Is there a funct to sum differences?
Assuming John's understanding is correct, you can also do this without for loops. It takes getting used to vector and matrix arithmetic, which you can read more about in the Introduction to R document that comes with R, or on R Exercises website [1]. You indicated having a problem with my last reproducible example... it did work, if you went through it one step at a time. If you skipped steps, you would have problems like you encountered. For completeness, I will give the whole reproducible example again here... don't mix in your own steps until you have worked through all the steps in this example... or at least if you do, go back and step through these steps one at a time if you change something that breaks it. [1] http://r-exercises.com/2015/11/28/matrix-exercises/ #########------ begin 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$nextone)/6.5*1000 rates$experiment1 <- rates$Int + c( rates$Int[ -1 ], NA ) rates$Int2 <- (1:11)^2 rates$experiment2 <- rates$Int2 + c( rates$Int2[ -1 ], NA ) # lag N <- 5 # see ?embed, or https://en.wikipedia.org/wiki/Embedding embed( c( rates$Int2, rep( NA, N ) ), N+1 ) # make a matrix of the same size as the embed result matrix( rep( rates$Int2, N+1 ), ncol=N+1 ) # subtract the first values embed( c( rates$Int2, rep( NA, N ) ), N+1 ) - rates$Int2 # or can rely on automatic replication ... depends on the # fact that the embed result is a matrix which is really just # a vector displayed in folded up form embed( c( rates$Int2, rep( NA, N ) ), N+1 ) - rates$Int2 # anyway, the result can be computed in one line (wrapped for readability) rates$experiment3 <- rowSums( embed( c( rates$Int2 , rep( NA, N ) ) , N+1 ) - rates$Int2 , na.rm=TRUE )
rates
Date Int thisone nextone lastone experiment1 Int2 experiment2 experiment3
1 Jan-1959 5 0 0 0 10 1 5 85
2 Feb-1959 5 0 0 0 10 4 13 115
3 Mar-1959 5 0 0 0 10 9 25 145
4 Apr-1959 5 0 0 0 10 16 41 175
5 May-1959 5 0 0 0 10 25 61 205
6 Jun-1959 5 0 0 0 10 36 85 235
7 Jul-1959 5 0 0 0 10 49 113 170
8 Aug-1959 5 0 0 0 10 64 145 110
9 Sep-1959 5 0 0 0 10 81 181 59
10 Oct-1959 5 0 NA NA 10 100 221 21
11 Nov-1959 5 NA NA NA NA 121 NA 0
#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), experiment1 = c(10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
NA), Int2 = c(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121), experiment2 =
c(5, 13, 25, 41, 61, 85, 113, 145, 181, 221, NA), experiment3 = c(85,
115, 145, 175, 205, 235, 170, 110, 59, 21, 0)), .Names = c("Date",
"Int", "thisone", "nextone", "lastone", "experiment1", "Int2",
"experiment2", "experiment3"), row.names = c(NA, -11L), class =
"data.frame")
#########------ end
On Sat, 24 Dec 2016, Fox, John wrote:
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) + }
head(diffs)
[,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 Canada L8S 4M4 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.
Thanks for the tip about 'reply all', I had no idea.
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" ) )
#your code
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$Int2[ -1 ], NA )
# dput(rates)
result <- structure(list(Date = c("Jan-1959", "Feb-1959", "Mar-1959", "Apr-
1959", "May-1959", "Jun-1959", "Jul-1959", "Aug-1959", "Sep-1959", "Oct-
1959", "Nov-1959"), Int = c(5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5), thisone = c(0, 0, 0, 0, 0,
0, 0, 0, 0, 0, NA), nextone = c(0, 0, 0, 0, 0, 0, 0, 0, 0, NA, NA), lastone = c(0, 0, 0,
0, 0, 0, 0, 0, 0, NA, NA), Int2 = c(1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121),
experiment1 = c(10, 10, 10, 10, 10, 10, 10, 10, 10, 10, NA), experiment2 = c(5,
13, 25, 41, 61, 85, 113, 145, 181, 221, NA)), .Names = c("Date", "Int",
"thisone", "nextone", "lastone", "Int2", "experiment1", "experiment2"),
row.names = c(NA, -11L), class = "data.frame")
On Sat, 24 Dec 2016, arthur brogard wrote:
Yes, sure, thanks for your interest. I apologise for not submitting in the
correct manner. I'll learn (I hope).
Here's the source - a spreadsheet with just two columns, date and 'Int'. Date Int Jan-1959 5 Feb-1959 5 Mar-1959 5 Apr-1959 5 May-1959 5 Jun-1959 5 Jul-1959 5 Aug-1959 5 Sep-1959 5 Oct-1959 5 Nov-1959 5 After processing it becomes this:
rates
Date Int thisone nextone lastone finalone
1 1959-01-01 5.00 0.00 0.00 0.000000 10
2 1959-02-01 5.00 0.00 0.00 0.000000 10
3 1959-03-01 5.00 0.00 0.00 0.000000 10
4 1959-04-01 5.00 0.00 0.00 0.000000 10
5 1959-05-01 5.00 0.00 0.00 0.000000 10
6 1959-06-01 5.00 0.00 0.00 0.000000 10
The one long column I'm referring to is the 'Int' column which R has imported.
The actual code is:
rates <- read.csv("Rates2.csv",header =
TRUE,colClasses=c("character","numeric"))
sapply(rates,class)
rates$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?
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______________________________________________ R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide 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