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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