# [R] Is there a funct to sum differences?

Fox, John jfox at mcmaster.ca
Sun Dec 25 16:53:16 CET 2016

```Dear Arthur,

Neither my nor Jeff Newmiller's solution uses any fancy math, just a little bit of programming. Here are the two solutions on a much larger simulated problem:

> set.seed(12345) # for reproducibility
> x <- rnorm(1e5)
> len <- length(x)
> maxlag <- 100
>
> # John:
> system.time(
+     {
+         diffs <- matrix(0, len, maxlag)
+         for (lag in 1:maxlag){
+             diffs[1:(len - lag), lag] <- diff(x, lag=lag)
+         }
+     }
+ )
user  system elapsed
0.22    0.19    0.41
[1] -34.39477 -48.65417  33.75448  67.30261 -39.10066 204.56559
>
> # Jeff:
> system.time(
+     diffs.2 <- embed(c(x, rep(NA, maxlag)), maxlag + 1) - x
+ )
user  system elapsed
0.36    0.04    0.39
[1] -34.39477 -48.65417  33.75448  67.30261 -39.10066 204.56559

My solution uses a loop, Jeff's uses the embed() function -- of which I was unaware -- which hides the loop in the function.

If you want to do this kind of simple data management in R, it helps to learn some R programming.

Best,
John

> -----Original Message-----
> From: arthur brogard [mailto:abrogard at yahoo.com]
> Sent: Saturday, December 24, 2016 4:24 PM
> To: Fox, John <jfox at mcmaster.ca>
> Cc: Jeff Newmiller <jdnewmil at dcn.davis.ca.us>; r-help at r-project.org
> Subject: Re: [R] Is there a funct to sum differences?
>
>
>
> Hello John,
>
>
> Here I am back again. Having learned no maths yet but I've looked over
> the results here and they are what I am after.
>
> Now I'll try to understand how you did it.
>
> :)
>
>
>
>
> ----- Original Message -----
> From: "Fox, John" <jfox at mcmaster.ca>
> To: arthur brogard <abrogard at yahoo.com>
> Cc: Jeff Newmiller <jdnewmil at dcn.davis.ca.us>; "r-help at r-project.org"
> <r-help at r-project.org>
> Sent: Sunday, 25 December 2016, 0:55
> Subject: RE: [R] Is there a funct to sum differences?
>
> Dear Arthur,
>
> Here's a simple script to do what I think you want. I've applied it to a
> contrived example, a vector of the squares of the integers 1 to 25, and
> have summed the first 5 differences, but the script is adaptable to any
> numeric vector and any maximum lag. You'll have to decide what to do
> with the last maximum-lag (in my case, 5) entries:
>
> -------------- snip ------------
> > (x <- (1:25)^2)
> [1]   1   4   9  16  25  36  49  64  81 100 121 144 169 196 225 256 289
> 324 361 400 441 484 529 576
> [25] 625
> > len <- length(x)
> > maxlag <- 5
> > diffs <- matrix(0, len, maxlag)
> > for (lag in 1:maxlag){
> +     diffs[1:(len - lag), lag] <- diff(x, lag=lag) }
>      [,1] [,2] [,3] [,4] [,5]
> [1,]    3    8   15   24   35
> [2,]    5   12   21   32   45
> [3,]    7   16   27   40   55
> [4,]    9   20   33   48   65
> [5,]   11   24   39   56   75
> [6,]   13   28   45   64   85
> > tail(diffs)
>       [,1] [,2] [,3] [,4] [,5]
> [20,]   41   84  129  176  225
> [21,]   43   88  135  184    0
> [22,]   45   92  141    0    0
> [23,]   47   96    0    0    0
> [24,]   49    0    0    0    0
> [25,]    0    0    0    0    0
> > rowSums(diffs)
> [1]  85 115 145 175 205 235 265 295 325 355 385 415 445 475 505 535 565
> 595 625 655 450 278 143  49
> [25]   0
> -------------- snip ------------
>
> The script could very simply be converted into a function if this is a
> repetitive task with variable inputs.
>
> I hope this helps,
> John
>
> -----------------------------
> John Fox, Professor
> McMaster University
> Hamilton, Ontario
> Web: socserv.mcmaster.ca/jfox
>
>
>
> > -----Original Message-----
> > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of arthur
> > brogard via R-help
> > Sent: December 24, 2016 12:29 AM
> > To: Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
> > Cc: r-help at r-project.org
> > Subject: Re: [R] Is there a funct to sum differences?
> >
> > Yes, sorry about that.  I keep making mistakes I shouldn't make.
> >
> >
> > You can ignore the finalone. I have been doing other work on this and
> > it comes from there. I took the example from the R screen after it had
> > run one of these other things that created the finalone.
> >
> > I guess I was thinking just seeing the data mentioned in the code was
> > be enough.
> >
> > I don't want a function to do the division and multiplication.
> >
> > It's a function that will ".. automatically sum the difference between
> > the first
> >
> >  and subsequent to the end of a list? "  that I am looking for.
> >
> > I will try to explain, I know I often don't make myself clear:
> >
> > I'm using this diff() function.
> >
> > This 'diff()' function finds the difference between two adjoining
> > entries and it applies itself to the whole list so that in an instant
> > I can have a list of differences between any two adjoining.
> >
> > Then I can have a list of differences between any two with any
> > specified gap - 'lag' it is called.
> > Using the same function.
> >
> > Now I have them and do that.  Then I add them together to find the
> 'lastone'
> > which is the total difference for the period.
> >
> >
> > Now here's the point:  that covers a period of two timespans, months,
> they are.
> >
> >  if I want to cover a span of 24 months, say, then I would have to
> > write this
> > diff() function 24 times.
> >
> >  what I'm doing is finding the difference between the starting point
> > and every other point and then adding them all together.  bit like
> > finding the area beneath the curve maybe.
> >
> >  And that's what I want to do.
> >
> >  :)
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > ----- Original Message -----
> > From: Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
> > To: arthur brogard <abrogard at yahoo.com>
> > Cc: r-help at r-project.org
> > Sent: Saturday, 24 December 2016, 15:34
> > Subject: Re: [R] Is there a funct to sum differences?
> >
> > You need to "reply all" so other people can help as well, and others
> > can learn from your questions.
> >
> > I am still puzzled by how you expect to compute "finalone". If you had
> > supplied numbers other than all 5's it might have been easier to
> > figure out what is going on.
> >
> > What is your purpose in performing this calculation?
> >
> > #### reproducible code
> > rates <- read.table( text =
> > "Date          Int
> > Jan-1959        5
> > Feb-1959        5
> > Mar-1959        5
> > Apr-1959        5
> > May-1959        5
> > Jun-1959        5
> > Jul-1959        5
> > Aug-1959        5
> > Sep-1959        5
> > Oct-1959        5
> > Nov-1959        5
> > ", header = TRUE, colClasses = c( "character", "numeric" ) )
> >
> > rates\$thisone <- c(diff(rates\$Int), NA) rates\$nextone <-
> > c(diff(rates\$Int, lag=2), NA, NA) rates\$lastone <- (rates\$thisone +
> > rates\$nextone)/6.5*1000 # I doubt there is a ready-built function that
> > knows you want to # divide by 6.5 or multiply by 1000
> >
> > # form a vector from positions 2:11 and append NA)
> > rates\$experiment1 <- rates\$Int + c( rates\$Int[ -1 ], NA ) # numbers
> > that are not all the same
> > rates\$Int2 <- (1:11)^2
> > rates\$experiment2 <- rates\$Int2 + c( rates\$Int2[ -1 ], NA )
> >
> > # dput(rates)
> > result <- structure(list(Date = c("Jan-1959", "Feb-1959", "Mar-1959",
> > "Apr- 1959", "May-1959", "Jun-1959", "Jul-1959", "Aug-1959",
> > "Sep-1959", "Oct- 1959", "Nov-1959"), Int = c(5, 5, 5, 5, 5, 5, 5, 5,
> > 5, 5, 5), thisone = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, NA), nextone =
> > c(0, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA), lastone = c(0, 0, 0, 0, 0, 0, 0,
> > 0, 0, NA, NA), Int2 = c(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121),
> > experiment1 = c(10, 10, 10, 10, 10, 10, 10, 10, 10, 10, NA),
> > experiment2 = c(5, 13, 25, 41, 61, 85, 113, 145, 181, 221, NA)),
> > .Names = c("Date", "Int", "thisone", "nextone", "lastone", "Int2",
> > "experiment1", "experiment2"), row.names = c(NA, -11L), class =
> > "data.frame")
> >
> > On Sat, 24 Dec 2016, arthur brogard wrote:
> >
> > >
> > >
> > > Yes, sure, thanks for your interest.  I apologise for not submitting
> > > in the
> > correct manner.  I'll learn (I hope).
> > >
> > > Here's the source - a spreadsheet with just two columns, date and
> 'Int'.
> > >
> > >
> > > Date    Int
> > > Jan-1959    5
> > > Feb-1959    5
> > > Mar-1959    5
> > > Apr-1959    5
> > > May-1959    5
> > > Jun-1959    5
> > > Jul-1959    5
> > > Aug-1959    5
> > > Sep-1959    5
> > > Oct-1959    5
> > > Nov-1959    5
> > >
> > >
> > > After processing it becomes this:
> > >
> > >
> > >> rates
> > > Date   Int thisone nextone     lastone finalone
> > > 1   1959-01-01  5.00    0.00    0.00    0.000000       10
> > > 2   1959-02-01  5.00    0.00    0.00    0.000000       10
> > > 3   1959-03-01  5.00    0.00    0.00    0.000000       10
> > > 4   1959-04-01  5.00    0.00    0.00    0.000000       10
> > > 5   1959-05-01  5.00    0.00    0.00    0.000000       10
> > > 6   1959-06-01  5.00    0.00    0.00    0.000000       10
> > >
> > > The one long column I'm referring to is the 'Int' column which R has
> imported.
> > >
> > > The actual code is:
> > >
> > >
> > > TRUE,colClasses=c("character","numeric"))
> > >
> > > sapply(rates,class)
> > >
> > > rates\$Date <- strptime(paste0("1-", rates\$Date), format="%d-%b-%Y",
> > > tz="UTC")
> > >
> > >
> > > rates\$thisone <- c(diff(rates\$Int), NA) rates\$nextone <-
> > > c(diff(rates\$Int, lag=2), NA, NA) rates\$lastone <- (rates\$thisone +
> > > rates\$nextone)/6.5*1000
> > >
> > >
> > > rates
> > >
> > >
> > >
> > > ab
> > >
> > >
> > >
> > > ----- Original Message -----
> > > From: Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
> > > To: arthur brogard <abrogard at yahoo.com>; arthur brogard via R-help
> > > <r-help at r-project.org>; "r-help at r-project.org"
> > > <r-help at r-project.org>
> > > Sent: Saturday, 24 December 2016, 13:25
> > > Subject: Re: [R] Is there a funct to sum differences?
> > >
> > > Could you make your example reproducible? That is, include some
> > > sample
> > input and output. You talk about a column of numbers and then you seem
> > to work with named lists and I can't reconcile your words with the
> code I see.
> > > --
> > > Sent from my phone. Please excuse my brevity.
> > >
> > >
> > > On December 23, 2016 3:40:18 PM PST, arthur brogard via R-help
> > > <r-help at r-
> > project.org> wrote:
> > >> I've been looking but I can't find a function to sum difference.
> > >>
> > >> I have this code:
> > >>
> > >>
> > >> rates\$thisone <- c(diff(rates\$Int), NA) rates\$nextone <-
> > >> c(diff(rates\$Int, lag=2), NA, NA) rates\$lastone <- (rates\$thisone +
> > >> rates\$nextone)
> > >>
> > >>
> > >> It is looking down one long column of numbers.
> > >>
> > >> It sums the difference between the first two and then between the
> > >> first and third and so on.
> > >>
> > >> Can it be made to automatically sum the difference between the
> > >> first and subsequent to the end of a list?
> > >>
> > >> ______________________________________________
> > >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > >> https://stat.ethz.ch/mailman/listinfo/r-help
> > >> http://www.R-project.org/posting-guide.html
> > >> and provide commented, minimal, self-contained, reproducible code.
> > >
> >
> > ----------------------------------------------------------------------
> -----
> > Jeff Newmiller                        The     .....       .....  Go
> Live...
> > DCN:<jdnewmil at dcn.davis.ca.us>        Basics: ##.#.       ##.#.  Live
> Go...
> >                                        Live:   OO#.. Dead: OO#..
> Playing
> > Research Engineer (Solar/Batteries            O.O#.       #.O#.  with
> > /Software/Embedded Controllers)               .OO#.       .OO#.
> rocks...1k
>
> >
> > ______________________________________________
> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help