I do not see the problem.
In base R
Min <- min(var, na.rm=T)
Max <- max(var, na.rm=T)
Span <- c(Min, Max)
Is the same output as
Span <- range(var, na.rm=T)
diff(range(var, na.rm=T)) returns Max - Min
range() returns a vector with two values, the minimum and the maximum.
What should the function
The end of file is a problem. In my case I have data files that can end in one
of several ways. A line can end with \r or \n.
1) No line feed at the end of the last row of data.
2) One line feed at the end of the last row of data.
3) Multiple line feeds at the end of the last row of data.
4) All o
more fun, log transform these variables before plotting so that parts
of the variable names are also part of the valid code. For me, coding is hard
enough without additional complexity.
Tim
-Original Message-
From: Jeff Newmiller
Sent: Wednesday, May 7, 2025 8:29 PM
To: r-help@r-pr
Some issues:
1) Variable names cannot have spaces. "merged 1" is not valid but "merged_1" is
a valid alternative.
2) You need to tell R what to merge by. It looks like you may be using data
tables rather than a data frame.
merged <- dataset2[dataset1, on = "id", nomatch = NA]
3) Alternatively: j
(adding slightly to Gregg's answer)
Why do professionals use both? Computer intensive methods (bootstrap,
randomization, jackknife) are data hungry. They do not work well if I have a
sample size of 4. One could argue that the traditional methods also have
trouble, but one could also think of the
ifficulty in asking the right questions as complexity
increases.
Tim
-Original Message-
From: Jeff Newmiller
Sent: Tuesday, April 15, 2025 3:16 AM
To: r-help@r-project.org; Ebert,Timothy Aaron ; Bert Gunter
; Duncan Murdoch
Cc: r-help@r-project.org
Subject: Re: [R] Drawing a sample ba
1) select only denied individuals from the original data. This is S.
2) There is a fixed sample size of exactly t
3) There is a fixed target sum T such that sum(t values from S) = T
You can reduce the problem. All large values where the max(S) + (t-1 smallest
values) > T can be eliminated from S
1) extract relevant observations to a new data frame. Either of these two
commands works.
filter(df, age > 24)
df[df$age > 24, ]
2) sample the new data frame
df[sample(nrow(df), 10, replace = TRUE), ]
change TRUE to FALSE if you do not want replacement.
Tim
-Original Message-
From: R-hel
I would use the approach that is most commonly used by my potential user base
unless I can define a clear set of reasons why I should do it differently. If
users will be unable to see the difference, then the decision is entirely up to
me based on ease of programming at all phases (writing, debu
The basic process is to make each tibble have one or more columns to merge by.
Then merge the tibbles. Can you show us a couple of tries and describe why or
how they failed? Note that you cannot merge using a row in one tibble and a
column in the other tibble without first reshaping the first ti
You could try writing code to write code. However, you need to tell the program
which variables are factor and which are continuous. If you are lucky and all
integer data are factor and all non-integers are continuous, then you can sort
for that. However, in a broad data perspective this is a ba
Your result data frame example makes no sense to me. The price and executed_qty
are the same for all symbols?
To get it all into one data frame you need a common variable that is used to
join the data frames.
My guess is that all_trade_sample$symbol has equivalents to the variables in
token_clo
variables. Rejoin the
dependent variable to the data frame. Pass that to reformulate.
Tim
-----Original Message-
From: Rui Barradas
Sent: Sunday, March 30, 2025 4:05 AM
To: Ebert,Timothy Aaron ; Bert Gunter ;
Richard M. Heiberger ; R-help
Subject: Re: [R] [External] Creating model formulas progr
I am confused. Richard's answer that Bert did not like did not use parse
explicitly. Richard pasted together a string that a function like lm() will
have to parse to run the analysis. However, the answers so far do not use
parse(). In the reply to Richard, Bert indicated we cannot use strings. E
The general formula is y ~ a + b + c + ...
There is this approach:
formula <- reformulate(independent_vars, response = "y")
model <- lm(formula, data = mydata)
summary(model)
It does not generate a string object, but the formula is still a string even if
it is of class formula. Also, in this app
The computer intensive approaches (randomization, permutation, bootstrap,
jackknife) are awesome when you have enough data. In this age we are all about
huge data sets. Yet basic agricultural research often does not come close. I
have three to ten replicates per treatment.
-Original Message
How about calculating a 95% confidence interval about the estimated proportion
in favor. The PooledInfRate package will do this for you. If confidence
intervals overlap then there is no significant difference.
-Original Message-
From: R-help On Behalf Of Kevin Zembower via
R-help
Sent:
Brian Manly wrote a nice book about computer intensive methods in data
analysis:
https://www.amazon.com/Randomization-Bootstrap-Methods-Biology-Statistical/dp/1584885416.
Therein there is a distinction between permutation tests and randomization
tests. A permutation test uses every permutation
This is fun.
In a stats class you are trying to deal with data. There is the underlying
distribution. This is the random number generator. I have a population that is
following the underlying distribution. In this case my population is 10,000
individuals with a true population mean of 40 and a s
This is a general problem across all software. A computer has finite memory but
a floating-point value can have an infinite number of digits. Consider trying
to store the exact value of pi. All computer software has a limit to available
precision, and that value is specific to each piece of soft
peers and professionalism should guide your
choices.
-Original Message-
From: Jeff Newmiller
Sent: Monday, December 30, 2024 11:23 PM
To: Ebert,Timothy Aaron ; r-help@r-project.org; Erin Hodgess
Subject: RE: [R] Citation for stock price data from Quantmod
[External Email]
The _point_ is
range or field
of possible/probable results.
-Original Message-
From: Jeff Newmiller
Sent: Monday, December 30, 2024 7:52 PM
To: r-help@r-project.org; Ebert,Timothy Aaron ; Erin Hodgess
; r-help@r-project.org
Subject: Re: [R] Citation for stock price data from Quantmod
[External Email
Yes, you need to cite the source of data and the method for retrieval. You also
need to cite programs used in manipulating the data, cleaning the data, and
analyzing the data. I should be able to read what you wrote, and using those
methods recover the data independently from you and finish all
POLWART
Sent: Friday, December 13, 2024 3:03 AM
To: Ebert,Timothy Aaron
Cc: Erin Hodgess ; Bill Dunlap
; r-help@R-project.org
Subject: Re: [R] [off-topic] crossword
[External Email]
Well to complicate things, I don't think RULES is the answer.
This is a cryptic crossword clue. They usually co
I do not understand the question and I do not understand the answer. Possibly
one confounds the other.
-Original Message-
From: R-help On Behalf Of Erin Hodgess
Sent: Thursday, December 12, 2024 11:56 AM
To: Bill Dunlap
Cc: r-help@R-project.org
Subject: Re: [R] [off-topic] crossword
[E
quot;, y = "") +
scale_fill_gradient2(low = "#E94A26", mid = "white", high = "#A1D385",
midpoint = 0.5) +
scale_x_continuous(
breaks = seq.Date(as.Date("2024-01-01"), as.Date("2024-06-01"), by =
"month"),
labels =
What happens if you switch the colors in this line:
scale_fill_gradient2(low = "#E94A26", mid = "white", high = "#A1D385",
midpoint = 0.5) +
to be the following
scale_fill_gradient2(low = "# A1D385", mid = "white", high = "# E94A26",
midpoint = 0.5) +
That said, a red-green heat map may be
Here is another version using for loops.
newdata3 <- olddata |>
dplyr::arrange(ID, date)
newdata3$firstday <- NA
for (i in 1:nrow(newdata3)) {
if (i == 1) {
dayz <- newdata3$date[i]
} else {
if (newdata3$ID[i] != newdata3$ID[i - 1]) {
dayz <- newdata3$date[i]
}
}
newdat
Very similar to what Oliver posted:
library(dplyr)
newdata <- olddata |>
group_by(ID) |>
mutate(firstdate = first(date))
newdata
1) I attached dplyr to the entire program. Oliver used dplyr::group_by() and
dplyr::mutate() to do the same thing.
2) I used the base R |> pipe while Oliver used
1) If you already have two data frames, one as shown, the other with Wind_Speed
and Wind_Dir, then you can use cbind() to join them if they have the same
number of rows. If not, then you will have to edit one of the data frames (to
get the same number of rows) or add column(s) so that both data
I had something like this problem where the program read files but some files
would be missing the last line of data. My solution was to write another
program that looked at the files and if a file did not end in a line feed to
add one. Possibly a bit primitive, but it worked.
Regards,
Tim
I might start by trying to figure out what they are trying to accomplish. Is R
a requirement for a GUI that requires no R coding? Would something like PAST
work? https://www.nhm.uio.no/english/research/resources/past/
Tim
-Original Message-
From: R-help On Behalf Of tgs77m--- via R-hel
Why must the answer use apply? It feels like there are elements of the problem
that are not explained.
-Original Message-
From: R-help On Behalf Of Ben Bolker
Sent: Friday, October 4, 2024 8:45 AM
To: r-help@r-project.org
Subject: Re: [R] apply
[External Email]
It's still hard to fi
If you can get a vector with all the package names (I do not know how to do
this) then you could do something like this;
# Function to check and install missing packages
install_if_missing <- function(pkg) {
if (!require(pkg, character.only = TRUE)) {
install.packages(pkg, dependencies = TRU
Say that you have several files from different places or times and you wanted
to run your program on all of them without reprogramming. You could start with
the readr package and use guess_encoding.
j <- 1
for (i in file_paths){
file_encoding[j] <- as.character(readr::guess_encoding(i)$encoding
Why not use na.omit() and then go from there? Unless one handles NA differently
in different groups there is no point in processing the data by groups to
remove NA even if later analysis steps do require group information.
Tim
-Original Message-
From: R-help On Behalf Of Rui Barradas
S
In proc glm, SAS will give both type I and type III sums of squares. You can
get type II if you ask. Other procedures in SAS have different output as
defaults.
Introductory textbooks may only cover type I SS. It is easy to calculate and
gets the idea across. In application one of the problems i
Here is one response: https://doi.org/10.1093/jisesa/iew092
Or paraphrased: yes.
Regards,
Tim
-Original Message-
From: R-help On Behalf Of Bert Gunter
Sent: Wednesday, July 24, 2024 10:44 AM
To: R-help
Subject: [R] OFF TOPIC: Nature article on File Drawer Problem in Reserach
[External
The desired result is odd.
1) It looks like the string is duplicated in the desired result. The first line
of data has "15, xc, Ab", and the desired result has "15, xc, Ab, 15, xc, Ab"
2) The example has S1 through S5, but the desired result has data for eight
variables in the first line (not fi
In a lm() model a significant intercept means that the line passes above or
below the intercept (x=0, y=0). A significant predictor means that the slope is
not zero. More generally the significant predictor means that the predictor
has some influence on the predicted. With nlme() the relationsh
NSims <- 4
Grps <- 5
DiffMeans <- matrix(nrow=NSims,ncol=1+Grps)
DiffMeans
#I have a problem when I try to name the columns of the matrix. I want the
first column to be NSims, #and the other columns to be something like Value1,
Value2, . . . Valuen where N=Grps
Colnames <- as.vector("NSims")
num
library(lubridate)
library(dplyr)
df1 <- read.table(text = "YEAR DOY HR IMF SW SSNDst f10.7
2012 214 0 3.4 403. 132-9 154.6
2012 214 1 3.7 388. 132 -10 154.6
2012 214 2 3.7 383. 132 -10 154.6
2012 214 3 3.7 391. 132-9 154.6
2012 215 4 5.1 371. 143-4 138.
Bert Gunter
Sent: Thursday, June 13, 2024 9:13 PM
To: Ebert,Timothy Aaron
Cc: Francesca PANCOTTO ; r-help@r-project.org
Subject: Re: [R] Create a numeric series in an efficient way
[External Email]
Nope. She would have wanted the 'each' argument = 84. See ?rep.
-- Bert
On Thu, Jun 13,
Maybe this was your solution?
blocC <- c(rep(x=c(1:13), times=84))
blocC <- arrange(.data = data.frame(blocC), blocC)
The second line sorts, but that may not be needed depending on application. The
object class is also different in the sorted solution.
Tim
-Original Message-
From: R-hel
Hi Yuan,
When you load some packages do you get messages saying that some function is
being masked? What can happen is that one package has a function with the same
name as another package. The conflict is resolved by having one package mask
the other. However, this can cause conflicts if you
Would this work?
xxxz$Average20 <- (xxxz$Low20 + xxxz$High20)/2
I tried this earlier but it does not appear to have gone through.
Tim
-Original Message-
From: R-help On Behalf Of avi.e.gr...@gmail.com
Sent: Saturday, June 8, 2024 2:16 PM
To: 'Sorkin, John' ; r-help@r-project.org
Subjec
Can this problem be made more direct?
xxxz$Average.20 <- (xxxz$Low20 + xxxz$High20)/2
That is literally the mean of two columns. Functions can be useful if there
will be more columns, but with just two this seems easier.
I will point out that the average daily temperature based on the midpoint
One could make dummy variables if the existing variables are otherwise. If
Female is a variable that includes other options (no-response, non-binary, ...)
then recode it using dummy.female and the others would be similarly named.
-Original Message-
From: R-help On Behalf Of peter dalgaa
https://www.r-bloggers.com/2021/06/reading-data-from-excel-files-xlsxlsxcsv-into-r-quick-guide/
Excel can hold a great quantity of data. However, I find that it is slow and
often crashes when I try to use Excel at large scale. It also grinds my entire
system to a halt. At the kb and mb scales I
You could have negative indices. There are two ways to do this.
1) provide a large offset.
Offset <- 30
for (i in -29 to 120) { print(df[i+Offset])}
2) use absolute values if all indices are negative.
for (i in -200 to -1) {print(df[abs(i)])}
Tim
-Original Message-
From: R-help On Be
Whatever you had as HTML was deleted. If that was data we did not get it.
1) manipulate wpi_def_nic2004 and wpi_def_nic2008 first so that the data are
compatible, then join them.
2) The full_join statement should explicitly state the columns to join by.
Using by=NULL joins by all the columns with
g_data)
values<-as.integer(levels(mydf$string_data))
for (i in 1:length(values)) {
assign(paste("VAR_", i, sep=""), values[i]) }
--- snip ---
Best,
Kimmo
to, 2024-03-28 kello 14:17 +, Ebert,Timothy Aaron kirjoitti:
> Here are some pieces of working code. I ass
Here are some pieces of working code. I assume you want the second one or the
third one that is functionally the same but all in one statement. I do not
understand why it is a factor, but I will assume that there is a current and
future reason for that. This means I cannot alter the string_data
You could declare a matrix much larger than you intend to use. This works with
a few megabytes of data. It is not very efficient, so scaling up may become a
problem.
m22 <- matrix(NA, 1:60, ncol=6)
It does not work to add a new column to the matrix, as in you get an error if
you try m22[ ,
Yea, that worked. Thank you. :)
From: jim holtman
Sent: Wednesday, February 28, 2024 12:52 PM
To: Ebert,Timothy Aaron
Cc: r-help@r-project.org
Subject: Re: [R] Trouble reading a UTF-16LE file
[External Email]
Try this:
> x <- file("C:\\Users\\Jim\\Downloads\\PV2-ch2 -
The earlier post had an attached text file that did not go through.
I hope this link works. I tested it with a coworker, but that is no guarantee.
https://uflorida-my.sharepoint.com/:u:/g/personal/tebert_ufl_edu/EXf5u_CtTwJCrhdfTBIPr7wBefZHx4P_suj4wAWb8i8HFA?e=iQawhh
Regards,
Tim
__
Dear R-help,
I am having trouble reading a UTF-16LE formatted file. The issue appears to
be a byte order mark at the beginning of the file. I have tried readLines(file,
encoding='utf-16LE') but that got me
[1]"\xff\xfe1" "" "" "" "" ""
This is a tab delimited text fil
Dear R-help,
I am having trouble reading a UTF-16LE formatted file. The issue appears to
be a byte order mark at the beginning of the file. I have tried readLines(file,
encoding='utf-16LE') but got me
[1]"\xff\xfe1" "" "" "" "" ""
Regards,
Tim
___
The data came through fine, the program was a miss. Can you paste the program
into a ".txt" document like a notepad file and send that? You could also paste
it into your email IF your email is configured to send text and NOT html.
TIm
-Original Message-
From: R-help On Behalf Of Pedro
Would something like this help?
library(ggplot2)
# Create a plot
p <- ggplot(mtcars, aes(x = wt, y = mpg)) +
geom_point() +
labs(title = "Scatter Plot", x = "Weight", y = "Miles Per Gallon")
# Add text at a specific location
p + annotate("text", x = min(mtcars$wt) + 23, y = max(mtcars$mpg) -
It would help to have reproducible code. Use dummy data for confidentiality (if
you care about that).
My guess is that you set margins somewhere and never returned them to a default
value. The first thing I would try is to open a new window in RStudio, copy the
smallest piece of code that will g
That depends on how exactly everything must match your primary question. The
ecology group might be helpful for how biodiversity changes with proximity to a
smokestack. They might have a better idea if the smokestack was from a coal
fired powerplant or oil refinery. The modeling process would be
Change year to a factor. Doing it in ggplot will not change the original data.
ggplot(df, aes(x = as.factor(year), y = score)) + geom_point() +
geom_smooth(method = "lm", formula = y ~ x) + labs(title = "Standard linear
regression for France", x = "Year", y = "PISA score in mathematics") +
scal
Look at the lubridate package in R.
Regards,
Tim
-Original Message-
From: R-help On Behalf Of Sorkin, John
Sent: Thursday, December 7, 2023 11:22 AM
To: r-help@r-project.org (r-help@r-project.org)
Subject: [R] Convert character date time to R date-time variable.
[External Email]
Collea
Would this work in general? Say I have a document with figures, special
equations, text, and tables. The text and tables are relatively easy. The
figures would need a conversion from pixels to lines, and the equations maybe
printed out, counted as a figure, and then added to the line count. It w
Your solution was educational. Thank you. I have two comments.
1) If you do not provide both options then you are forcing people to conform to
your approach. In general I disapprove, but for specific cases I can see
advantages.
2) Without reading the relevant papers (and possibly understanding th
The "problem" goes away if you use
x$C <- y[1,]
If you have another row in your x, say:
x <- data.frame(A=c(1,4), B=c(2,5), C=c(3,6))
then your code
x$C <- y[1]
returns an error.
If y has the same number of rows as x$C then R has the same outcome as in your
example.
It looks like your code te
Is that a method where a program that I write today would still run without
changes in 10 years?
Tim
-Original Message-
From: R-help On Behalf Of Richard O'Keefe
Sent: Wednesday, October 11, 2023 8:08 AM
To: Uwe Ligges
Cc: r-help@r-project.org
Subject: Re: [R] Problem with compatible li
I would answer "local files only," but with sufficient motive it is possible
for some people to abuse a system. Base R does not download any of your data.
The packages that I know about do not download data. You can add a layer of
protection by only downloading directly from the source rather th
In this sort of post it would help if we knew the package that was being used
for the example. I found one option.
https://cran.r-project.org/web/packages/pivottabler/vignettes/v00-vignettes.html
There may be a way to create a custom data type that would be a date but
restricted to a -mm for
NA))
or
c(1,-1)[match(side, c("BUY", "SELL"))]
or
vals <- c(BUY=1, SELL = -1)
vals[side]
On 2023-09-29 9:21 a.m., Ebert,Timothy Aaron wrote:
> Does this work?
> mynewdf$side <- as.numeric(mynewdf$side)
>
> This code would be the next line after your muta
Does this work?
mynewdf$side <- as.numeric(mynewdf$side)
This code would be the next line after your mutate.
TIm
-Original Message-
From: R-help On Behalf Of Enrico Schumann
Sent: Thursday, September 28, 2023 3:13 AM
To: arnaud gaboury
Cc: r-help
Subject: Re: [R] replace character by
An update please:
Collectively we have suggested removing commas from the "E..coli" column,
checking for different forms of "NA", and looking outside the dataset for
e-trash (spaces, text, or other content). For removing commas, I would use
global replace to ensure that all commas were removed f
I tend to keep data in Excel. The reason is that I can keep data and analysis
output in one file. A part of this is that I tend to use SAS where I get
abundant output.
One way that this type of result happens is with junk in the file. Someone
might put a space in a cell or a period. Such charact
Why insist on agricolae?
Here is an example using multcompiew
https://r-graph-gallery.com/84-tukey-test.html
You have the same question posted to stackoverflow.
https://stackoverflow.com/questions/77090467/graph-in-r-with-grouping-letters-from-the-tukey-lsd-duncan-test-with-agricolae
I searched i
t the date (say Day). Group_by the day and apply a max
function to the grouped data. Then plot the result.
Tim
-Original Message-
From: Kevin Zembower
Sent: Wednesday, September 13, 2023 3:26 PM
To: Ebert,Timothy Aaron ; Richard O'Keefe
Cc: r-help@r-project.org
Subject: Re: [R] Hel
larger cities (Duluth, International
Falls, Thunder Bay) and take a metal average. There is a lake effect for two of
these more than the other.
All good?
Tim
-Original Message-
From: Kevin Zembower
Sent: Wednesday, September 13, 2023 2:05 PM
To: Ebert,Timothy Aaron
: Wednesday, September 13, 2023 1:22 PM
To: Richard O'Keefe ; Ebert,Timothy Aaron
Cc: r-help@r-project.org
Subject: Re: [R] Help with plotting and date-times for climate data
[External Email]
Tim, Richard, y'all are reading too much into this. I believe that TMAX is the
high temperature
+"turkey test +"mean comparison" 84 hits in google scholar.
There is an aphid "Aphis gossypii." Some people have changed this to "Apis
gossypii." "Apis" is a genus for bees, and there is no critter named "Apis
gossypii." However there are 45 papers in google scholar suffering from this
malady.
I had the same question.
However, I can partly answer the off-topic question. Min and max can be
important as lower and upper development thresholds. Below the min no growth or
development occur because reaction rates are too slow to enable such. Above
max, temperatures are too hot. Protein func
Say I have one machine that produces 15 million widgets per day. Every day a
few widgets are defective. Is the proportion increasing?
The data analyst needs to know what time span is of interest. I assume that
there is some day-to-day variability. Is today's defect rate greater or less
than yest
The example is the example in the documentation for the method. There were no
details.
https://www.rdocumentation.org/packages/stats/versions/3.6.2/topics/prop.trend.test
More documentation would be useful. Answering questions like what are these
numbers?
As shown, I see a cluster of three valu
gt; g(4, LETTERS[1:5], "B", "E", repeats = FALSE)
> > >> [1] "BCAE" "BDAE" "BEAE" "BABE" "BCBE" "BDBE" "BEBE" "BACE"
> > >> [9] "BDCE" "BECE" "BADE"
Does this work for you?
t0<-t1<-t2<-LETTERS[1:5]
al2<-expand.grid(t0, t1, t2)
al3<-paste(al2$Var1, al2$Var2, al2$Var3)
al4 <- gsub(" ", "", al3)
head(al3)
Tim
-Original Message-
From: R-help On Behalf Of Eric Berger
Sent: Monday, September 4, 2023 10:17 AM
To: Christofer Bogaso
Cc: r-h
Add a break. Something like:
If (is.character(x)) break
If you have nested loops then a similar statement is needed for each level.
"break" only exits the innermost loop.
Tim
-Original Message-
From: R-help On Behalf Of Jeff Reichman
Sent: Wednesday, August 30, 2023 9:46 PM
To: r-help@
When I installed R no personally identifiable information was given, none was
requested. There is no login, so no password or user name. R resides on the
host computer, so security is the responsibility of the owner(s) of that
computer. There are packages and other third party programs, but thes
I have entered values into Excel, and sorted them. I am assuming you are asking
why the value 3 in x2 is ranked 4.5 versus in x5 it has a rank of 5.
X2 looks like this
Value RankOrder
1 1.5 1
1 1.5 2
2 3 3
3 4.5 4
3 4.5 5
4 6 6
You could use a for loop in a brute force approach.
-Original Message-
From: R-help On Behalf Of Rui Barradas
Sent: Sunday, August 6, 2023 7:37 PM
To: Iris Simmons ; Steven Yen
Cc: R-help Mailing List
Subject: Re: [R] Stacking matrix columns
[External Email]
Às 01:15 de 06/08/2023, Ir
I did a google search for "R books" and looked at the R in the titles. There is
no consistency except it is always an upper case letter.
Serif, white R on red: "R for Data Science" and again "R Packages." Given a
common author this is a Hadley Wickham style.
Sans-serif, black R on cream and a sty
7;Jim Lemon' ; Ebert,Timothy Aaron
Cc: 'R-help'
Subject: RE: [R] Off-topic: ChatGPT Code Interpreter
[External Email]
Jim,
I am not sure what your example means but text to image conversion can be done
quite easily in many programming environments and does not need an AI unless
This seems spot on, given that Paul Bernal just posted a request to fit all
possible models and compare them using r-squared. The all models approach does
not promote understanding about what your model is saying about the data and
how the system works. Nor does it facilitate fitting the methods
At least on my system string has a single value of " xxx1 xxx2" not "xxx1" and
"xxx2".
The variable zzz has two values: "J K xxx1" and "J K xxx2"
What you want is "J", "K", "xxx1", "xxx2"
If I cheat everything works. So then the goal is to rewrite the program so
cheating is not needed.
# cr
Reposting the data did not help. We do not like to guess, and doing so takes a
great deal of time that is likely wasted.
Rows are observations.
Columns are variables.
In Excel, the first row will be variable names and all subsequent rows will be
observations.
Income is the first variable. It has
Another suggestion:
The statistics does not care where the numbers come from. The values 1, 2,
3 have a mean of 2 no matter if these are weights of a bird, plant heights, or
concrete tensile strength. Your interpretation might change, but the mean is
still 2.
Try synthetic data.
X<-rnorm(1
Start with defining your dependent variable and independent variable(s). As an
equation like y equals some function of x, the y is the dependent variable. It
is often continuous, but does not have to be.
If your continuous variable is the dependent variable and you have one
categorical independ
Reorganize the data so that you have three columns
Something more like this:
Date Country Value
2005-01-03 Crepub1.21
You ggplot statement has a mistake. The geom_line() should be outside the
ggplot() call.
You might then have a ggplot statement like
g
Are the numbers you provided multiple estimates of one coefficient or are they
one estimate each of over 20 coefficients?
-Original Message-
From: R-help On Behalf Of Michael Dewey
Sent: Wednesday, April 26, 2023 5:14 AM
To: bharat rawlley ; r-help@R-project.org
Subject: Re: [R] Error on
Thank you for sharing the paper.
Tim
-Original Message-
From: R-help On Behalf Of Hadley Wickham
Sent: Tuesday, April 25, 2023 9:44 AM
To: Sigbert Klinke
Cc: r-help@r-project.org
Subject: Re: [R] diamonds data set from ggplot2
[External Email]
On Tue, Apr 25, 2023 at 4:09 AM Sigbert Kl
1) If you do not need it do not plot it. However, also consider how others will
use your content. Might it be a trivial piece of information for you, but a
critical piece of information for someone trying to use your content. A meta
analysis, or just wanting to try to relate your outcomes to the
Sometimes outliers happen. No matter the sample size there is always the
possibility that one or more values are correct though highly improbable.
-Original Message-
From: R-help On Behalf Of Richard O'Keefe
Sent: Friday, April 21, 2023 7:31 PM
To: AbouEl-Makarim Aboueissa
Cc: R mailing
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