On 07/07/17 13:55, θ ” wrote:
hi:
I want to know the math behind the "tm" package findAssocs().
I have found that someone had asked the question before, and have a good
explanation by Rick.
�]http://r.789695.n4.nabble.com/findAssocs-td3845751.html�^
But I still don't understand how to calcu
No, that would remap B to A. Convert to character before doing this, then back
to factors.
--
Sent from my phone. Please excuse my brevity.
On July 6, 2017 4:43:00 PM PDT, Ista Zahn wrote:
>Untested, but I expect that setting the levels to be the same across
>the
>two factors
>
>levels(tmp$R1)
Untested, but I expect that setting the levels to be the same across the
two factors
levels(tmp$R1) <- levels(tmp$R2) <- LETTERS[1:6]
and proceeding as before should be fine.
Best,
Ista
On Jul 6, 2017 6:54 PM, "Gang Chen" wrote:
Thanks a lot, Ista! I really appreciate it.
How about a slightl
Thanks a lot, Ista! I really appreciate it.
How about a slightly different case as the following:
set.seed(1)
(tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace =
TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE)))
x R1 R2
1 C B
2 B B
3 C E
4 E C
5
Looks good, but there are some ways in which I think you’re making it more
complicated than necessary.
Two calls to plot(), with plot.new() in between, is definitely not needed.
Recalculating LAI_simulation$Date three times should not be needed.
## create example data
day0 <- as.Date('2009-10-7'
Hello R Community;
Just posted a question about Bayes Factor. But, just answered my own question.
What is needed is to convert a 'dplyr' data frame into a standard data frame.
competition2 <- data.frame(competition), where 'competition' is a
'dplyr' data frame with the following class structure.
How about
foo <- with(list(r1 = tmp$R1,
r2 = tmp$R2,
swapme = (as.numeric(tmp$R1) - as.numeric(tmp$R2)) %% 2 != 0),
{
tmp[swapme, "R1"] <- r2[swapme]
tmp[swapme, "R2"] <- r1[swapme]
tmp
})
Best,
Ista
On Thu, Jul 6, 2017 at 4:06 PM, Gang Chen wrote:
Suppose that we have the following dataframe:
set.seed(1)
(tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace =
TRUE), R2 = sample(LETTERS[1:5], 10, replace = TRUE)))
x R1 R2
1 1 B B
2 2 B A
3 3 C D
4 4 E B
5 5 B D
6 6 E C
7 7 E D
8 8 D E
9 9
Hello R Community,
Subject: Bayes Factor
A Bayesian ANOVA of the form:
competitionBayesOut <- anovaBF(biomass ~ clipping, data = competition)
Returns the following Error message:
Error in (function (classes, fdef, mtable) :
unable to find an inherited method for function ‘compare’ for
signature
Dear all,
I wanted to compare Bonferroni vs TukeyHSD correction over a range of groups
and group sizes, and wanted to use the function qtukey.
In the help documentation it says
qtukey(p, nmeans, df, nranges = 1, lower.tail = TRUE, log.p = FALSE)
Arguments
q
vector of quantiles.
p
vector of pr
Yes, definitely. However, this is so close to being legal R code that I feel
you have not made any effort to translate it yourself, and this is the "R-help"
mailing list, not the "R-do-my-work-for-me" mailing list. Is this homework?
Have you read the "Introduction to R" document that is supplie
Hello,
Is it possible to generate simulated data that look like the attached figure?
The curve in the figure can be obtained from this equation respecting some
conditions:
if(Temperature > T_min & Temperature < T_max){
a*( Temperature -T_min)*( Temperature -T_max)
} else 0
T_min ~ Unif
Don't know what your data looks like, but you might try:
p <- Scenario1+guides(fill = guide_colorbar(bar width = 1.5, barheight
= unit(10, "mm")))
print(p)
On 7/5/2017 5:13 PM, Kristi Glover wrote:
Hi R Users,
I tried to increase the legend height in ggplot2, but it did not respond at al
Thanks it worked for me. I wanted to plot days since planting on
x-axis 1 and years on x-axis 3.
LAI_simulation$Date <- as.Date( LAI_simulation$Date, '%Y/%m/%d')
LAI_simulation$Date <- as.integer(LAI_simulation$Date - as.Date("2009-10-07"))
plot(LAI~Date,data=LAI_simulation,xlab="Days since Oct, 7
Dear community,
I'm performing svm-regression with svm at library e1071.
As I wrote in another post: "svm e1071 call - different results", I get
different results if I use the svm.default rather than the svm.formula, being
better the ones at svm.formula
I've debugged both options.
While
Hi
I hope I can use this R mailing list for this but it seems appropriate.
I started writing a small package which enables me to create plantuml graphs
from within R. It is working and available on GitHub at
https://github.com/rkrug/plantuml
It is at the moment very rudimentary and can either
Please keep the conversation on the list: Others may be able to help you
better than I can.
On 2017-07-06 10:38, SEB140004 Student wrote:
Ya. I had successfully got the result. Thank you very much. :)
Isn't possible for me to obtain the network from the correlation matrix?
I know nothing abo
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