This was already clear from Rich Heiberger's reply. But my point was not
that the as.matrix() coercion was necessary, but that it would be wise, as
operations with matrices are generally (often much) more efficient than
with data frames. Of course, other considerations may exist, but that was
my po
Hello,
The answer is given but there is no need to coerce to matrix first, as
long as the columns are numeric.
From ?exp, right at the beginning of section Details:
Details
All except logb are generic functions: methods can be defined for them
individually or via the Math group generic.
Fo
Hello,
exp() is better, see the 2nd identical:
identical(
exp(tmp),
2.718281828^tmp
)
#[1] FALSE
identical(
exp(tmp),
exp(1)^tmp
)
#[1] FALSE
all.equal(
exp(tmp),
2.718281828^tmp
)
#[1] TRUE
all.equal(
exp(tmp),
exp(1)^tmp
)
#[1] TRUE
Às 18:16 de 14/10/21, Richard M. Heibe
Thank you so much!
On Thu, Oct 14, 2021 at 12:17 PM Bert Gunter wrote:
> As all of your columns are numeric, you should probably convert your df to
> a matrix. Then use exp() on that, of course:
> exp(as.matrix(b))
>
> see ?exp
>
> Bert Gunter
>
> "The trouble with having an open mind is that pe
As all of your columns are numeric, you should probably convert your df to
a matrix. Then use exp() on that, of course:
exp(as.matrix(b))
see ?exp
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 hi
> tmp <- data.frame(a=1:3,b=4:6)
> exp(tmp)
a b
1 2.718282 54.59815
2 7.389056 148.41316
3 20.085537 403.42879
> 2.718281828^tmp
a b
1 2.718282 54.59815
2 7.389056 148.41316
3 20.085537 403.42879
> On Oct 14, 2021, at 13:10, Ana Marija wrote:
>
>> 2.718
Hi All,
I have a data frame like this:
> head(b)
LRET02LRET04LRET06LRET08LRET10LRET12LRET14
1 0 0.6931472 . 1.0986123 1.0986123 1.0986123 0.6931472
2 2.1972246 2.4849066 2.4849066 . 2.5649494 2.6390573 2.6390573
3 1.6094379 1.7917595 1.6094379
We are very happy to announce that
Sebastian Meyer (http://www.imbe.med.uni-erlangen.de/ma/S.Meyer ;
Twitter @bastistician; also https://github.com/bastistician/)
has joined the R core team yesterday (Oct 13). He has been an
active contributor notably in handling and fixing R
I presumed there was some reason why the 'optimise' function did not suit you.
optimize(function (x) a*dnorm(x, x1, s1) + (1-a)*dnorm(x, x2, s2),
range(c(x1, x2)))
should do the trick.
On Thu, 14 Oct 2021 at 23:42, Luigi Marongiu wrote:
>
> Thank you, I hoped there might be an automatic method
Thank you, I hoped there might be an automatic method more than
function analysis...
On Thu, Oct 14, 2021 at 9:06 AM Richard O'Keefe wrote:
>
> Do you really want the minimum?
> It sounds as though your model is a*N(x1,s1) + (1-a)*N(x2,s2) where
> you use mixtools to estimate
> the parameters. F
Hello,
If the sub-df has more than 2 rows,
tail(df[(df$Y>0.2) & (df$X<10),], 1)
Hope this helps,
Rui Barradas
Às 08:51 de 14/10/21, Luigi Marongiu escreveu:
Hello,
I have selected a subset of a dataframe with the vector syntax (if
this is the name):
```
df[(df$Y>0.2) & (df$X<10),]
Got it, thanks!
On Thu, Oct 14, 2021 at 9:59 AM Eric Berger wrote:
>
> df[(df$Y>0.2) & (df$X<10),][2,]
>
> On Thu, Oct 14, 2021 at 10:52 AM Luigi Marongiu
> wrote:
>>
>> Hello,
>> I have selected a subset of a dataframe with the vector syntax (if
>> this is the name):
>> ```
>> > df[(df$Y>0.2)
df[(df$Y>0.2) & (df$X<10),][2,]
On Thu, Oct 14, 2021 at 10:52 AM Luigi Marongiu
wrote:
> Hello,
> I have selected a subset of a dataframe with the vector syntax (if
> this is the name):
> ```
> > df[(df$Y>0.2) & (df$X<10),]
> YX
> 10 0.2200642 1.591589
> 13 0.2941828 1.48595
You're missing a comma, this should fix it
df[(df$Y>0.2) & (df$X<10),][2, ]
On Thu, Oct 14, 2021, 03:51 Luigi Marongiu wrote:
> Hello,
> I have selected a subset of a dataframe with the vector syntax (if
> this is the name):
> ```
> > df[(df$Y>0.2) & (df$X<10),]
> YX
> 10 0
Hello,
I have selected a subset of a dataframe with the vector syntax (if
this is the name):
```
> df[(df$Y>0.2) & (df$X<10),]
YX
10 0.2200642 1.591589
13 0.2941828 1.485951
```
How can I select only the second row? I only managed to get the whole
of whole columns:
```
> df[(df
Do you really want the minimum?
It sounds as though your model is a*N(x1,s1) + (1-a)*N(x2,s2) where
you use mixtools to estimate
the parameters. Finding the derivative of that is fairly
straightforward calculus, and solving for the
derivative being zero gives you extrema (you want the one between
16 matches
Mail list logo
