On 12-10-15 06:00 AM, r-help-requ...@r-project.org wrote:
Jeff,
My understanding is that the lag command will lag an entire time series. That
isn't what I'm looking for.
I just want, for example, today, and 5 entries back.
for exmple:
iter <- '2011-05-18'
observations[iter] # works fin
Hi Noah,
On Sun, Oct 14, 2012 at 12:14 PM, Noah Silverman wrote:
> Hello,
>
>
> I have a time series object (xts) that I iterate over in a loop. Works fine.
>
> My challenge is that I want to be able to reference other entries in the
> series by math. i.e. For today's observation, what were t
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Noah Silverman
> Sent: Sunday, October 14, 2012 7:14 PM
> To: R-help@r-project.org
> Subject: [R] Date Math
>
> Hello,
>
>
> I have
Jeff,
My understanding is that the lag command will lag an entire time series. That
isn't what I'm looking for.
I just want, for example, today, and 5 entries back.
for exmple:
iter <- '2011-05-18'
observations[iter] # works fine, returns the row at that date.
index(observations[iter[)
?embed
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Jeff,
My understanding is that the lag command will lag an entire time series. That
isn't what I'm looking for.
I just want, for example, today, and 5 entries back.
for exmple:
iter <- '2011-05-18'
observations[iter] # works fine, returns the row at that date.
index(observations[iter[)
There are a few ways. The xts package has a "lag" function. So does
"zoo". Pay careful attention to the conventions used for specifying
relative time in these various packages. You can also infill your missing
data to create a regularly-spaced time series. There is no shortage of
web informa
Hello,
I have a time series object (xts) that I iterate over in a loop. Works fine.
My challenge is that I want to be able to reference other entries in the series
by math. i.e. For today's observation, what were the last 5 observations? If
indexed numerically, it is trivial, but I can fi
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