invalid 'tz' value, problems with time zone - r

Question

I'm working with minute data of NASDAQ, it has the index "2015-07-13 12:05:00 EST". I adjusted the system time with Sys.setenv(TZ = 'EST').
I want to program a simple buy/hold/sell strategy, therefore I create a vector of flat positions as a foundation.
pos_flat <- xts(rep(0, nrow(NASDAQ)), index(NASDAQ))
Then I want to apply a constraint, that in a certain time window, positions are bound to be flat, which in my case means equal to 1.
pos_flat["T13:41/T14:00"] <- 1
And this returns the error:
"Error in as.POSIXlt.POSIXct(.POSIXct(.index(x)), tz = indexTZ(x)) :invalid 'tz' value".
I also get this error doing other calculations, I just used this example because it is easy and shows the problem.
As extra information:
> Sys.timezone
function (location = TRUE)
{
tz <- Sys.getenv("TZ", names = FALSE)
if (nzchar(tz))
return(tz)
if (location)
return(.Internal(tzone_name()))
z <- as.POSIXlt(Sys.time())
zz <- attr(z, "tzone")
if (length(zz) == 3L)
zz[2L + z$isdst]
else zz[1L]
}
<bytecode: 0x03648ff4>
<environment: namespace:base>
I don't understand the problem with the tz value... Any ideas?

Answer 1

The source of your "invalid 'tz' value" error is because, for whatever reason, R doesn't accept tz = df$var. If you set tz = 'America/New_York' or some other character value, then it will work.
Better answer (instead of using force_tz below) for converting UTC times to various timezones based on location. It is also simpler and better than looping through or using a nested ifelse. I subset and change tz based on a timezone column (which my data already has, if not you can create it). Just make sure you account for all timezones in your data
(unique(df$timezone))
df$datetime2[df$timezone == 'America/New_York'] <- format(df$datetime, tz="America/New_York")[df$timezone == 'America/New_York']
df$datetime2[df$timezone == 'America/Chicago'] <- format(df$datetime, tz="America/Chicago")[df$timezone == 'America/Chicago']
df$datetime2[df$timezone == 'America/Denver'] <- format(df$datetime, tz="America/Denver")[df$timezone == 'America/Denver']
df$datetime2[df$timezone == 'America/Los_Angeles'] <- format(df$datetime, tz="America/Los_Angeles")[df$timezone == 'America/Los_Angeles']
Previous solution: Converting to Local Time in R - Vector of Timezones
require(lubridate)
require(dplyr)
df = data.frame(timestring = c("2015-12-12 13:34:56", "2015-12-14 16:23:32"), localzone = c("America/Los_Angeles", "America/New_York"), stringsAsFactors = F)
df$moment = as.POSIXct(df$timestring, format="%Y-%m-%d %H:%M:%S", tz="UTC")
df = df %>% rowwise() %>% mutate(localtime = force_tz(moment, localzone))
df

Answer 2

You are getting errors because "EST" is not a valid timezone specification. It's an abbreviation that's often used when printing and displaying timezones.
The index is printed as "2015-07-13 12:05:00 EST" because "EST" probably represents Eastern Standard Time in the United States. If you want to set the TZ environment variable to that timezone, you should use Sys.setenv() with Country/City notation:
Sys.setenv(TZ = "America/New_York")
You can also set the timezone in the xts constructor:
pos_flat <- xts(rep(0, nrow(NASDAQ)), index(NASDAQ), tzone = "America/New_York")

Answer 3

Your error occurs because of a misinterpretation of the time object. You need to have UNIX timestamps in order to use something like
pos_flat["T13:41/T14:00"] <- 1
Try a conversion of your indices by doing something like this:
index(NASDAQ) <- as.POSIXct(strptime(index(NASDAQ), "%Y-%m-%d %H:%M:%S"))
As you want to use EST, you have to change your environment variables (if you are not living in EST timezone). So all in all, this should work:
Sys.setenv(TZ = 'EST')
#load stuff
#...
index(NASDAQ) <- as.POSIXct(strptime(index(NASDAQ), "%Y-%m-%d %H:%M:%S"))
pos_flat <- xts(rep(0, nrow(NASDAQ)), index(NASDAQ))
pos_flat["T13:41/T14:00"] <- 1
For further information, have a look at the POSIXct and POSIXlt structures in R.
Best regards