
with my real data. I have several million vscan records and several thousand
garmin records. Is there a better way to do this?
My function and test data:
myvscan-data.frame(c(1,NA,1.5),times(c(12:00:00,12:14:00,12:20:00)))
names(myvscan)-c(Latitude,DateTime)
mygarmin-data.frame(c(20,30,40),times((12:00:00,12:10:00,12:15:00)))
names(mygarmin)-c(Latitude,DateTime)
minute.diff-1/24/12 #Time diff is in days, so this is 5 minutes
for (k in 1:nrow(myvscan))
{
if (is.na(myvscan$Latitude[k]))
{
if ((min(abs(mygarmin$DateTime-myvscan$DateTime[k]))) minute.diff )
{
index.min.date-which.min(abs(mygarmin$DateTime-myvscan$DateTime[k]))
myvscan$Latitude[k]-mygarmin$Latitude[index.min.date]
}}}
I appreciate your help and advice.
Aloha,
Tim
Tim Clark
Department of Zoology
University of Hawaii
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