Combined, the answers are close to an example in the documentation for
microbenchmark where they (in part) look at execution times for c() versus
append(). Here is the code, but c() has the shortest execution time in this
example. If I remove a3 and a4 then c() is significantly shorter than appe
I may be missing something but using the plain old c() combine function
seems to work fine:
df <- data.frame(left = 1:5, right = 6:10)
df.combined <- data.frame(comb = c(df$left, df$right))
df
left right
11 6
22 7
33 8
44 9
5510
df.combined
comb
1
Jeff Newmiller wrote/hat geschrieben on/am 03.04.2023 18:26:
unname(unlist(NamesWide))
Why not:
NamesWide <- data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Curly"))
NamesLong <- data.frame(Names=with(NamesWide, c(Name1, Name2)))
On April 3, 2023 8:08:59 AM PDT, "Sparks, John" wrote:
Hi
IMHO The difference makes sense for paired samples. For unpaired samples it is
just the difference in means. I think presenting the means (and assuming the
audience can perform subtraction) conveys more information. The interpretation
of a difference of 5 might be influenced if I have means of -
Hi
Thanks for your feedback. I didn't think about that.
Still, the mean difference is computed for paired, not because there are
two samples. IMHO, the help should be updated.
Best,
Samuel
Le 2023-04-03 à 12:10, PIKAL Petr a écrit :
Hi
You need to use paired option
t.test(x=0:4, y=sample
unname(unlist(NamesWide))
On April 3, 2023 8:08:59 AM PDT, "Sparks, John" wrote:
>Hi R-Helpers,
>
>Sorry to bother you, but I have a simple task that I can't figure out how to
>do.
>
>For example, I have some names in two columns
>
>NamesWide<-data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Cu
My first thought was pivot_longer, and stack() is new to me.
How about append(c1,c2) as another solution? Or data.frame(append(c1,c2)) if
you want that form.
Tim
-Original Message-
From: R-help On Behalf Of Marc Schwartz via
R-help
Sent: Monday, April 3, 2023 11:44 AM
To: Sparks, John
Hi,
You were on the right track using stack(), but you just pass the entire data
frame as a single object, not the separate columns:
> stack(NamesWide)
values ind
1 Tom Name1
2 Dick Name1
3 Larry Name2
4 Curly Name2
Note that stack also returns the index (second column of 'ind' value
pivot_longer()
Sent from my iPhone
> On 3 Apr 2023, at 18:09, Sparks, John wrote:
>
> Hi R-Helpers,
>
> Sorry to bother you, but I have a simple task that I can't figure out how to
> do.
>
> For example, I have some names in two columns
>
> NamesWide<-data.frame(Name1=c("Tom","Dick"),Name2
Hi R-Helpers,
Sorry to bother you, but I have a simple task that I can't figure out how to do.
For example, I have some names in two columns
NamesWide<-data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Curly"))
and I simply want to get a single column
NamesLong<-data.frame(Names=c("Tom","Dick",
Hallo,
you are probably right that
"the estimated mean or difference in means depending on whether it was a
one-sample test or a two-sample test"
should be rephrased to
"the estimated mean or difference in means depending on whether it was a
one-sample test, two-sample test or two sample pai
Dear Nick,
Looking at dput(pheno::seqMK),
dput(pheno::tau), and [1], I think you
need to change the function pheno::tau
to change the confidence level ... R
[1]
https://mannkendall.github.io/about.html#application-of-the-seasonal-mann-kendall-test
Hi
You need to use paired option
> t.test(x=0:4, y=sample(5:9), paired=TRUE)$estimate
mean difference
-5
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Samuel Granjeaud
> IR/Inserm
> Sent: Sunday, April 2, 2023 11:39 PM
> To: r-help@r-project.org
> Subject: [
Hi,
Not important, but IMHO the estimate component of the t.test holds an
estimate of mean of each group, never a difference. The doc says
"estimate the estimated mean or difference in means depending on
whether it was a one-sample test or a two-sample test."
> t.test(0:4)$estimate
mean o
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