# [R] Calculate daily means from 5-minute interval data

Rich Shepard r@hep@rd @end|ng |rom @pp|-eco@y@@com
Thu Sep 2 20:16:16 CEST 2021

```On Mon, 30 Aug 2021, Richard O'Keefe wrote:

>> x <- rnorm(samples.per.day * 365)
>> length(x)
>  105120
>
> Reshape the fake data into a matrix where each row represents one
> 24-hour period.
>
>> m <- matrix(x, ncol=samples.per.day, byrow=TRUE)

Richard,

Now I understand the need to keep the date and time as a single datetime
column; separately dplyr's sumamrize() provides daily means (too many data
points to plot over 3-5 years). I reformatted the data to provide a
sampledatetime column and a values column.

If I correctly understand the output of as.POSIXlt each date and time
element is separate, so input such as 2016-03-03 12:00 would now be 2016 03
03 12 00 (I've not read how the elements are separated). (The TZ is not
important because all data are either PST or PDT.)

> Now we can summarise the rows any way we want.
> The basic tool here is ?apply.
> ?rowMeans is said to be faster than using apply to calculate means,
> so we'll use that.  There is no *rowSds so we have to use apply
> for the standard deviation.  I use ?head because I don't want to
> post tens of thousands of meaningless numbers.

If I create a matrix using the above syntax the resulting rows contain all
recorded values for a specific day. What would be the syntax to collect all
values for each month?

This would result in 12 rows per year; the periods of record for the five
variables availble from that gauge station vary in length.

Regards,

Rich

```