On Tue, Mar 16, 2021 at 11:35 PM Richard M. Heiberger wrote:
>
> library(lattice)
> library(latticeExtra)
>
> barchart(matrix(c(1:6, 5:6)), main="unanticipated left axis labels",
> ylab="unanticipated inside labels") +
> latticeExtra::layer(panel.axis("left", half=FALSE, labels=1:8))
S
Hi,
stringr::str_replace() treats the 2nd argument ('pattern') as a regular
expression and some characters have a special meaning when they are used
in a regular expression. For example the dot plays the role of a
wildcard (i.e. it means "any character"):
> str_replace("aaXcc", "a.c", "ZZ"
I prefer using regular expressions directly, so this may not satisfy you:
> a <-"Women's footwear (excluding athletic)"
> b <- gsub("(.*) \\(.*$","\\1",a)
> b
[1] "Women's footwear"
There are, of course other ways to do this with regex's or even substring()
Bert Gunter
"The trouble with having
Not sure what you mean by a horizontal line, Greg.
I change one of my plots to add a path between corresponding values of the
sequence variable but obviously those lines are mostly not horizontal. Look
at geom_path and similar geoms like geom_line to connect endpoints grouped
whatever way you spec
I have a problem with the str_replace() function in the stringr package.
Please refer to my reprex below.
I start with a vector of strings, called x. Some of the strings contain
apostrophes and brackets. I make a simple replacement as with x1, and
there is no problem. I make another simple rep
Thank you very much.
In addition to what your did, for event 1, I would like to draw a horizontal
line connecting from day 1 to day 2 to day 3 to day 4.
Then, for event 2, I would like to draw a horizontal line connecting from day 1
to day 2 to day 3 to day 4.
Similarly for events 3, and 4. Is th
Dan, Thank you for this guidance.
Unfortunately, I do not have the library lubridate, and I do not at this moment
know where to go to get this library for an Apple MacBook.
> library(lubridate)
Error in library(lubridate) : there is no package called ‘lubridate’
Greg Coats
Reston, Virginia USA
It sounds to me like you want to take your data and extract one column for
JUST the date and another column for just some measure of the time, such as
the number of seconds since midnight or hours in a decimal format where
12:45 PM might be 12.75.
You now can graph date along the X axis and time a
On 3/16/2021 3:32 PM, Gregory Coats via R-help wrote:
Thank you. Let me redefine the situation.
Each time an event starts, I record the date and time.
Each day there are 4 new events. Time is the only variable.
I would like to graphically show how the time for events 1, 2, 3, and 4 for the
curre
Thank you. Let me redefine the situation.
Each time an event starts, I record the date and time.
Each day there are 4 new events. Time is the only variable.
I would like to graphically show how the time for events 1, 2, 3, and 4 for the
current day compare to the times for events 1, 2, 3, and 4 fo
Hi Greg,
This example may give you a start:
myDat<-read.table(text=
"2021-03-11 10:00:00
2021-03-11 14:17:00
2021-03-12 05:16:46
2021-03-12 09:17:02
2021-03-12 13:31:43
2021-03-12 22:00:32
2021-03-13 09:21:43",
sep=",",
stringsAsFactors=FALSE)
myDat$datetime<-strptime(myDat$X,for
Hello,
Yes, you can ask R-Help questions on R code, that's what it is intended for.
But please read the posting guide, it's link is at the bottom of this
and of any R-Help mail.
Good luck, enjoy R!
Rui Barradas
Às 18:19 de 16/03/21, Jonathan Lim escreveu:
Hi
I found your email on a websit
https://www.r-project.org/mail.html
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Tue, Mar 16, 2021 at 12:51 PM Jonathan Lim wrote:
> Hi
>
> I found your em
Hello,
I don't really understand what is to be plotted, just the time of the
event? But what event?
Anyway, with the data read with Sarah's code, maybe
library(ggplot2)
ggplot(myDat, aes(x = datetime, y = 1)) +
geom_linerange(aes(ymin = 0, ymax = 1), linetype = "dotted") +
geom_point()
Hi
I found your email on a website
Can I ask some questions about R please
Many thanks
Jonathan
[[alternative HTML version deleted]]
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https://stat.ethz.ch/mailman/listinfo/r
Dear Greg,
Coordinate plots typically have a horizontal (x) and vertical (y) axis.
The command
ggplot(myDat, aes(x=datetime, y = datetime)) + geom_point()
works, but I doubt that it produces what you want.
You have only one variable in your data set -- datetime -- so it's not
obviou
I need a plot that shows the date and time that each event started.
This ggplot command was publicly given to me via this R Help Mailing LIst.
But the result of issuing the ggplot command is an Error in FUN message.
ggplot(myDat, aes(x=datetime, y = Y_Var)) + geom_point()
Error in FUN(X[[i]], ...)
So what do you want quantity on the y-axis to be?
On March 16, 2021 11:45:32 AM PDT, Gregory Coats wrote:
>I want to plot the date and time of the event, as reflected in data.
>2021-03-11 10:00:00
>Greg Coats
>
>> On Mar 16, 2021, at 2:23 PM, Jeff Newmiller
> wrote:
>>
>> You don't seem to have
I want to plot the date and time of the event, as reflected in data.
2021-03-11 10:00:00
Greg Coats
> On Mar 16, 2021, at 2:23 PM, Jeff Newmiller wrote:
>
> You don't seem to have a Y_Var in your data. What is it that you want to plot?
>
> On March 16, 2021 9:21:05 AM PDT, Gregory Coats via R-h
You don't seem to have a Y_Var in your data. What is it that you want to plot?
On March 16, 2021 9:21:05 AM PDT, Gregory Coats via R-help
wrote:
>Sarah, Thank you. Yes, now as.POSIXct works.
>But the ggplot command I was told to use yields an Error message, and
>there is no output plot.
>Please
Dear Greg,
There is no variable named Y_Var in your data set. I suspect that it's
intended to be a generic specification in the recipe you were apparently
given. In fact, there appears to be only one variable in myDat and
that's datetime. What is it that you're trying to do?
A more general c
The length of the mean vector must match the number of rows and
columns of the sigma matrix. Once you give 3 entries in the mean
vector you will run into the problem that the sigma you are using is
not positive (semi-)definite - a variance must be the product of a
matrix and its transpose.
-Bill
A 2 dim distribution must have a 2 x 2 covariance matrix. Your mean in b)
specifies 2 dim, but your covariance matrix is 3x3.
If you haven't just made a typo and you don't know what this means, then
either consult statistics references or find someone to help you.
Cheers,
Bert Gunter
"The troub
library(lattice)
library(latticeExtra)
barchart(matrix(c(1:6, 5:6)), main="unanticipated left axis labels",
ylab="unanticipated inside labels") +
latticeExtra::layer(panel.axis("left", half=FALSE, labels=1:8))
barchart(matrix(c(1:6, 5:6)), main="ok 1", ylab="anticipated") +
latticeEx
Code a is working. But code b is given error like given below. How can I
write code b?
> a<-rmvnorm(750, mean=c(0, 0),
+sigma=matrix(c(1, .3, .3, 1), ncol=2))
> head(a)
[,1][,2]
[1,] -0.97622921 -0.87129405
[2,] 0.54763494 0.16080131
[3,] -1.16627647
Sarah, Thank you. Yes, now as.POSIXct works.
But the ggplot command I was told to use yields an Error message, and there is
no output plot.
Please help me. Greg
> library(ggplot2)
> myDat <- read.table(text =
+ "datetime
+ 2021-03-11 10:00:00
+ 2021-03-11 14:17:00
+ 2021-03-12 05:16:46
+ 2021-03-1
Hello there,
This error happens if you did not finish
reading from the pipe.
E.g. I have this function in my local
zshrc, to read tsv files with less in
terminal width ...
readdelim() { Rscript -e "options(width=$COLUMNS); read.delim('$1',
check.names=FALSE, na='N/A')" | less }
..
Hi,
```R
> sessionInfo()
R version 4.0.3 (2020-10-10)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Debian GNU/Linux bullseye/sid
Matrix products: default
BLAS: /usr/lib/x86_64-linux-gnu/openblas-pthread/libblas.so.3
LAPACK: /usr/lib/x86_64-linux-gnu/openblas-pthread/libopenblasp-r0.3.1
Hi,
It doesn't have anything to do with having a Mac - you have POSIX.
It's because something is wrong with your data import. Looking at the
head() output you provided, it looks like your data file does NOT have
a header, because there's no datetime column, and the column name is
actually X2021.0
My computer is an Apple MacBook. I do not have POSIX.
The command
myDat$datetime <- as.POSIXct(myDat$datetime, tz = "", format = "%Y-%M-%d
%H:%M:%OS")
yields the error
Error in `$<-.data.frame`(`*tmp*`, datetime, value = numeric(0)) :
replacement has 0 rows, data has 13
Please advise, How to p
Just an update:
I tried with desmos and the fitting looks good. Desmos calculated the
parameters as:
Fmax = 11839.8
Chalf = 27.1102 (with matches with my estimate of 27 cycles)
k = 2.76798
Fb = -138.864
I forced R to accept the right parameters using a single named list
and re-written the formula (
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