-----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 ------------
[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
[,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
[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
which is the total difference for the period.
Now here's the point: that covers a period of two timespans, months,
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
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:
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
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$nextone)/6.5*1000
rates
ab
----- Original Message -----
From: Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
To: arthur brogard <abrogard at yahoo.com>; arthur brogard via R-help
<r-help at r-project.org>; "r-help at r-project.org"
<r-help at r-project.org>
Sent: Saturday, 24 December 2016, 13:25
Subject: Re: [R] Is there a funct to sum differences?
Could you make your example reproducible? That is, include some
sample
input and output. You talk about a column of numbers and then you seem
to work with named lists and I can't reconcile your words with the
--
Sent from my phone. Please excuse my brevity.
On December 23, 2016 3:40:18 PM PST, arthur brogard via R-help
<r-help at r-
I've been looking but I can't find a function to sum difference.
I have this code:
rates$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?