How to remove a row from zoo/xts object, given a timestamp - r

Question

I was happily running with this code:
z=lapply(filename_list, function(fname){
read.zoo(file=fname,header=TRUE,sep = ",",tz = "")
})
xts( do.call(rbind,z) )
until Dirty Data came along with this at the end of one file:
Open High Low Close Volume
2011-09-20 21:00:00 1.370105 1.370105 1.370105 1.370105 1
and this at the start of the next file:
Open High Low Close Volume
2011-09-20 21:00:00 1.370105 1.371045 1.369685 1.3702 2230
So rbind.zoo complains about a duplicate.
I can't use something like:
y <- x[ ! duplicated( index(x) ), ]
as they are in different zoo objects, inside a list. And I cannot use aggregate, as suggested here because they are a list of zoo objects, not one big zoo object. And I can't get one big object 'cos of the duplicates. Catch-22.
So, when the going gets tough, the tough hack together some for loops (excuse the prints and a stop, as this isn't working code yet):
indexes <- do.call("c", unname(lapply(z, index)))
dups=duplicated(indexes)
if(any(dups)){
duplicate_timestamps=indexes[dups]
for(tix in 1:length(duplicate_timestamps)){
t=duplicate_timestamps[tix]
print("We have a duplicate:");print(t)
for(zix in 1:length(z)){
if(t %in% index(z[[zix]])){
print(z[[zix]][t])
if(z[[zix]][t]$Volume==1){
print("-->Deleting this one");
z[[zix]][t]=NULL #<-- PROBLEM
}
}
}
}
stop("There are duplicate bars!!")
}
The bit I've got stumped on is assigning NULL to a zoo row doesn't delete it (Error in NextMethod("[<-") : replacement has length zero). OK, so I do a filter-copy, without the offending item... but I'm tripping up on these:
> z[[zix]][!t,]
Error in Ops.POSIXt(t) : unary '!' not defined for "POSIXt" objects
> z[[zix]][-t,]
Error in `-.POSIXt`(t) : unary '-' is not defined for "POSIXt" objects
P.S. While high-level solutions to my real problem of "duplicates rows across a list of zoo objects" are very welcome, the question here is specifically about how to delete a row from a zoo object given a POSIXt index object.
A small bit of test data:
list(structure(c(1.36864, 1.367045, 1.370105, 1.36928, 1.37039,
1.370105, 1.36604, 1.36676, 1.370105, 1.367065, 1.37009, 1.370105,
5498, 3244, 1), .Dim = c(3L, 5L), .Dimnames = list(NULL, c("Open",
"High", "Low", "Close", "Volume")), index = structure(c(1316512800,
1316516400, 1316520000), class = c("POSIXct", "POSIXt"), tzone = ""), class = "zoo"),
structure(c(1.370105, 1.370115, 1.36913, 1.371045, 1.37023,
1.37075, 1.369685, 1.36847, 1.367885, 1.3702, 1.36917, 1.37061,
2230, 2909, 2782), .Dim = c(3L, 5L), .Dimnames = list(NULL,
c("Open", "High", "Low", "Close", "Volume")), index = structure(c(1316520000,
1316523600, 1316527200), class = c("POSIXct", "POSIXt"), tzone = ""), class = "zoo"))
UPDATE: Thanks to G. Grothendieck for the row-deleting solution. In the actual code I followed the advice of Joshua and GSee to get a list of xts objects instead of a list of zoo objects. So my code became:
z=lapply(filename_list, function(fname){
xts(read.zoo(file=fname,header=TRUE,sep = ",",tz = ""))
})
x=do.call.rbind(z)
(As an aside, please note the call to do.call.rbind. This is because rbind.xts has some serious memory issues. See https://stackoverflow.com/a/12029366/841830 )
Then I remove duplicates as a post-process step:
dups=duplicated(index(x))
if(any(dups)){
duplicate_timestamps=index(x)[dups]
to_delete=x[ (index(x) %in% duplicate_timestamps) & x$Volume<=1]
if(nrow(to_delete)>0){
#Next line says all lines that are not in the duplicate_timestamp group
# OR are in the duplicate timestamps, but have a volume greater than 1.
print("Will delete the volume=1 entry")
x=x[ !(index(x) %in% duplicate_timestamps) | x$Volume>1]
}else{
stop("Duplicate timestamps, and we cannot easily remove them just based on low volume.")
}
}

Answer 1

If z1 and z2 are your zoo objects then to rbind while removing any duplicates in z2:
rbind( z1, z2[ ! time(z2) %in% time(z1) ] )
Regarding deleting points in a zoo object having specified times, the above already illustrates this but in general if tt is a vector of times to delete:
z[ ! time(z) %in% tt ]
or if we knew there were a single element in tt then z[ time(z) != tt ] .

Answer 2

rbind.xts will allow duplicate index values, so it could work if you convert to xts first.
x <- lapply(z, as.xts)
y <- do.call(rbind, x)
# keep last value of any duplicates
y <- y[!duplicated(index(y),fromLast=TRUE),]

Answer 3

I think you'll have better luck if you convert to xts first.
a <- structure(c(1.370105, 1.370105, 1.370105, 1.370105, 1), .Dim = c(1L,
5L), index = structure(1316570400, tzone = "", tclass = c("POSIXct",
"POSIXt")), .indexCLASS = c("POSIXct", "POSIXt"), tclass = c("POSIXct",
"POSIXt"), .indexTZ = "", tzone = "", .Dimnames = list(NULL,
c("Open", "High", "Low", "Close", "Volume")), class = c("xts",
"zoo"))
b <- structure(c(1.370105, 1.371045, 1.369685, 1.3702, 2230), .Dim = c(1L,
5L), index = structure(1316570400, tzone = "", tclass = c("POSIXct",
"POSIXt")), .indexCLASS = c("POSIXct", "POSIXt"), tclass = c("POSIXct",
"POSIXt"), .indexTZ = "", tzone = "", .Dimnames = list(NULL,
c("Open", "High", "Low", "Close", "Volume")), class = c("xts",
"zoo"))
(comb <- rbind(a, b))
# Open High Low Close Volume
#2011-09-20 21:00:00 1.370105 1.370105 1.370105 1.370105 1
#2011-09-20 21:00:00 1.370105 1.371045 1.369685 1.370200 2230
dupidx <- index(comb)[duplicated(index(comb))] # indexes of duplicates
tail(comb[dupidx], 1) #last duplicate
# now rbind the last duplicated row with all non-duplicated data
rbind(comb[!index(comb) %in% dupidx], tail(comb[dupidx], 1))