Quizás podrías tener 2 columnas, una numérica sobre la que se base el gráfico y
otra tipo carácter, con los separadores para cuando necesites usar esa
presentación.
Saludos,
Juan
De: juan manuel dias
Enviado el: jueves, 14 de octubre de 2021 20:45
Para: JCMld
CC: Lista R
Asunto: Re:
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
Claro entiendo! ...en algunos casos pude guardar como character y mostrar
la los números con los separadores, pero hay tablas donde necesito además
de mostrar el número hacer un gráfico con esa misma variable...pero cuándo
modifico a character "monto total de la factura" me arroja error porque no
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.
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.
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
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
> 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:
>
>>
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
Hola Juan Manuel,
Si el dato es numérico no puede guardarse con separadores, a no ser que lo
convirtieras a character, ya que el separador es un tema de formato de
salida.
Entonces tienes 2 opciones, o lo guardas como numérico, pero a la hora de
presentarlo lo imprimes con la función format, o
Muy buena también! gracias Carlos!
El mié, 13 oct 2021 a las 3:58, Carlos Ortega ()
escribió:
> Hola,
>
> Aún cerrada la duda, dejo aquí otra posible alternativa:
>
> #---
>
> > library(stringr)
> > a <- c("7/6/2020 7:55:38 p.m.", "7/3/2020 1:08:36 p.m.", "7/3/2020
> 6:08:35 p.m.")
>
Hola. Como andan!
Consulta, del siguiente data frame la columna "importe $" es de tipo
numérico y querría que siga siendo numérica pero agregando separadores de
miles.
Proveedores`Importe en $` Porcentaje
1
I presumed there was some reason why the 'optimise' function did not suit you.
optimize(function (x) a*dnorm(x, x1, s1) + (1-a)*dnorm(x, x2, s2),
range(c(x1, x2)))
should do the trick.
On Thu, 14 Oct 2021 at 23:42, Luigi Marongiu wrote:
>
> Thank you, I hoped there might be an automatic method
Thank you, I hoped there might be an automatic method more than
function analysis...
On Thu, Oct 14, 2021 at 9:06 AM Richard O'Keefe wrote:
>
> Do you really want the minimum?
> It sounds as though your model is a*N(x1,s1) + (1-a)*N(x2,s2) where
> you use mixtools to estimate
> the parameters.
Hello,
If the sub-df has more than 2 rows,
tail(df[(df$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
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
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:
```
>
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
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