Hi Spencer,
The an1 syntax is adding regression coefficients (or NAs where a regression
could not be done) to the downloaded and processed data, which ends up a
matrix. The cbind function adds the regression coefficients to the last column
of the matrix (i.e. bind the columns of the inputs in t
Hallo Steve.
Thank you for pointing to ImageJ, I will try to inspect it. I expected that
somebody in R comunity automated such procedure within some R package, but it
is probably too specialised.
S pozdravem | Best Regards
RNDr. Petr PIKAL
Vedoucí Výzkumu a vývoje | Research Manager
PRECHEZA a.
Mr. Fowler,
Thank you! This information is most helpful. So from my understanding, I
can use the regression coefficients shown (via the coding I originally
sent, to generate a continuous distribution with what is essentially a line
of best fit? The data added here had some 30,000 variables (it is
I am not sure I would use the word ‘accounted’, more like discounted (tossed
out).
From: Spencer Brackett
Sent: February 14, 2019 9:21 AM
To: Fowler, Mark
Cc: R-help ; Sarah Goslee ;
Caitlin Gibbons ; Jeff Newmiller
Subject: Re: R Data
Mr. Fowler,
Thank you! This information is most helpfu
I am having trouble finding the mean of a specific part of my dataset. Here is
a sample of it:
plot lai leaf
1 104 82 1
2 104 167 2
3 104 248 3
4 104 343 4
5 104 377 5
6 105 64 1
7 105 139 2
8 105 211 3
9 105 296 4
10 105 348 5
11 106 94 1
12 106 167 2
13 106 243 3
14 106 281 4
15 106 332 5
16 10
Hi Isaac,
I am sure you will get lots of answers to this. Here is one using the dplyr
package.
Assuming that your data frame is called 'a', then
library(dplyr)
b <- dplyr::group_by(a,plot) %>% dplyr::summarise( mean(lai) )
b
# A tibble: 4 x 2
plot `mean(lai)`
1 104243.
2 10
Hi,
You might try your hand at the tidyverse collection of tools which are veddy
nice for this kind of wrangling. https://www.tidyverse.org/
Does this do the trick?
## START
library(readr)
library(dplyr)
txt <- "row plot lai leaf
1 104 82 1
2 104 167 2
3 104 248 3
4 104 343 4
5 104 377 5
6 105
On Feb 14, 2019, at 9:31 AM, Isaac Barnhart wrote:
>
> I am having trouble finding the mean of a specific part of my dataset. Here
> is a sample of it:
>
> plot lai leaf
> 1 104 82 1
> 2 104 167 2
> 3 104 248 3
> 4 104 343 4
> 5 104 377 5
> 6 105 64 1
> 7 105 139 2
> 8 105 211 3
> 9 105 296 4
>
Dear R-helpers,
We have recently upgraded from R-3.3.1 to R-3.5.2.
It seems there has been a change in behaviour of `lapply` and the `POSIXlt`
class that I cannot find explicitly documented.
In R-3.3.1:
> lapply(as.POSIXlt(Sys.Date()), length)
$sec
[1] 1
$min
[1] 1
$hour
[1] 1
$mday
[1] 1
$mo
Hi,
try
library("dplyr")
plot <- c(104, 104 ,104 ,104 ,104 ,105 ,105 ,105 ,105 ,105,106,
106,106, 106, 106,108, 108,108,108,108)
lai <- c(82, 167, 248, 343, 377, 64, 139, 211, 296, 348,
94, 167,243,281,332,83, 382,320,146,129)
leaf <- c(1,2, 3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5)
d
Somewhere between R-3.3.3 and R-3.5.2 a POSIXlt method for as.list() was
added, and lapply probably calls as.list().
> RCompare(methods("as.list"))
R version 3.3.3 (2017-03-06)| R version 3.5.1
(2018-07-02)
[1] as.list.data.frame as.list.Date| [1]
as.list.d
Dear to whom it may concern on the R team,
I am having an issue with creating a code in which i can hold information such
as the author of a paper, the year of publication, and the title. Also would
like to add into this data frame a logical variable which would show some
keywords I used to
Thanks for the help!
Isaac
From: Marc Schwartz
Sent: Thursday, February 14, 2019 11:05:23 AM
To: Isaac Barnhart
Cc: R-help
Subject: Re: [R] Finding the Mean of a Specific Set of Columns
On Feb 14, 2019, at 9:31 AM, Isaac Barnhart wrote:
>
> I am having troub
There's a bibtex parser for R: you could adapt that for your use,
rather than trying to reinvent the equivalent tool.
https://cran.r-project.org/web/packages/bibtex/index.html
Sarah
On Thu, Feb 14, 2019 at 5:57 PM Gouresh Kamble via R-help
wrote:
>
> Dear to whom it may concern on the R team,
>
Dear List,
I have a simple code with which I convert year, month, and day to a date format.
My data looks like:
67 01 2618464
67 01 2618472
67 01 2618408
67 01 2618360
67 01 2618328
67 01 2618320
67 01 2618296
while my code is:
data <- read.table("CALG.txt", col.names
The Date class is not designed to handle time... you need to use the
ISOdatetime function and convert to POSIXct instead of Date. Just be sure to
set your timezone to some appropriate value before you convert any times into
datetime types.
Sys.setenv( TZ="GMT" )
# avoid using `data` as that is
Dear Jeff,
Thank you so much.
I ran the code but got an error message. I then try to run them line by line.
The problem is in:
dta$datetime <- with( dta, as.POSIXct(ISOdatetime(year, month,day,hour,0,0)))
Error in with(dta, as.POSIXct(ISOdatetime(year, month, day, hour, 0, 0))) :
object 'dta'
Dear Jeff,
Please hold.
It is begging to work. There was an error somewhere. One ")" is
missing and as I went back to check the lines one by one with cursor,
I stubbed on non matching bracket.
I completed, run the code again and got some result.
Will get back to you once I am through.
Thanks in
Dear Jeff,
I am alright now
Please accept my indebtedness!!!
Warmest regards
Ogbos
On Fri, Feb 15, 2019 at 8:25 AM Ogbos Okike wrote:
>
> Dear Jeff,
>
> Please hold.
> It is begging to work. There was an error somewhere. One ")" is
> missing and as I went back to check the lines one by one w
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