-----Original Message-----
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of arthur
brogard via R-help
Sent: December 24, 2016 12:29 AM
To: Jeff Newmiller <jdnew...@dcn.davis.ca.us>
Cc: r-help@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 <jdnew...@dcn.davis.ca.us>
To: arthur brogard <abrog...@yahoo.com>
Cc: r-help@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$nextone)/6.5*1000
rates
ab
----- Original Message -----
From: Jeff Newmiller <jdnew...@dcn.davis.ca.us>
To: arthur brogard <abrog...@yahoo.com>; arthur brogard via R-help
<r-help@r-project.org>; "r-help@r-project.org" <r-help@r-project.org>
Sent: Saturday, 24 December 2016, 13:25
Subject: Re: [R] Is there a funct to sum differences?
Could you make your example reproducible? That is, include some sample
input and output. You talk about a column of numbers and then you seem to
work with named lists and I can't reconcile your words with the code I see.
--
Sent from my phone. Please excuse my brevity.
On December 23, 2016 3:40:18 PM PST, arthur brogard via R-help <r-help@r-
project.org> wrote:
I've been looking but I can't find a function to sum difference.
I have this code:
rates$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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