Hi is there a faster way to "extract" rows of a matrix many times to for a
longer matrix based in a vector or for indices than M[ V, ]
I need to "expand" ( rather than subset) a matrix M of 10-100,000 rows x
~50 columns to produce a matrix with a greater number (10^6-10^8) of rows
using a vector
On Mon, 13 Sep 2021, Bert Gunter wrote:
If you are interested in extracting seasonal patterns from time series,
you might wish to check out ?stl (in the stats package). Of course, there
are all sorts of ways in many packages to fit seasonality in time series
that are more sophisticated, but prob
If you are interested in extracting seasonal patterns from time
series, you might wish to check out ?stl (in the stats package). Of
course, there are all sorts of ways in many packages to fit
seasonality in time series that are more sophisticated, but probably
also more complicated, than your manua
On Mon, 13 Sep 2021, Avi Gross via R-help wrote:
Just FYI, Rich, the way the idiom with pipeline works does allow but not
require the method you used:
...
But equally valid are forms that assign the result at the end:
Avi,
I'll read more about tidyverse and summarize() in R and not just i
Just FYI, Rich, the way the idiom with pipeline works does allow but not
require the method you used:
Yours was
RESULT <-
DATAFRAME %>%
FN1(args) %>%
...
FNn(args)
But equally valid are forms that assign the result at the end:
DATAFRAME %>%
FN1(args) %>%
...
On Mon, 13 Sep 2021, Avi Gross via R-help wrote:
As Eric has pointed out, perhaps Rich is not thinking pipelined. Summarize()
takes a first argument as:
summarise(.data=whatever, ...)
But in a pipeline, you OMIT the first argument and let the pipeline supply an
argument silently.
Av
Dear Brian,
On 2021-09-13 9:33 a.m., Brian Lunergan wrote:
Hi folks:
I'm running Linux Mint 19.3 on my machine. Tried to install a more
recent edition of R but I couldn't seem to get it working so I pulled it
off and went with a good, basic install of the edition available through
the software
I think we wandered away into a package rather than base R, but the request
seems easy enough.
Just FYI, Rich, as you seem not to have incorporated the advice we gave yet
about the first argument, your use of group_by() is a tad odd.
disc %>%
group_by(hour) %>%
group_by(day) %>%
As Eric has pointed out, perhaps Rich is not thinking pipelined. Summarize()
takes a first argument as:
summarise(.data=whatever, ...)
But in a pipeline, you OMIT the first argument and let the pipeline supply an
argument silently.
What I think summarize saw was something like:
summari
On Tue, 14 Sep 2021, Eric Berger wrote:
This code is not correct:
disc_by_month %>%
group_by(year, month) %>%
summarize(disc_by_month, vol = mean(cfs, na.rm = TRUE))
It should be:
disc %>% group_by(year,month) %>% summarize(vol=mean(cfs,na.rm=TRUE)
Eric/Avi:
That makes no difference:
This code is not correct:
disc_by_month %>%
group_by(year, month) %>%
summarize(disc_by_month, vol = mean(cfs, na.rm = TRUE))
It should be:
disc %>% group_by(year,month) %>% summarize(vol=mean(cfs,na.rm=TRUE)
On Tue, Sep 14, 2021 at 12:51 AM Rich Shepard
wrote:
> On Mon, 13 Sep 2
On Mon, 13 Sep 2021, Rich Shepard wrote:
That's what I thought I did. I'll rewrite the script and work toward the
output I need.
Still not the correct syntax. Command is now:
disc_by_month %>%
group_by(year, month) %>%
summarize(disc_by_month, vol = mean(cfs, na.rm = TRUE))
and result
On Mon, 13 Sep 2021, Avi Gross via R-help wrote:
Did I miss something?
Avi,
Probably not.
The summarise() command is telling you that you had not implicitly
grouped the data and it made a guess. The canonical way is:
... %>% group_by(year, month, day, hour) %>% summarise(...)
After sendi
Rich,
Did I miss something? The summarise() command is telling you that you had not
implicitly grouped the data and it made a guess. The canonical way is:
... %>% group_by(year, month, day, hour) %>% summarise(...)
You decide which fields to group by, sometimes including others so they are in
I changed the data files so the date-times are in five separate columns:
year, month, day, hour, and minute; for example,
year,month,day,hour,min,cfs
2016,03,03,12,00,149000
2016,03,03,12,10,15
2016,03,03,12,20,151000
2016,03,03,12,30,156000
2016,03,03,12,40,154000
2016,03,03,12,50,15
2016
On 9/13/2021 11:28 PM, Andrew Simmons wrote:
> In the example you gave : r(x) <- 1
> r(x) is never evaluated, the above calls `r<-`,
> in fact r does not even have to be an existing function.
I meant:
'*tmp*' <- x; # "x" is evaluated here;
'r<-' is called after this step, which makes sense in
In the example you gave : r(x) <- 1
r(x) is never evaluated, the above calls `r<-`,
in fact r does not even have to be an existing function.
On Mon, Sep 13, 2021, 16:18 Leonard Mada wrote:
> Hello,
>
>
> I have found the evaluation: it is described in the section on subsetting.
> The forced eval
Hello,
I have found the evaluation: it is described in the section on
subsetting. The forced evaluation makes sense for subsetting.
On 9/13/2021 9:42 PM, Leonard Mada wrote:
>
> Hello Andrew,
>
>
> I try now to understand the evaluation of the expression:
>
> e = expression(r(x) <- 1)
>
> # pa
On 13/09/2021 9:33 a.m., Brian Lunergan wrote:
Hi folks:
I'm running Linux Mint 19.3 on my machine. Tried to install a more
recent edition of R but I couldn't seem to get it working so I pulled it
off and went with a good, basic install of the edition available through
the software manager. So..
e = expression(r(x) <- 1)
lapply(e, as.list)
[[1]]
[[1]][[1]]
`<-`
[[1]][[2]]
r(x)
[[1]][[3]]
[1] 1
###
lapply(e[[1]], as.list)
[[1]]
[[1]][[1]]
`<-`
[[2]]
[[2]][[1]]
r
[[2]][[2]]
x
[[3]]
[[3]][[1]]
[1] 1
However, I would urge you not to go down this rabbit hole unless
Hi,
Instead of ggsave(), use save_plot() from the "cowplot" package:
library(ggplot2)
library(cowplot)
x <- 1:10
y <- x^2
df <- data.frame(x, y)
p <- ggplot(df, aes(x, y)) + geom_point()
save_plot("/tmp/plot.png", p, base_aspect_ratio = 1, base_width = 5,
base_height = NULL)
--
Regards,
Adam W
Hi folks:
I'm running Linux Mint 19.3 on my machine. Tried to install a more
recent edition of R but I couldn't seem to get it working so I pulled it
off and went with a good, basic install of the edition available through
the software manager. So... I'm running version 3.4.4.
Mucking about with
Hi List,
I want to append some rows from R into sql server. So, I submitted the code
below. there is not any error message:
dbWriteTable(conn = con, name = "PMDB._Alias_A", value = try1, overwrite=FALSE,
append=TRUE, row.names = FALSE)
But when I try to query the data from the Sql server, I can n
Hello Andrew,
I try now to understand the evaluation of the expression:
e = expression(r(x) <- 1)
# parameter named "value" seems to be required;
'r<-' = function(x, value) {print("R");}
eval(e, list(x=2))
# [1] "R"
# both versions work
'r<-' = function(value, x) {print("R");}
eval(e, list(x=2
R's parser doesn't work the way you're expecting it to. When doing an
assignment like:
padding(right(df)) <- 1
it is broken into small stages. The guide "R Language Definition" claims
that the above would be equivalent to:
`<-`(df, `padding<-`(df, value = `right<-`(padding(df), value = 1)))
Hello Andrew,
this could work. I will think about it.
But I was thinking more generically. Suppose we have a series of functions:
padding(), border(), some_other_style();
Each of these functions has the parameter "right" (or the group of
parameters c("right", ...)).
Then I could design a fun
I think you're trying to do something like:
`padding<-` <- function (x, which, value)
{
which <- match.arg(which, c("bottom", "left", "top", "right"),
several.ok = TRUE)
# code to pad to each side here
}
Then you could use it like
df <- data.frame(x=1:5, y = sample(1:5, 5))
padding(df, "
I try to clarify the code:
###
right = function(x, val) {print("Right");};
padding = function(x, right, left, top, bottom) {print("Padding");};
'padding<-' = function(x, ...) {print("Padding = ");};
df = data.frame(x=1:5, y = sample(1:5, 5)); # anything
### Does NOT work as expected
'right<-' =
On 13/09/2021 9:38 a.m., Leonard Mada wrote:
Hello,
I can include code for "padding<-"as well, but the error is before that,
namely in 'right<-':
right = function(x, val) {print("Right");};
# more options:
padding = function(x, right, left, top, bottom) {print("Padding");};
'padding<-' = funct
Hello,
I can include code for "padding<-"as well, but the error is before that,
namely in 'right<-':
right = function(x, val) {print("Right");};
# more options:
padding = function(x, right, left, top, bottom) {print("Padding");};
'padding<-' = function(x, ...) {print("Padding = ");};
df = dat
On 12/09/2021 10:33 a.m., Leonard Mada via R-help wrote:
How can I avoid evaluation?
right = function(x, val) {print("Right");};
padding = function(x) {print("Padding");};
df = data.frame(x=1:5, y = sample(1:5, 5));
### OK
'%=%' = function(x, val) {
x = substitute(x);
}
right(padding(df))
How can I avoid evaluation?
right = function(x, val) {print("Right");};
padding = function(x) {print("Padding");};
df = data.frame(x=1:5, y = sample(1:5, 5));
### OK
'%=%' = function(x, val) {
x = substitute(x);
}
right(padding(df)) %=% 1; # but ugly
### Does NOT work
'right<-' = function(x
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